Given that Lenny and George started with two rabbits and during the first month those rabbits have two babies.
Every month thereafter, each pair of rabbits has two babies. Let $r(n)$ represents the number of rabbits in $n^text{th}$ month then the situation can be represented by the equation as shown below.
begin{align*}
r(0)&=2\
r(1)&=2times t(0)= 2times 2=2^2=2^{1+1}\
r(2)&=2times t(1)= 2times 2^2=2^3=2^{2+1}\
r(3)&=2times t(2)= 2times 2^3=2^4=2^{3+1}\
r(n)&=2times t(n-1)=2times 2^{n}=2^{n+1}
end{align*}
Therefore, the number of rabbits at the end of $n^text{th}$ month is given by the equation
$$boxed{r(n)=2^{n+1}}$$
$bullet$ The number of rabbits Lenny and George will have after 12 months can be found out by putting $n=12$ in the equation obtained above.
begin{align*}
r(n)&=2^{n+1}\
r(12)&=2^{12+1}\
r(12)&=2^{13}\
r(12)&=8192
end{align*}\
$bullet$ The pattern grows such as the number becomes two times of whatever the number of rabbits in the last month.\
The table for the growth is shown below where $n$ represents the number of months and $r(n)$ represents the number of rabbits.\\
begin{tabular}{ |p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
$n$ &0 &1&2&3&4&n \
hline
$r(n)$ &2&4&8&16&32&$2^{n+1}$\
hline
end{tabular}

$bullet$ The assumption made while calculating the number of rabbits at the end of each month is that not a single rabbit died during any period of consideration.

$bullet$ This can be compared to any other pattern which grows two times at the end of each month and with the starting number is 2.
Most of the pattern we investigated earlier were linear and grew with the constant number but this pattern grows non linearly.

The number of rabbits at the end of $n^text{th}$ month is given by the equation

$$
r(n)=2^{n+1}
$$

For detailed explanation of each part see inside.

$textbf{(a)}$ The pattern of growth of rabbits is shown by the diagram attached below where a box represent a rabbit and $r(n)$ represents the number of rabbits at the end of the $n^text{th}$ month.

The number of rabbits at the end of $3^text{th}$ month is 16.

textbf{(b)}
The table for the growth is shown below where $n$ represents the number of months and $r(n)$ represents the number of rabbits.\\
begin{tabular}{ |p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
$n$ &0 &1&2&3&4&5&6&n \
hline
$r(n)$ &2&4&8&16&32&64&128&$2^{n+1}$\
hline
end{tabular}\\
The pattern grows such as the number becomes two times of whatever the number of rabbits in the last month.\\
textbf{(c)} Let $r(n)$ represents the number of rabbits in $n^text{th}$ month then the situation can be represented by the equation as shown below.
begin{align*}
r(0)&=2\
r(1)&=2times t(0)= 2times 2=2^2=2^{1+1}\
r(2)&=2times t(1)= 2times 2^2=2^3=2^{2+1}\
r(3)&=2times t(2)= 2times 2^3=2^4=2^{3+1}\
r(n)&=2times t(n-1)=2times 2^{n}=2^{n+1}
end{align*}
Therefore, the number of rabbits at the end of $n^text{th}$ month is given by the equation
$$boxed{r(n)=2^{n+1}}$$
Hence,the number of rabbits Lenny and George will have after 12 months can be found out by putting $n=12$ in the equation obtained above.
begin{align*}
r(n)&=2^{n+1}\
r(12)&=2^{12+1}\
r(12)&=2^{13}\
&boxed{r(12)=8192}
end{align*}\
textbf{(d)} Most of the pattern we investigated earlier were linear and grew with the constant number but this pattern grows non linearly.\

$textbf{(a)}$ The number of rabbits at the end of $3^text{th}$ month is 16.

$textbf{(b)}$ The pattern grows such as the number becomes two times of whatever the number of rabbits in the last month.

$textbf{(c)}$ 8192

$textbf{(d)}$ Most of the pattern we investigated earlier were linear and grew with the constant number but this pattern grows non linearly.

textbf{(a)} The table for the growth for different cases is shown below where $n$ represents the number of months and $r(n)$ represents the number of rabbits.\
$bullet$ Case 2:\\
begin{tabular}{ |p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
$n$ &0 &1&2&3&4&5 \
hline
$r(n)$ &10&20&40&80&160&320\
hline
end{tabular}\\
$bullet$ Case 3:\\
begin{tabular}{ |p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
$n$ &0 &1&2&3&4&5 \
hline
$r(n)$ &2&8&32&128&512&2048\
hline
end{tabular}\\
$bullet$ Case 4:\\
begin{tabular}{ |p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}|p{1cm}| }
hline
$n$ &0 &1&2&3&4&5 \
hline
$r(n)$ &2&12&72&432&2592&15552\
hline
end{tabular}\\
Therefore, a higher growth pattern will yield largest population rather than higher starting number of rabbits (if it is not very large starting value).\

$textbf{(b)}$ $bullet$ Case 2:

Let $r(n)$ represents the number of rabbits in $n^text{th}$ month then the situation can be represented by the equation as shown below.

$$
begin{align*}
r(0)&=10=2times 5\
r(1)&=2times t(0)= 2times 2times 5= 2^2 times 5=2^{1+1}times 5\
r(2)&=2times t(1)= 2times 2^2 times 5=2^3times 5=2^{2+1}times 5\
r(n)&=2times t(n-1)=2times 2^{n}times 5 =2^{n+1}times 5
end{align*}
$$

Therefore, the number of rabbits at the end of $n^text{th}$ month is given by the equation

$$
boxed{r(n)=2^{n+1}times 5}
$$

Hence,the number of rabbits Lenny and George will have after 12 months can be found out by putting $n=12$ in the equation obtained above.

$$
begin{align*}
r(n)&=2^{n+1}times 5\
r(12)&=2^{12+1}times 5\
r(12)&=2^{13}times 5\
&boxed{r(12)=40960}
end{align*}
$$

$bullet$ Case 3:

Let $r(n)$ represents the number of rabbits in $n^text{th}$ month then the situation can be represented by the equation as shown below.

$$
begin{align*}
r(0)&=2\
r(1)&=r(0)times 4= 2times 4= 2times 4^1\
r(2)&=r(1)times 4= 2times 4^1 times 4= 2times 4^2\
r(n)&=r(n-1)times 4= 2times 4^(n-1) times 4= 2times 4^n\
end{align*}
$$

Therefore, the number of rabbits at the end of $n^text{th}$ month is given by the equation

$$
boxed{r(n)=2times 4^n }
$$

Hence,the number of rabbits Lenny and George will have after 12 months can be found out by putting $n=12$ in the equation obtained above.

$$
begin{align*}
r(n)&=2times 4^n\
r(12)&=2times 4^12\
r(12)&=2times 16777216\
&boxed{r(12)=33554432}
end{align*}
$$

$bullet$ Case 4:

Let $r(n)$ represents the number of rabbits in $n^text{th}$ month then the situation can be represented by the equation as shown below.

$$
begin{align*}
r(0)&=2\
r(1)&=r(0)times 6= 2times 6= 2times 6^1\
r(2)&=r(1)times 6= 2times 6^1 times 6= 2times 6^2\
r(n)&=r(n-1)times 6= 2times 6^(n-1) times 6= 2times 6^n\
end{align*}
$$

Therefore, the number of rabbits at the end of $n^text{th}$ month is given by the equation

$$
boxed{r(n)=2times 6^n }
$$

Hence,the number of rabbits Lenny and George will have after 12 months can be found out by putting $n=12$ in the equation obtained above.

$$
begin{align*}
r(n)&=2times 6^n\
r(12)&=2times 6^12\
r(12)&=2times 2176782336\
&boxed{r(12)=4353564672}
end{align*}
$$

$textbf{(a)}$ A higher growth pattern will yield largest population rather than higher starting number of rabbits (if it is not very large starting value). The table is attached inside.

$textbf{(b)}$ Case 2: $rightarrow$ 40960

Case 3: $rightarrow$ 33554432

Case 4: $rightarrow$ 4353564672

$textbf{(a)}$ The patterns are increasing non linearly with the powers of a constant numbers.

$textbf{(b)}$ The graph for case 1: is attached below.
The graph is growing exponentially with the equation of the function
$$
r(n)=2^{n+1}
$$
where, $r(n)$ is the number of rabbits and $n$ is the number of months.

$textbf{(c)}$ $bullet$ The domain for the graph is whole numbers $(0,1,2,3,4,cdots)$

The range of the graph is discrete points which are $(2,4,8,16,32,64,128,cdots)$

$bullet$ The graph is discrete as we are assuming that number are multiplying at the end of each month not at every point of time.

$bullet$ Yes, it is a discrete function, as corresponding to a value of input there is only single value of output.

$textbf{(a)}$ The patterns are increasing non linearly with the powers of a constant numbers.

$textbf{(b)}$ The graph is growing exponentially with the equation of the function
$$
r(n)=2^{n+1}
$$
where, $r(n)$ is the number of rabbits and $n$ is the number of months.

$textbf{(c)}$ The domain for the graph is whole numbers $(0,1,2,3,4,cdots)$

The range of the graph is discrete points which are $(2,4,8,16,32,64,128,cdots)$

The graph is discrete as we are assuming that number are multiplying at the end of each month not at every point of time.

Yes, it is a discrete function, as corresponding to a value of input there is only single value of output.

Exponential functions have graphs that curve upward or downward. One part of its graph approaches a value, represented by a horizontal line, but never actually reach it. As $x$ increases, $y$ either increases or decreases continuously. By observing a table, the $y$-values are multiplied by the same factor for consecutive integer $x$-values.

textbf{(a)} We can observe the given values in the table and try to write the general form of the equation so that we can write the number of rabbits for any required month.\ Let $n$ represents the number of months and $r(n)$ represents the number of rabbits in the $n^text{th}$ month.
begin{align*}
r(0)&=4\
r(1)&=12=3times 4=3times r(0)=3^1times 4 \
r(2)&=36=3times 12=3times r(1)=3^2times 4 \
r(n)&=3times r(n-1)=3^{n}times 4
end{align*}
Therefore, from the starting number of 4 rabbits we can find the number of rabbits after any $n^text{th}$ month by the equation $$boxed{r(n)=3^{n}times 4 } $$
For example, Number of rabbits in the $3^text{rd}$ month is given by
begin{align*}
r(n)&=3^{n}times 4 \
r(3)&=3^{3}times 4 \
r(3)&=27times 4 \
r(3)&=108
end{align*}
The table of other values is included below.
begin{center}
begin{tabular}{ |p{2cm}|p{2cm}| }
hline
Months & Rabbits \
hline
0 & 4 \
hline
1 & 12 \
hline
2 & 36 \
hline
3&108\
hline
4&324\
hline
end{tabular}
end{center}

textbf{(b)} We can observe the given values in the table and try to write the general form of the equation so that we can write the number of rabbits for any required month.\ Let $n$ represents the number of months and $r(n)$ represents the number of rabbits in the $n^text{th}$ month.
begin{align*}
r(0)&=6\
r(2)&=24= 4times 6=2^2times 6 \
r(4)&=96=16times 6=2^4times 6 \
r(n)&=2^n times 6
end{align*}
Therefore, from the starting number of 6 rabbits we can find the number of rabbits after any $n^text{th}$ month by the equation $$boxed{r(n)=2^n times 6 } $$
For example, Number of rabbits in the $1^text{st}$ month is given by
begin{align*}
r(n)&=2^n times 6 \
r(1)&=2^1 times 6 \
r(1)&=2times 6\
r(1)&=12
end{align*}
Similarly, the number of rabbits in the $3^text{rd}$ month is given by
begin{align*}
r(n)&=2^n times 6 \
r(3)&=2^3 times 6 \
r(3)&=8times 6\
r(3)&=48
end{align*}
The table of other values is included below.
begin{center}
begin{tabular}{ |p{2cm}|p{2cm}| }
hline
Months & Rabbits \
hline
0 & 6 \
hline
1 & 12 \
hline
2 & 24 \
hline
3&48\
hline
4&96\
hline
end{tabular}
end{center}

$textbf{(a)}$ From the starting number of 4 rabbits, we can find the number of rabbits after any $n^text{th}$ month by the equation
$$
r(n)=3^{n}times 4
$$

$textbf{(b)}$ From the starting number of 6 rabbits, we can find the number of rabbits after any $n^text{th}$ month by the equation
$$
r(n)=2^{n}times 6
$$

$textbf{(a)}$ We can write the equation of line by choosing any two pair of points.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Lets choose the points $(3,4)$ and $(3,2)$. Let $(x_1,y_1)=(3,4)$ and $(x_2,y_2)=(3,2)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{2-4}{3-3}\
&=dfrac{-2}{0}\
&approx infty
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $infty$ passing through a point $(3,4)$ is given by

$$
begin{align*}
y-4&= infty cdot (x-3)\
dfrac{y-4}{infty}&= dfrac{infty}{infty}cdot (x-3)\
0&=1cdot (x-3)\
0&=x-3\
boxed{x=3}
end{align*}
$$

Therefore, the equation of line is is $x=3$ and coordinates of the other points on it are marked in the red in the graph attached below.

$textbf{(b)}$ We can write the equation of line by choosing any two pair of points.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Lets choose the points $(-3,-1)$ and $(1,-1)$. Let $(x_1,y_1)=(-3,-1)$ and $(x_2,y_2)=(1,-1)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-1-(-1)}{1-(-3)}\
&=dfrac{-1+1}{1+3}\
&=dfrac{0}{4}\
&=0
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $0$ passing through a point $(-3,-1)$ is given by

$$
begin{align*}
y-(-1)&= 0 cdot (x-(-3))\
y+1&=0\
y&=-1\
boxed{y=-1}
end{align*}
$$

Therefore, the equation of line is is $y=-1$ and coordinates of the other points on it are marked in the red in the graph attached below.

$textbf{(c)}$ The $x$ coordinate of the points on the $y$ axis will be 0. We can find the equation of line ($x=0$) just like in part (a).
The graph is attached below.

$textbf{(a)}$ $x=3$

$textbf{(b)}$ $y=-1$

$textbf{(c)}$ $x=0$

The domain of the any function is defined as the set of values that the input variable can take for which the function $g(x)$ does not become not defined. Given function is $g(x)=dfrac{x+2}{x-1}$ and $x$ is the input variable here.

Now, when we put $x=1$ into the $g(x)$ function we get

$$
begin{align*}
g(x)&=dfrac{x+2}{x-1}\
g(1)&=dfrac{1+2}{1-1}\
g(1)&=dfrac{3}{0}\
g(1)&=text{not defined}
end{align*}
$$

Therefore, $x=1$ cannot be in the domain of the function $g(x)$.

$x=1$ cannot be in the domain of the function $g(x)$.

We know that the 1 kg is equivalent to 1000 grams. Therefore, 3.295 kg is equivalent to

$$
begin{align*}
&3.295text{ kg} cdot( dfrac{1000 text{ g}}{1 text{kg}})\
&=3.295 1 cdot 1000 text{ g} cdot ( dfrac{1 text{ kg}}{1 text{ kg}})\
&=3.295cdot 1000 text { g}\
&=3295 text { g}
end{align*}
$$

Given that there are 28.3 grams in 1 ounce. Therefore, 3295 grams can be converted into ounce as shown below

$$
begin{align*}
&3295 text{ g} cdot dfrac{ 1 text{ ounce}}{28.3 text{ g}}\
&=dfrac{3295}{28.3}cdot dfrac{text{ g}}{text{ g}}cdot text{ ounce}\
&=dfrac{3295}{28.3}cdot text{ ounce}\
&=116.43 text{ ounce}
end{align*}
$$

Now, as we know that there are 16 ounces in 1 pound. Therefore, 116.43 ounces can be rewritten as

$$
dfrac{116.43}{16}=7. 2768text{ pounds}
$$

This can be also written as 7 pounds and $0.2768times 16=4.4288text { ounces}$

Hence, 3.295 kg is equivalent to 7 pounds and 4.4288 ounces.

$textbf{(a)}$ The coordinates after performing the function $(xrightarrow -x,yrightarrow y)$ is given below

$$
begin{align*}
A(1,1)&rightarrow A'(-1,1)\
B(-1,-1)&rightarrow B'(1,-1)\
C(4,-1)&rightarrow C'(-4,-1)
end{align*}
$$

Therefore, the coordinates of the vertex points of $Delta$A’B’C’ are $A'(-1,1)$, $B'(1,-1)$ and $C'(-4,-1)$.

The resultant figure $Delta$A’B’C’ is the result of reflection of $Delta$ABC across the line $x=0$.

$textbf{(b)}$ The coordinates of the resultant figure $Delta$A”B”C” are

$$
begin{align*}
A(1,1)&rightarrow A”(5,6)\
B(-1,-1)&rightarrow B”(3,4)\
C(4,-1)&rightarrow C”(8,4)
end{align*}
$$

$textbf{(a)}$ The coordinates after performing the function $(xrightarrow -x,yrightarrow y)$ are given below

$$
begin{align*}
A(1,1)&rightarrow A'(-1,1)\
B(-1,-1)&rightarrow B'(1,-1)\
C(4,-1)&rightarrow C'(-4,-1)
end{align*}
$$
$textbf{(b)}$ The coordinates of the resultant figure $Delta$A”B”C” are

$$
begin{align*}
A(1,1)&rightarrow A”(5,6)\
B(-1,-1)&rightarrow B”(3,4)\
C(4,-1)&rightarrow C”(8,4)
end{align*}
$$

In the first sentence, one more variable which can be the actual cause is $textbf{bad diet}$.

In the second sentence, one more variable which can be the actual cause is $textbf{smoking cigarettes}$.

textbf{(a)} The graph for this situation should be curved as the increase is not by the constant amount. The bacteria cells are growing such that number becomes 2 times of the number of previous day.\\
textbf{(b)}
begin{center}
begin{tabular}{|p{3cm}|p{4cm}|}
hline
Number of days & Number of bacteria\
hline
0 & 1000\
hline
1 & 2000\
hline
2 &4000\
hline
3 &8000\
hline
n & $2^ntimes 1000$\
hline
end{tabular}
end{center}

$textbf{(c)}$ The graph is exponential which has domain as whole numbers $(0,1,2,3,4,cdots)$ and the range is the discrete points $(1000,2000,4000,8000,cdots)$
The points are connected on the graph to show the shape of the curve . It does imply that the function takes all those values. Yes, it is a Discrete function.

$textbf{(a)}$ The graph for this situation should be curved as the increase is not by the constant amount. The bacteria cells are growing such that number becomes 2 times of the number of previous day.

$textbf{(b)}$ The number of bacteria on $n^text{th} text{ day is }2^ntimes 1000$

$textbf{(c)}$ The graph is exponential which has domain as whole numbers $(0,1,2,3,4,cdots)$ and the range is the discrete points $(1000,2000,4000,8000,cdots)$
The points are connected on the graph to show the shape of the curve . It does imply that the function takes all those values. Yes, it is a Discrete function.

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
frac{5^{723}}{5^{721}}& {=}quad : 5^{723-721}\
&=5^2 \
&={color{#c34632}25}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:x^{a-b}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
frac{3^{300}}{3^{249}}& {=}quad : 3^{300-249}\
&={color{#c34632}3^{51}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:x^{a-b}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
left(frac{34^3}{3^{-2}cdot :4^{-7}}right)^0& {=}quad : color{#c34632}1\
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:rule}:a^0=1,:ane :0
$$

$$
text{ }
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
left(frac{4cdot :10^3}{10^{-2}}right)^2& {=}quad : left(4cdot :10^5right)^2\
&=4^2left(10^5right)^2 \
&=color{#c34632}4^2cdot :10^{10} \
end{align*}
$$

$$
color{#4257b2} text{ a) }25
$$

$$
color{#4257b2} text{ b) }3^{51}
$$

$$
color{#4257b2}text{ c) }1
$$

$$
color{#4257b2}text{ d) }4^2*10^{10}
$$

Jackie made the mistake while opening the square bracket. We know that
$$
(a+b)^2=a^2+2ab+b^2
$$

But during solving for $(x+4)^2$ she wrote $(x+4)^2=x^2+16$ which is wrong. Also, in the right side of equation, while opening $(x-1)^2$ she wrote $(x-1)^2=x^2+1$.

The correct solution is shown below

$$
begin{align*}
(x + 4)^2 -2 x -5 &= (x -1 )^2\
(x^2+8x+16)-2x-5&=x^2-2x+1\
x^2+6x+11 &=x^2-2x+1\
6x+2x&=1-11\
8x&=-10\
x&=-dfrac{10}{8}\
x&=-dfrac{5}{4}\
end{align*}
$$

The correct answer is $-dfrac{5}{4}$

$textbf{(a)}$ The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-2$ passing through a point $(0,7)$ is given by

$$
begin{align*}
y&-7=-2cdot (x-0)\
y&-7=-2x\
&boxed{y=-2x+7}
end{align*}
$$

$textbf{(b)}$ The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-dfrac{3}{2}$ passing through a point $(4,0)$ is given by

$$
begin{align*}
y-0&=-dfrac{3}{2}cdot (x-4)\
y&=-dfrac{3}{2}cdot x + (-dfrac{3}{2} cdot -4) \
y&=-dfrac{3}{2}cdot x +dfrac{12}{2}\
&boxed{y=-dfrac{3}{2}cdot x+6}
end{align*}
$$

$textbf{(c)}$ Two perpendicular lines have product of their slope as $-1$. Therefore, a line of slope $m$ which is perpendicular to a line of slope $-dfrac{3}{2}$, which implies

$$
begin{align*}
mtimes -dfrac{3}{2}&=-1\
m&=dfrac{-1}{(-dfrac{3}{2})}\
m&=dfrac{-1times 2}{-3}\
m&=dfrac{2}{3}
end{align*}
$$

Therefore, the equation of a line of slope $dfrac{2}{3}$ passing through a point $(4,0)$ is given by

$$
begin{align*}
y-0&=dfrac{2}{3}cdot (x-4)\
y&=dfrac{2}{3}cdot x + (dfrac{2}{3} cdot -4) \
&boxed{y=dfrac{2}{3}cdot x -dfrac{8}{3}}
end{align*}
$$

$textbf{(a)}$ $y=-2x+7$

$textbf{(b)}$ $y=-dfrac{3}{2}cdot x+6$

$textbf{(c)}$ $y=dfrac{2}{3}cdot x -dfrac{8}{3}$

$textbf{(c)}$ Individual areas of the divided figures are mentioned in the Image attached.The total area is given by sum of all the three parts, which is
$$
text{Area}= dfrac{3}{2}+6+6 = dfrac{27}{2} text { unit}^2
$$

$textbf{(b)}$ A”$=(1,2)$ and C”$=(-2,5)$

$textbf{(c)}$ $dfrac{27}{2} text{ unit}^2$

We can evaluate the expression $dfrac{1}{4}k^5-3k^3+k^2-k$ by putting $k=2$ in the expression as solving further.

$$
begin{align*}
&dfrac{1}{4}k^5-3k^3+k^2-k\
&=dfrac{1}{4}(2)^5-3(2)^3+(2)^2-(2)\
&=dfrac{1}{4}(32)-3(8)+(4)-(2)\
&=8-24+2\
&=boxed{-14}
end{align*}
$$

The value of expression for $k=2$ is $-14$

For comparing the bounciness of the different balls we must convert it into the same unit and find the rebound height for per unit dropping height. Lets covert all the data in Centimeters only.

$textbf{(a)}$ Tennis balls rebound approximately 111 cm when dropped from 200 cm.
Therefore, when dropped from 1 cm it will rebound $dfrac{111}{200}=0.555$ cm.

$textbf{(b)}$ Soccer balls rebound approximately 120 cm when dropped from 200 cm.
Therefore, when dropped from 1 cm it will rebound $dfrac{120}{200}=0.6$ cm.

$textbf{(c)}$ Basketballs rebound approximately 53.5 inches when dropped from 72 inches. 1 inch is equivalent to 2.54 cm. Therefore 53.5 inches= 135.89 cm and 72 inches= 182.88 cm.
Therefore, when a basketball is dropped from 1 cm it will rebound $dfrac{135.89}{182.88}=0.743$ cm.

$textbf{(d)}$ Squash balls rebound approximately 29 .5 inches when dropped from 100 inches. 29.5 inches= 74.93 cm and 100 inches= 254 cm.
Therefore, when a basketball is dropped from 1 cm it will rebound $dfrac{74.93}{254}=0.295$ cm.

Comparing the rebound height of each ball when it is dropped from 1 cm, we see that basketball rebounds highest of all which is 0.743 cm. Therefore, basketball is the bounciest of all.

$textbf{(a)}$ The rebound height of the ball is dependent on the dropping height.

Rebound height $rightarrow$ Dependent variable

Dropping height $rightarrow$ Independent variable

$textbf{(b)}$ The graph will be discrete as not all the values of height can be obtained by dropping the ball.

$textbf{(c)}$ The graph may vary according to different types of ball (re bouncing characteristics). But the line will always pass through origin as if the dropping height is 0 then the rebound height will also be 0.

$textbf{(d)}$ Some of the example lines are shown in the graph below. We can generalise that the rebound height is always less than the dropping height. Therefore, on the graph the lines obtained will always lie below the line drawn in red ($y=x$)

Also, the as negative height is not possible, so the line will not be valid for the negative values of any of the $x$ and $y$ variables.

$textbf{(a)}$ The rebound height of the ball is dependent on the dropping height.

Rebound height $rightarrow$ Dependent variable

Dropping height $rightarrow$ Independent variable

$textbf{(b)}$ The graph will be discrete as not all the values of height can be obtained by dropping the ball.

$textbf{(c)}$ The graph may vary according to different types of ball (re bouncing characteristics). But the line will always pass through origin as if the dropping height is 0 then the rebound height will also be 0.

$textbf{(d)}$ Some of the example lines are shown in the graph attached inside.

$bullet$ The value of the rebound ratio will vary for the different group. It will depend upon the material of the ball.

$bullet$ Rebound ratio is given by the slope of the line of the best fit.

$bullet$ In the general form equation of line which is given by $y=mx+c$. The $m$ represents the slope of line and hence, here in this context $m$ is equivalent to rebound ratio.

$bullet$ The rebound ratio which is equivalent to slope of the line of best fit can be calculated by choosing coordinates of any two set of points on the line as shown below.

$$
begin{align*}
text{Rebound ratio}&=dfrac{y_2-y_1}{x_2-x_1}\
&=dfrac{Delta y}{Delta x}
end{align*}
$$

$textbf{(a)}$
$$
begin{align*}
2(x-2)&=-6tag{divide each side by 2}\
dfrac{2(x-2)}{2}&=dfrac{-6}{2}\
x-2&=-3tag{add 2 to each side}\
x-2+2&=-3+2\
&boxed{x=-1}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
2(x+1)+3&=3(x-1)tag{solve brackets using distributive property}\
2x+2+3&=3x-3tag{subtract $2x$ from each side}\
2x-2x+5&=3x-2x-3\
5&=x-3tag{add 3 to the each side}\
5+3&=x-3+3\
8&=xtag{interchange side}\
&boxed{x=8}
end{align*}
$$

$textbf{(a)}$ $x=-1$

$textbf{(b)}$ $x=8$

$textbf{(a)}$ Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have given points as $(6,-8)$ and $(3,-4)$. Let $(x_1,y_1)=(3,-4)$ and $(x_2,y_2)=(6,-8)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-8-(-4)}{6-3}\
&=dfrac{-8+4}{3}\
&=dfrac{-4}{3}
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $dfrac{-4}{3}$ passing through a point $(3,-4)$ is given by

$$
begin{align*}
y-(-4)&=dfrac{-4}{3}cdot (x-3)\
y+4&=dfrac{-4}{3}cdot x + dfrac{-4}{3}cdot (-3)\
y+4&=dfrac{-4}{3}cdot x +4 tag{subtract 4 from each side}\
y+4-4&=dfrac{-4}{3}cdot x +4-4\
&boxed{y=dfrac{-4}{3}cdot x}
end{align*}
$$

$textbf{(b)}$ For checking if the given point $(-3,4)$ lies on the obtained line. We should plug the values of given $x$ and $y$ into the equation of line. If it satisfies the line $y=-dfrac{4}{3}x$ then it lies on the line otherwise it does not.

$$
begin{align*}
y&=-dfrac{4}{3}x\
(4)&=-dfrac{4}{3}(-3)\
4&=-1cdot dfrac{-3}{3}cdot 4\
4&=-1cdot -1 cdot 4\
4&=4tag{true always}
end{align*}
$$

The point $(-3,4)$ satisfies the line $y=-dfrac{4}{3}x$ and hence it lies on it.

$textbf{(a)}$ $y=dfrac{-4}{3}cdot x$

$textbf{(b)}$ Yes, the point $(-3,4)$ lies on the line.

$textbf{(a)}$ The given value of correlation coefficient $r$ is $-0.903$ R-squared value represent the coefficient of determination which can be found out by squaring the $r$ value.

$Rightarrow$
$$
R^2=(-0.903)^2=0.8154
$$

The value of coefficient of determination ($R^2$) is $0.8154approx 0.81$ which signifies that the approximately 81 percent variation in fuel efficiency can be predicted by weight of the cars. Rest 19 percent variation is dependent on the other factors.

$textbf{(b)}$ The given equation is $y=49-8.4x$

The slope of this line represent the rate of change in the fuel efficiency (miles per gallon) of the car per thousands pound change in the weight of a car.

Here slope is negative which shows that if the weight of the car increases then the efficiency of the car will decrease.

$textbf{(a)}$ The value of coefficient of determination ($R^2$) is $0.8154approx 0.81$ which signifies that the approximately 81 percent variation in fuel efficiency can be predicted by weight of the cars. Rest 19 percent variation is dependent on the other factors.

$textbf{(b)}$

The slope of this line represent the rate of change in the fuel efficiency (miles per gallon) of the car per thousands pound change in the weight of a car.

Here slope is negative which shows that if the weight of the car increases then the efficiency of the car will decrease.

$textbf{(a)}$ Given that, AAA Packages Plus sends packages overnight for $5 plus $0.25 per ounce.\
Let$w$represent the weight of the package in ounces, then the cost of sending a package via AAA Packages Plus is given by$ $text{Cost (via AAA)}= 5+0.25w$ $United Packages charges $2 plus $0.35 per ounce.Therefore,$ $text{Cost (via United)}= 2+0.35w$ $Given that, Molinari’s package would cost the same to mail using either service.$Rightarrow$5+0.25w=2+0.35w$$

$textbf{(b)}$ weight of the Molinari’s package can be found out by solving the final equation in the previous part.

$$
begin{align*}
5+0.25w&=2+0.35w\
5-2&=(0.35-0.25)w\
3&=0.1w\
w&=dfrac{3}{0.1}\
w&=3 times 10\
w&=30 text{ ounces}
end{align*}
$$

$textbf{(a)}$ $5+0.25w=2+0.35w$ where, $w$ is the weight of the package in ounces.

$textbf{(b)}$ 30 ounces.

$textbf{(c)}$ From the table we can see that time taken by her to reach her friend’s house decreases as Meredith increases her speed.
The formula $text{Time}=dfrac{text{Distance}}{text{Speed}}$ explains it well.

$textbf{(d)}$ Yes, this is a proportional relationship as the formula $text{Time}=dfrac{text{Distance}}{text{Speed}}$ contains a unit power to the each terms.

$textbf{(a)}$24 minutes and 12 minutes.

$textbf{(c)}$ From the table we can see that time taken by her to reach her friend’s house decreases as Meredith increases her speed.

$textbf{(d)}$ Yes, this is a proportional relationship.

The equation for the tile pattern is $y = 10x – 8$. for finding the number of tiles in the Figure 12 of this pattern, we need to put $x=12$ in this equation.

$$
begin{align*}
y &= 10times(12) – 8\
y&=120-8\
y&=112
end{align*}
$$

By solving the given equation, we got that the Figure 12 of this pattern contains 112 tiles whereas Kalil thinks there should be 108 tiles in it. Therefore, kalil is not correct.

$textbf{(a)}$ The rebound ratio will be different for different teams. For example let our team got soccer ball. Then rebound ratio for this ball will
$$
text{rebound ratio}=dfrac{120}{200}=0.6
$$

$textbf{(b)}$ No, the dropping height of the ball will not affect the ratio.

$textbf{(c)}$ For finding rebound height for any given dropping height we multiply the ratio with the dropping height.
As rebound ratio is given by
$$
text{rebound ratio}=dfrac{text{rebound height}}{text{dropping height}}
$$

$Rightarrow$
$$
text{rebound height}=text{rebound ratio} times text{dropping height}
$$

$textbf{(a)}$ The rebound ratio will be different for different teams. For example, for team with soccer ball the value of rebound ratio will be $0.6$

$textbf{(b)}$ No, the dropping height of the ball will not affect the ratio.

$$
textbf{(c)}text{rebound height}=text{rebound ratio} times text{dropping height}
$$

textbf{(a)} We can solve the question by assuming any value for the ratio and then generalise the result for all the cases. Lets assume the bouncing ratio of the ball is 0.48 and starting height is given 200 cm.\
The rebound height after each bounce can be found by the formula
begin{align*}
text{RH}_n=200times ( dfrac{48}{100})^n
end{align*}
where, $text{RH}_n$ represents the rebound height after $n$ rebounds.\
begin{center}
begin{tabular}{|p{4cm}|p{3cm}|}
hline
Number of bounces &Rebound height\
hline
1&96\
hline
2&46.08\
hline
3&22.118\
hline
4&10.616\
hline
n& $200times ( dfrac{48}{100})^n$\
hline
end{tabular}
end{center}

$textbf{(b)}$ Independent variable $rightarrow$ Dropping height

Dependent variable $rightarrow$ Rebound height

$textbf{(c)}$ The graph of the predicted rebound heights is attached below.

$textbf{(d)}$ The graph is discrete as it is not possible to obtain all the values of rebounding heights via bouncing ball.

$textbf{(a)}$ $text{RH}_n=200times ( dfrac{48}{100})^n$

$textbf{(b)}$ Independent variable $rightarrow$ Dropping height

Dependent variable $rightarrow$ Rebound height

$textbf{(c)}$ The graph of the predicted rebound heights is attached below.

$textbf{(d)}$ The graph is discrete as it is not possible to obtain all the values of rebounding heights via bouncing ball.

$bullet$ The experimental data is different than the predicted data because there may be various factor that can affect the experimental data, such as measuring error, friction, air drag, relation may not represented via ratio etc.

$bullet$ Graph and model represents the model with some errors (negative residues) which is tolerable upto some extent.The negative residues between actual and predicted data signifies that the actual values of rebound height is smaller than the predicted height after every bounce

$textbf{a.}$

The model for successive bounces is not linear so we cannot use $y=mx+b$ to model the data. The data is also not quadratic since the bounces are never higher the the previous bounce. So, the model for the data is exponential.

$textbf{b.}$

The pattern of growth is that each successive bounce height is the previous bounce height multiplied by the same factor that is less than 1. This is the same as the rabbit problem where the factor that is being multiplied is greater than 1.

We have calculated the formula of rebound height after every bounce (rebound ratio 48%) which is

$$
text{RH}_n=200(dfrac{48}{100})^n
$$

Therefore, the rebound height of ball after twelth bounce will be

$$
text{RH}_{12}=200(dfrac{48}{100})^{12}
$$

From the equation we can see, the rebound value will only become 0 when $n$ reaches to infinity which means that theoretically ball never stops. But, due to friction and air drag ball stops after some time.

From the equation of rebound height we can see that the value of rebound height will only become 0 when $n$ reaches to infinity which means that theoretically ball never stops. But, due to friction and air drag ball stops after some time.

The exponential function is:

$$
f(x)=ab^x
$$

Where $b>0$.

For $b=1$, this function is constant.

For $a>0$ and $b>1$, this function is increasing.

For $b<1$, this function is decreasing.

$$
f(x)=ab^x
$$

$textbf{a.}$

To find the rebound ratio, divide the rebound height by the drop height. We do this for all data and then estimate:

$$
begin{align*}
dfrac{124}{150}&approx 0.83 & dfrac{83}{100}&= 0.83\
dfrac{59}{70}&approx 0.84 & dfrac{92}{110}&= 0.84\
dfrac{100}{120}&approx 0.83 & dfrac{33}{40}&= 0.83
end{align*}
$$

A good estimate for the rebound ratio for their ball is $text{textcolor{#c34632}{0.83}}$.

$textbf{b.}$

To find the rebound height, multiply the drop height by the rebound ratio:

$$
275text{ cm}cdot 0.83approx color{#c34632}228text{ cm}
$$

$textbf{c.}$

To find the drop height, divide the rebound height by the rebound ratio:

$$
dfrac{60text{ cm}}{0.83}approx color{#c34632}72text{ cm}
$$

$textbf{d.}$

In this case, the drop height is 200 m. To find the rebound height, multiply the drop height by the rebound ratio:

$$
200text{ m}cdot 0.83approx color{#c34632}166text{ m}
$$

$textbf{e.}$

Continue multiplying the drop height by the rebound ratio to find the required rebound heights:

$$
begin{align*}
&text{after the second bounce: }166text{ m}cdot 0.83approx color{#c34632}138text{ m}\
&text{after the third bounce: }138text{ m}cdot 0.83approx color{#c34632}115text{ m}
end{align*}
$$

The rebound ratio for tennis ball is $dfrac{111}{200}=0.555$

$textbf{(a)}$ Given that the dropping height of the tennis ball is 10 feet. The rebound height of the ball can be found by multiplying the dropping height by rebound ratio as shown below.

$$
begin{align*}
text{rebound height}&=10times 0.555\
&=5.55 text{ feet}
end{align*}
$$

$textbf{(b)}$ For finding the rebound height after the second bounce, we consider the new dropping height as 5.55 feet and multiply it with the rebound ratio.

$$
begin{align*}
text{rebound height after second bounce}&=5.55times 0.555\
&=3.08 text{ feet}
end{align*}
$$

$textbf{(c)}$ Either we can keeps on multiplying rebound ratio five times as done above or we can put the $n=5$ in the general form equation, which is given by

$$
begin{align*}
text{rebound height}_n&=10times (0.555)^n\
text{rebound height}_5&=10times (0.555)^5\
&=10times0.0526\
&=0.526
end{align*}
$$

$textbf{(a)}$ $5.5$ feet

$textbf{(b)}$ $3.08$ feet

$textbf{(c)}$ $0.526$ feet

We can write the equation of line by picking any two pair of given points in the table with the procedure explained below.

First we can consider the given variable $f(x)$ as output variable $y$.
Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we can choose two points as $(3,4)$ and $(5,10)$. Let $(x_1,y_1)=(3,4)$ and $(x_2,y_2)=(5,10)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{10-4}{5-3}\
&=dfrac{6}{2}\
&=3
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $3$ passing through a point $(3,4)$ is given by

$$
begin{align*}
y-(4)&=3cdot (x-3)\
y-4&=3cdot x+3cdot (-3)\
y&=3cdot x-9+4\
y&=3cdot x-5
end{align*}
$$

Now, again replace the $y$ as $f(x)$ then the equation becomes

$$
boxed{f(x)=3cdot x-5}
$$

We can check the validity of the equation by putting any other point from the table and see if it satisfies the equation or not.

Lets check the validity of the equation for the point $(12,31)$

$$
begin{align*}
f(x)&=3cdot x-5\
31&=3cdot (12)-5\
31&=36-5\
31&=31tag{true always}
end{align*}
$$

Hence, our equation for the line is valid.

$$
f(x)=3cdot x-5
$$

$textbf{(a)}$
$$
begin{align*}
dfrac{2x-8}{10}&=6tag{multiply each side by 10}\
(dfrac{2x-8}{10})cdot 10&=6cdot 10\
2x-8&=60tag{add 8 to the each side}\
2x-8+8&=60+8\
2x&=68tag{ divide each side by 2}\
dfrac{2x}{2}&=dfrac{68}{2}\
&boxed{x=34}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
9^x&=3^{40} tag{replace $9$ with $3^2$}\
(3^2)^x&=3^{40}tag{use $(a^m)^n=a^{mcdot n}$}\
3^{2 cdot x}&=3^40 tag{equate the powers as base is same}\
2x&=40tag{divide each side by 2}\
dfrac{2x}{2}&=dfrac{40}{2}\
&boxed{x=20}
end{align*}
$$

$textbf{(a)}$ $34$

$textbf{(b)}$ $20$

$textbf{(a)}$ The shape created by the reflection of segment AB across the line $x=3$ and connecting the point $A$ to $A’$ is found to be a triangle. The coordinates of triangle formed are $A(6,2)$, $B(3,5)$ and $A'(0,2)$.

$textbf{(b)}$ The resultant polygon is a rectangle. The reflected line is marked in red and segments drawn to connect the vertex points are in green colour. See the figure attached below.

(a) The shape created by the reflection of segment AB across the line $x=3$ is a triangle.

(b) The resultant polygon is a rectangle.

$textbf{a.}$

The sequence generator multiplies the input by 3:

$$
begin{align*}
2cdot 3&=6\
6cdot 3&=18\
18cdot 3&=54
end{align*}
$$

$textbf{b.}$

By continuing to multiply the input by 3, as what we found the sequence generator does, the next two terms are:

$$
begin{align*}
54cdot 3&=color{#c34632}162\
162cdot 3&=color{#c34632}486
end{align*}
$$

$textbf{c.}$

The first term is 5 so the next four terms are:

$$
begin{align*}
5cdot 3&=color{#c34632}15\
15cdot 3&=color{#c34632}45\
45cdot 3&=color{#c34632}135\
135cdot 3&=color{#c34632}405
end{align*}
$$

First, we find a pattern for the given sequences to know which ones must be in the same family of sequences. The next 3 terms will also be given:

$textbf{a.}$

Each term is found by adding a constant number to the previous term which is 3:

$$
-4,-1,2,5,color{#c34632}8,11,14
$$

$textbf{b.}$

Each term is found by multiplying a constant number to the previous term which is 2:

$$
1.5,3,6,12,color{#c34632}24,48,96
$$

$textbf{c.}$

The terms of the sequence are the squares of the whole numbers:

$$
0^2,1^2,2^2,3^2,dots
$$

So, the next 3 terms are:

$$
0^2,1^2,2^2,3^2, 4^2,5^2,6^2
$$

or

$$
0,1,4,9, color{#c34632}16,25,36
$$

$textbf{d.}$

Each term is found by adding a constant number to the previous term which is 1.5:

$$
2,3.5, 5,6.5,color{#c34632}8,9.5, 11
$$

$textbf{e.}$

Starting from the third term, each term is found by adding the previous two terms:

$$
1,1,2,3,5,8,color{#c34632}13,21,34
$$

$textbf{f.}$

Each term is found by adding a constant number to the previous term which is $-2$:

$$
9,7,5,3,color{#c34632}1,-1,-3
$$

$textbf{g.}$

Each term is found by multiplying a constant number to the previous term which is $dfrac{1}{2}$:

$$
48,24,12,color{#c34632}6,3,dfrac{3}{2}
$$

$textbf{h.}$

Each term is found by multiplying a constant number to the previous term which is $dfrac{1}{3}$:

$$
27,9,3,1,color{#c34632}dfrac{1}{3}, dfrac{1}{9},dfrac{1}{27}
$$

$textbf{i.}$

The terms of the sequence are the squares of the integers from $-2$, multiplied by 2:

$$
2(-2)^2, 2(-1)^2, 2(0)^2, 2(1)^2, 2(2)^2, 2(3)^2,dots
$$

So, the next 3 terms are:

$$
2(-2)^2, 2(-1)^2, 2(0)^2, 2(1)^2, 2(2)^2, 2(3)^2, 2(4)^2, 2(5)^2, 2(6)^2
$$

or

$$
8,2,0,2,8,18,color{#c34632}32,50,72
$$

$textbf{j.}$

Each term is found by multiplying a constant number to the previous term which is $2$:

$$
dfrac{5}{4},dfrac{5}{2},5,10,color{#c34632}20,40,80
$$

(1)

Based on the results, the following sequences belong to the same family where you add a constant number to find the next term:
a, d, and f
The following sequences belong to the same family where you multiply a constant number to find the next term:
b, g, h, and j
The following sequences belong to the same family where you square a number:
c and i
Only sequence e does not belong to any kind of family.

(2)

The next terms were already given above (which should be written on the strips).

(3) and (4)

The Resource Page for (3) is a table and a graph for (4) which is shown below. All sequences should be discrete (that is, just points), since we only took natural numbers from $n=1$.

The family that I used is the sequence where you add a constant number to the previous term to find the next term:

* $bullet$ There is a common difference between consecutive terms.
* $bullet$ When graphed, the points representing the terms lie on the same line.
* $bullet$ The growth is linear so we can compare it to a linear function and thus, a pattern rule can be found using the form: $y=mx+b$ or in this case, $t(n)=mn+t(0)$.

$textbf{(a)}$ The sequences which falls into the family of arithmetic sequence are (a) , (d) and (f)

The sequence (f) is $9, 7, 5, 3,cdots$

We can assume that the sequence generator adds $-2$ to the each previous term. Hence it should be included to the family of arithmetic sequence.

$textbf{(b)}$ The sequences which falls into the family of geometric sequence are (b) , (g) , (h) and (j)

The sequence (f) is $27, 9, 3, 1,cdots$

We can assume that the sequence generator multiply $dfrac{1}{3}$ to the each previous term. Hence it should be included to the family of geometric sequence.

$textbf{a.}$

The (extended) sequences for (c) and (i) are:

$$
begin{align*}
&text{(c): }0,1,4,9, 16,25,36\
&text{(i): }8,2,0,2,8,18, 32,50,72
end{align*}
$$

As what we have found, the similarity is that these two sequences are in the same family in a sense that squaring is involved in finding the terms: (c) is the perfect squares and (d) is twice the perfect squares.

$textbf{b.}$

The terms of the Fibonacci sequence rely on the previous terms unlike the other three sequences which can be found using their term numbers.

$textbf{(a)}$ Let $a_n$ represents the $n^text{th}$ term then we can represent these data points as shown below.

$$
begin{align*}
a_1&=475\
a_2&=290\
a_3&=175\
a_4&=100\
a_5&=60
end{align*}
$$

Lets find out the common difference between the consecutive terms to check for the arithmetic sequence.

$$
begin{align*}
a_2-a_1&=475-290=185\
a_3-a_2&=290-175=115\
a_4-a_3&=175-100=75\
a_5-a_4&=100-60=40
end{align*}
$$

We can see that the common difference between the terms is not a constant number, hence it does belong to family of arithmetic sequence.

Lets find the common ratio of consecutive terms to check for the geometric sequence.

$$
begin{align*}
dfrac{a_1}{a_2}&=dfrac{475}{290}=1.637\
dfrac{a_2}{a_3}&=dfrac{290}{175}=1.657\
dfrac{a_3}{a_4}&=dfrac{175}{100}=1.75\
dfrac{a_4}{a_5}&=dfrac{100}{60}=1.667
end{align*}
$$

We can see that the common ratio is also not the same. hence this sequence also not belongs to the family of geometric sequence.

$textbf{(b)}$ By using the argument of the previous option we can see that the given sequence is neither arithmetic nor geometric and belongs to something else.

The average distance from the earth to the moon is $3.844times 10^8$ meters and the length of the average pencil is $1.8 times 10^{-1}$ meters. For finding the number of pencils that needs to put together to reach the moon we need to divide the distance of earth from the moon with the length of the pencil.

$$
begin{align*}
text{no. of pencils required}&=dfrac{3.844times 10^8}{1.8 times 10^{-1}}\
&=dfrac{3.844}{1.8}times dfrac{10^8}{10^{-1}}tag{use $dfrac{a^m}{a^n}=a^{m-n}$}\
&=2.135times 10^{8-(-1)}\
&=2.135times 10^{8+1}\
&boxed{2.135times 10^{9}}
end{align*}
$$

Therefore, we will require approximately $2.136times 10^9$ pencils to cover the distance between earth and moon.

we will require approximately $2.136times 10^9$ pencils to cover the distance between earth and moon.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have given points as $(-2,1)$ and $(2,-11)$. Let $(x_1,y_1)=(-2,1)$ and $(x_2,y_2)=(2,-11)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-11-1}{2-(-2)}\
&=dfrac{-12}{2+2}\
&=dfrac{-12}{4}\
&=-3
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-3$ passing through a point $(-2,1)$ is given by

$$
begin{align*}
y-(1)&=-3cdot (x-(-2))\
y-1&=-3cdot x -3cdot (+2)\
y-1&=-3cdot x-6\
y=-3cdot x-6+1\
&boxed{y=-3 x-5}
end{align*}
$$

$$
y=-3x-5
$$

3 dozen muffins requires 16 ounces of chocolate chips. Therefore, ounces of Chocolate chips required per dozen muffins is given by $dfrac{16}{3}$ ounces.

Now, the chocolate chips required to make 8 dozen muffins

$$
begin{align*} &=text{ounces per dozen} times text{8 dozen}\
&=dfrac{16}{3} times 8\
&=dfrac{16times 8}{3}\
&approx 43 text{ ounces}
end{align*}
$$

$43$ ounces

For calculating the area we need to see first if the triangle formed by the given vertices is a right angled triangle or not.
If it is a right angled triangle then we can easily find the length of base and height of the triangle and easily calculate the area of the triangle as
$$
text{Area}=dfrac{1}{2}cdot text{Base}times text{perpendicular height}
$$

Now, for calculating the length of the line segment joining 2 vertices we can apply the formula to calculate the distance between the coordinates points on the graph.

The distance between the two points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{distance}=sqrt{(y_2-y_1)^2+(x_2-x_1)^2}
$$

From the figure we can say that the segment AC is making the perpendicular distance of the triangle which is

$$
begin{align*}
text{length}_{AC}&=sqrt{(260-(-10))^2+(525-525)^2}\
&=sqrt{270^2+0^2}\
&=sqrt{270^2}\
&=270
end{align*}
$$

Similarly, The segment AB is making the base of the triangle.

$$
begin{align*}
text{length}_{AB}&=sqrt{(260-260)^2+(725-525)^2}\
&=sqrt{0^2+200^2}\
&=sqrt{200^2}\
&=200
end{align*}
$$

Therefore, the area of the $Delta$ABC is given by

$$
begin{align*}
text{Area}&=dfrac{1}{2}cdottext{Base}times text{perpendicular height}\
&=dfrac{1}{2}cdot 200times 270\
&=dfrac{54000}{2}\
&boxed{27000} text{ unit}^2
end{align*}
$$

Area$=27000 text{ unit}^2$

$textbf{(a)}$We can represent the amount of water at the end of each day by multiplying the the amount of water present in the tank at each previous day with $dfrac{1}{2}$.

Let $w(n)$ represents the amount of liters of water in the tank at the end of $n^text{th}$ day.

$$
begin{align*}
w(0)&=8000\
w(1)&=8000cdot dfrac{1}{2}=4000\
w(2)&=8000cdot dfrac{1}{2}=2000\
w(3)&=8000cdot dfrac{1}{2}=1000\
w(4)&=8000cdot dfrac{1}{2}=500
end{align*}
$$

Therefore, the amount of water in the tank on the $4^text{th}$ day will be 500 liters.

$textbf{(b)}$ Lets calculate the general representation of amount of water in the tank at the end of $n^text{th}$ day

$$
begin{align*}
w(0)&=8000\
w(1)&=8000cdot dfrac{1}{2}=4000=8000cdot(dfrac{1}{2})^1 \
w(2)&=8000cdot dfrac{1}{2}=2000=8000cdot(dfrac{1}{2})^2\
w(3)&=8000cdot dfrac{1}{2}=1000=8000cdot(dfrac{1}{2})^3 \
w(n)&=8000cdot dfrac{1}{2}=8000cdot(dfrac{1}{2})^n \
end{align*}
$$

Hence, we can calculate the amount of water at the end of the $8^text{th}$ day by putting value of $n$ equal to 8 in the equation $w(n)=8000cdot(dfrac{1}{2})^n$

$$
begin{align*}
w(8)&=8000cdot(dfrac{1}{2})^8\
&=8000cdot (dfrac{1}{256})\
&=dfrac{8000}{256}\
&boxed{31.25}
end{align*}
$$

$textbf{(a)}$ The amount of water in the tank on the $4^text{th}$ day will be 500 liters.

$textbf{(b)}$ The amount of water at the end of the $8^text{th}$ day will be 31.25 liters.

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
y+0.03y& {=}quad :color{#c34632}1.03y\
end{align*}
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
z-0.2z& {=}quad :color{#c34632} 0.8z\
end{align*}
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
x+0.002x& {=}quad : color{#c34632}1.002x\
end{align*}
$$

$$
color{#4257b2} text{ a) }1.03y
$$

$$
color{#4257b2} text{ b) }0.8z
$$

$$
color{#4257b2} text{ c) }1.002x
$$

The slope triangle is drawn in the figure attached below.

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, we have given points as $(2,150)$ and $(4,130)$. Let $(x_1,y_1)=(2,150)$ and $(x_2,y_2)=(3,130)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{130-150}{4-2}\
&=dfrac{-20}{2}\
&=-10
end{align*}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-10$ passing through a point $(2,150)$ is given by

$$
begin{align*}
y-(150)&=-10cdot (x-2)\
y-150&=-10cdot x -10cdot -2\
y-150&=-10cdot x +20\
y&=-10 cdot x + 20+150\
&boxed{y=-10 x + 170}
end{align*}
$$

$$
y=-10 x + 170
$$

$textbf{(a)}$ $f(c-4)$ is ” the output for the input that is 4 less than $x$”.

$textbf{(b)}$ $f(0.5b)$ is ” the output for the input that is half of $b$ “.

$textbf{(c)}$ $f(d)+12$ is ” the output which is 12 more than the output for input $d$”

$textbf{(a)}$ $f(c-4)$ is ” the output for the input that is 4 less than $x$”.

$textbf{(b)}$ $f(0.5b)$ is ” the output for the input that is half of $b$ “.

$textbf{(c)}$ $f(d)+12$ is ” the output which is 12 more than the output for input $d$”

Let the $y$ represents the number of lily pads in the pond and $t$ represents time in hour then we can represent the given condition of Harry’s ponds as shown below

$$
y_1=20-4t
$$

we can represent the given condition of Heinrick’s ponds as shown below

$$
y_2=29-7t
$$

$textbf{(a)}$ We need to find the time when the number of lily pads remaining in each pond will be same. This means we need to equate both $y_1$ and $y_2$ and solve for $t$

$$
begin{align*}
Rightarrow\
20-4t&=29-7ttag{add $7t$ to the each side}\
20-4t+7t&=29-7t+7t\
20+3t&=29tag{subtract 20 from each side}\
20-20+3t&=29-20\
3t&=9tag{divide each side by 3}\
t&=3
end{align*}
$$

Therefore, after 3 hours there will be same number of lily pads in the both the ponds.

$textbf{(b)}$ we can put $t=3$ in any of the equation above to find the number of lily pads after 3 hours when number of the lily pads are the same.
$$
begin{align*}
y_1&=20-4t\
&=20-4(3)\
&=20-12\
&=8
end{align*}
$$

Therefore, there will be 8 lily pads in the each ponds at the time when the number becomes equal.

$textbf{(a)}$ After 3 hours there will be same number of lily pads in the both the ponds.

$textbf{(b)}$ There will be 8 lily pads in the each ponds at the time when the number becomes equal.

#### (a)

An expression which represents the total number of birds is:

$$
G+D+C+E+S+B+L+F
$$

#### (b)

An expression which represents the percentage of birds they saw at the lake is:

$$
dfrac{100text{L}}{G+D+C+E+S+B+L+F}
$$

#### (c)

An expression which represents the fraction of birds that were geese or ducks is:

$dfrac{E}{G+D+C+E+S+B+L+F}$ or $dfrac{D}{G+D+C+E+S+B+L+F}$

a) $G+D+C+E+S+B+L+F$; b) $dfrac{100text{L}}{G+D+C+E+S+B+L+F}$; c) $dfrac{E}{G+D+C+E+S+B+L+F}$ or $dfrac{D}{G+D+C+E+S+B+L+F}$

textbf{(a)} The given sequence $-9,-5,-1,3,7,cdots $ is an arithmetic sequence as each next term can be obtained by adding a constant term (4) in the previous term.\\
textbf{(b)} The first term of the sequence is $-9$ which is represented by $t(1)$.\\
textbf{(c)} The sequence generator is a function which takes each previous term of a sequence as input and provides the next term of sequence as output. For example in case of a arithmetic sequence a sequence generator takes each term as input then add a constant value to it to give the next term of the sequence which will go to the another sequence generator as input which will add a constant value and give the next term of the sequence. This process continued as long as we want to generate the terms of this sequence.\\
textbf{(d)}
begin{center}
begin{tabular}{|p{2cm}|p{2cm}|}
hline
$n$&$t(n)$\
hline
0&-13\
hline
1&-9\
hline
2&-5\
hline
3&-1\
hline
4&3\
hline
5&7\
hline
6&11\
hline
7&15\
hline
end{tabular}
end{center}
textbf{(e)} The graph is attached below. The graph is discrete as the sequence does not take all values of numbers\

$textbf{(f)}$ The common difference is $4$ which is the difference of the $y$ coordinates of the consecutive points on the graph. This does make sense because difference of the $y$ coordinates of any two consecutive points represents the common difference which should be constant and same for every difference of consecutive terms.

$textbf{(g)}$
$$
begin{align*}
t(1)&=-9\
t(2)&=-5=-9+4=t(1)+4times(1)=-9+4times(2-1)\
t(3)&=-1=-5+4=t(2)+4=-9+4+4=-9+4times (3-1)\
&boxed{t(n)=-9+4times (n-1)}
end{align*}
$$

$textbf{(h)}$ As the sequence can start from term number 0 (which gives its zeroth term which is not listed generally) and can have only positive integral value of term number. Therefore, the domain for the sequence is set of whole numbers ($0,1,2,3,4,5,cdots$)

$textbf{(i)}$ $t(n)$ represents the zeroth term of the sequence when $n=0$ which can be found out by putting $n=0$ into general form equation
$$
t(0)=-9+4times(0-4)=-9-4=-13
$$

$textbf{(a)}$ Arithmetic sequence

$textbf{(b)}$ $-9$

$textbf{(c)}$ The sequence generator is a function which takes each previous term of a sequence as input and provides the next term of sequence as output.

$textbf{(d)}$ Table is attached inside.

$textbf{(e)}$ See the graph inside.

$textbf{(f)}$ The common difference is $4$ which is the difference of the $y$ coordinates of the consecutive points on the graph. This does make sense because difference of the $y$ coordinates of any two consecutive points represents the common difference which should be constant and same for every difference of consecutive terms.

$textbf{(g)}$ $t(n)=-9+4times (n-1)$

$textbf{(h)}$ As the sequence can start from term number 0 (which gives its zeroth term) and will have only positive integral value of term number. Therefore, the domain for the sequence is set of whole numbers ($0,1,2,3,4,5,cdots$)

$textbf{(i)}$ $t(n)$ represents the zeroth term of the sequence when $n=0$ which is $-13$ in this case.

Given that $t(n)=-4, -1, 2, 5, cdots$

$textbf{(a)}$ The first term of this sequence is $-4$.
For the common difference we can find difference between any two consecutive terms in order of higher term number subtracted from lower term number. This gives common difference as $-1-(-4)=3$

Now. zeroth term can be found by subtracting common difference from the first term as it is a previous term than the first. which implies
$$
t(0)=-4-(3)=-7
$$

$textbf{(b)}$ The general term of an arithmetic sequence is given by
$$
t(n)=t(1)+(n-1)cdot d
$$
where $t(1)$ is the first term of the sequence and $d$ is the common difference and $n$ represents the term number. For this sequence , we have $t(1)=-4$ and $d=3$ so equation for $t(n)$ can be found out by putting these value in the general form which gives

$$
t(n)=-4+(n-1)cdot 3
$$

Now, we can check if the given terms of the sequence satisfy it or not.

$bullet$ Checking for $t(2)=-1$

$$
begin{align*}
t(n)&=-4+(n-1)cdot 3\
t(2)&=-4+(2-1)cdot 3\
-1&=-4+1cdot 3\
-1&=-1tag{true}
end{align*}
$$

$bullet$ Checking for $t(3)=2$

$$
begin{align*}
t(3)&=-4+(3-1)cdot 3\
2&=-4+2cdot 3\
2&=-4+6\
2&=2 tag{true}
end{align*}
$$

$bullet$ Checking for $t(4)=5$

$$
begin{align*}
t(4)&=-4+(4-1)cdot 3\
5&=-4+3cdot 3\
5&=-4+9\
5&=5 tag{true}
end{align*}
$$

$textbf{(c)}$ For this we put $t(n)=42$ in the equation of the sequence and solve for the value of $n$ at which the equation satisfies this. If this value of $n$ comes as positive integer value, then we say that 42 is a part of the sequence otherwise it is not.

$$
begin{align*}
t(n)&=-4+(n-1)cdot 3\
42&=-4+ncdot 3-1cdot 3\
42&=-4-3+3n\
42&=-7+3ntag{add 7 to the each side}\
42+7&=-7+7+3n\
49&=3ntag{divide each side by 3}\
n&=dfrac{49}{3}
end{align*}
$$

The value of $n$ obtained is not integral and hence 42 cannot be a term of of the sequence.

$textbf{(d)}$ Yes, given function $f(x)=3x-7$ can obtain a value 42 because it can take all real values for input $x$. Unlike the sequence equation it is not constrained to only take integral values of inputs. Hence it will take the value 42 corresponding to $x=dfrac{49}{3}$

$textbf{(e)}$ The $t(n)$ is a discrete function which can take only whole number as input and have only discrete outputs. whereas $f(x)$ is a continuous function which can take any real value of $x$ as input and have all real values as output.

$textbf{(a)}$ -7 is the zeroth term, -4 is the first term and the common difference is $3$

$textbf{(b)}$ $t(n)=-4+(n-1)d$

$textbf{(c)}$ No

$textbf{(d)}$ yes

$textbf{(e)}$ The $t(n)$ is a discrete function which can take only whole number as input and have only discrete outputs. whereas $f(x)$ is a continuous function which can take any real value of $x$ as input and have all real values as output.

Since the graph of an arithmetic sequence is a line when the terms are connected, then we can find an equation for the arithmetic sequence using the slope-intercept form of a line:

$$
y=mx+b
$$

or in this case,

$$
t(n)=mn+t(0)
$$

Representing the two terms as points, we have $(5,11)$ and $(50,371)$ so the slope is:

$$
m=dfrac{Delta y}{Delta x }=dfrac{371-11}{50-5}=dfrac{360}{45}=8
$$

This is also the sequence generator.

To find $t(0)$, use the slope and either of the two points. I used $(5,11)$:

$$
begin{align*}
11&=8(5)+t(0)\
11&=40+t(0)\
-29&=t(0)\
color{#c34632}t(0)&= color{#c34632}-29
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=8n-29
$$

sequence generator: 8

$t(0)=-29$

$$
t(n)=8n-29
$$

$textbf{(a)}$ The initial value or the zeroth term is represented by $t(0)$ and this number is not listed as a part of the sequence. The sequence always start with first term which is represented by $t(1)$. Hence, there were $t(1)$ cards in the original box.

We know that the general term of an arithmetic progression is given by

$$
t(n)=a+(n-1)d
$$

where, $t(n)$ represents the $n^text{th}$ term of the sequence , $a$ represents the first term of the sequence, $d$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.

Given that $t(3)=52$ and $t(7)=108$
We can put the given known value ot terms into the equations and solve for the values of $a$ and $d$ as shown below.

$$
begin{align*}
t(n)&=a+(n-1)d\
t(3)&=a+(3-1)d\
52&=a+2d
end{align*}
$$

$$
begin{align*}
t(n)&=a+(n-1)d\
t(7)&=a+(7-1)d\
108&=a+6d
end{align*}
$$

So, we have 2 unknown and 2 equation in two variables, which are

$$
begin{align}
52&=a+2d\
108&=a+6d
end{align}
$$

To solve the above equation, first subtract eq(1) from the eq(2), which will result in

$$
begin{align*}
108-52&=(a-a)+(6-2)d\
56&=4dtag{divide each side by 4}\
dfrac{56}{4}&=dfrac{4d}{4}\
14&=d
end{align*}
$$

Now, we can put $d=14$ in any of the equation to solve for $a$.
begin{align*}
52&=a+2d\
52&=a+2(14)\
52&=a+28tag{subtract 28 from each side}\
52-28&=a+28-28\
24&=a
end{align*}
Now, we can put $a=24$ and $d=14$ to write the final equation.
begin{align*}
boxed{t(n)=24+(n-1)14}
end{align*}
So we can fin the terms of the sequence by putting value of $n$ in the general equation. The table of values is attached below.\
begin{center}
begin{tabular}{|p{1.5cm}|p{2cm}|}
hline
term no. & term value\
hline
1&24\
hline
2&38\
hline
3&52\
hline
4&66\
hline
5&70\
hline
end{tabular}
end{center}

$textbf{(b)}$ For finding the number of cards in Kodi’s collection after 10 years from now, we need to find the value of $t(10+7)=t(17)$

$$
begin{align*}
t(n)&=24+(n-1)14\
t(17)&=24+(17-1)14\
t(17)&=24+16cdot 14\
t(17)&=24+224\
t(17)&=248
end{align*}
$$

Therefore, there will be 248 cards in his collection after 10 years from now.

$textbf{(c)}$ We have already determined the equation in the part (a).

$$
boxed{t(n)=24+(n-1)14}
$$

where, $t(n)$ represents the $n^text{th}$ term of the sequence , $24$ represents the first term of the sequence, $14$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.

$textbf{(a)}$ $(24, 38, 52, 66, 70,cdots)$

$textbf{(b)}$ $248$ cards

$textbf{(c)}$ $t(n)=24+(n-1)14$

Since the graph of an arithmetic sequence is a line when the terms are connected, then we can find an equation for the arithmetic sequence using the slope-intercept form of a line:

$$
y=mx+b
$$

or in this case,

$$
t(n)=mn+t(0)
$$

We are given that $m=-17$ and representing the term as a point, we have $(16,93)$, we can find $t(0):$
$$
begin{align*}
93&=-17(16)+t(0)\
93&=-272+t(0)\
365&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=-17n+365
$$

$$
t(n)=-17n+365
$$

$textbf{(a)}$ We know that the general term of an arithmetic progression is given by

$$
t(n)=a+(n-1)d
$$

where, $t(n)$ represents the $n^text{th}$ term of the sequence , $a$ represents the first term of the sequence, $d$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.

We are given $t(15)=735$ and common difference $d=20$
So, we can put the given known value ot terms into the equations and solve for the value of
$a$ which in our case represent the price at the beginning of the day.

$$
begin{align*}
t(n)&=a+(n-1)d\
t(15)&=a+(15-1)d\
735&=a+14(20)\
735&=a+280tag{subtract 280 from each side}\
735-280&=a+280-280\
455&=a
end{align*}
$$

Therefore, the price at the beginning of the day was $455.\\
textbf{(b)} \Now, it is given that the contest always starts with $100 which means$t(1)=100$and we need to find the value of term number$n$corresponding to$t(n)=1360$

$$
begin{align*}
t(n)&=a+(n-1)d\
1360&=100+(n-1)20\
1360&=100+(n)20-20\
1360&=80+20ntag{subtract 80 from each side}\
1360-80&=80-80+20n\
1280&=20ntag{divide each side by 20}\
dfrac{1280}{20}&=dfrac{20n}{20}\
64&=n
end{align*}
$$

Therefore, we can say that 64th person was the winner and hence, 63 people must have answered the questions incorrectly.

$textbf{(a)}$ $$ 455$

$textbf{(b)}$ 63 people answered the questions incorrectly.

The slope of 22 is also the sequence generator. Given two terms, we can check if we can obtain $t(13)=300$ from $t(8)=164$ by continually adding:

$$
begin{align*}
t(9)&=t(8)+22=164+22=186\
t(10)&=t(9)+22=186+22=208\
t(11)&=t(10)+22=208+22=230\
t(12)&=t(11)+22=230+22=252\
t(13)&=t(12)+22=252+22=274ne 300
end{align*}
$$

Since we did not obtain $t(13)=300$ using the slope, then it is not possible to create an arithmetic sequence to fit her information.

Since the graph of an arithmetic sequence is a line when the terms are connected, then we can find an equation for the arithmetic sequence using the slope-intercept form of a line:

$$
y=mx+b
$$

or in this case,

$$
t(n)=mn+t(0)
$$

$textbf{a.}$

Find the slope (or sequence generator) using two points, say $(3,10)$ and $(7,54)$:

$$
m=dfrac{Delta y}{Delta x }=dfrac{54-10}{7-3}=dfrac{44}{4}=11
$$

Using any point, say $(3,10)$, solve for $t(0)$:

$$
begin{align*}
10&=11(3)+t(0)\
10&=33+t(0)\
-23&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=11n-23
$$

$textbf{b.}$

Find the slope (or sequence generator) using two points, $(100,10)$ and $(70,100)$:

$$
m=dfrac{Delta y}{Delta x }=dfrac{100-10}{70-100}=dfrac{90}{-30}=-3
$$

Using any point, say $(100,10)$, solve for $t(0)$:

$$
begin{align*}
10&=-3(100)+t(0)\
10&=-300+t(0)\
310&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=-3n+310
$$

a. $t(n)=11n-23$

b. $t(n)=-3n+310$

447 is a term of the sequence if you can solve for $n$ such that it is a positive integer. If it is, then 447 is the $n$th term. Recall that in a sequence, $n$ are positive integers.

$textbf{a.}$

Solve for $t(n)=447$:

$$
begin{align*}
5n-3&=447\
5n&=450\
n&=90
end{align*}
$$

Since $90$ is a positive integer, then 447 is a term of this sequence and it is the $text{textcolor{#c34632}{90th term}}$.

$textbf{b.}$

Solve for $t(n)=447$:

$$
begin{align*}
24-5n&=447\
-5n&=423\
n&=-84.6
end{align*}
$$

Since $-84.6$ is not a positive integer, then 447 is not a term of this sequence.

$textbf{c.}$

Solve for $t(n)=447$:

$$
begin{align*}
-6+3(n-1)&=447\
3(n-1)&=453\
n-1&=151\
n&=152
end{align*}
$$

Since $152$ is a positive integer, then 447 is a term of this sequence and it is the $text{textcolor{#c34632}{152nd term}}$.

$textbf{d.}$

Solve for $t(n)=447$:

$$
begin{align*}
14-3n&=447\
-3n&=433\
n&=-dfrac{433}{3}
end{align*}
$$

Since $-dfrac{433}{3}$ is not a positive integer, then 447 is not a term of this sequence.

$textbf{e.}$

Solve for $t(n)=447$:

$$
begin{align*}
-8-7(n-1)&=447\
-7(n-1)&=455\
n-1&=-65\
n&=-64
end{align*}
$$

Since $-64$ is not a positive integer, then 447 is not a term of this sequence.

Lets find out the term number $n$ for which the sequence attain the value 447

$$
begin{align*}
t(n)&=14-3n\
447&=14-3n tag{subtract 14 from each side}\
447-14&=14-14 -3n\
433&=-3ntag{divide each side by -3}\
dfrac{433}{-3}&=dfrac{-3n}{-3}\
n&=dfrac{433}{-3}
end{align*}
$$

Now, As n represents the term number so it cannot take negative values and fractional value. In fact $n$ should be a whole number $(0, 1, 2, 3, 4,cdots)$

For the sequence $t(n)=14-3n$ we got the fractional term number corresponding to the term 447. hence it cannot be a term of the sequence.

As n represents the term number so it cannot take negative values and fractional value. In fact $n$ should be a whole number $(0, 1, 2, 3, 4,cdots)$

For the sequence $t(n)=14-3n$ we got the fractional term number corresponding to the term 447. hence it cannot be a term of the sequence.

Since the graph of an arithmetic sequence is a line when the terms are connected, then we can find an equation for the arithmetic sequence using the slope-intercept form of a line:

$$
y=mx+b
$$

or in this case,

$$
t(n)=mn+t(0)
$$

$textbf{a.}$

The slope (or sequence generator) is the difference of two conseuctive terms:

$$
m=7-4=10-7=13-10=3
$$

Since $t(1)=4$ and the sequence generator is $3$, then:

$$
begin{align*}
t(1)&=t(0)+3\
4&=t(0)+3\
1&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=3n+1
$$

$textbf{b.}$

The slope (or sequence generator) is the difference of two conseuctive terms:

$$
m=8-3=13-8=5
$$

Since $t(1)=3$ and the sequence generator is $5$, then:

$$
begin{align*}
t(1)&=t(0)+5\
3&=t(0)+5\
-2&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=5n-2
$$

$textbf{c.}$

The slope (or sequence generator) is the difference of two conseuctive terms:

$$
m=19-24=14-19=-5
$$

Since $t(1)=24$ and the sequence generator is $-5$, then:

$$
begin{align*}
t(1)&=t(0)-5\
24&=t(0)-5\
29&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=-5n+29
$$

$textbf{d.}$

The slope (or sequence generator) is the difference of two conseuctive terms:

$$
m=9.5-7=12-9.5=2.5
$$

Since $t(1)=7$ and the sequence generator is $2.5$, then:

$$
begin{align*}
t(1)&=t(0)+ 2.5\
7&=t(0)+ 2.5\
4.5&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=2.5n+4.5
$$

a. $t(n)=3n+1$

b. $t(n)=5n-2$

c. $t(n)=-5n+29$

d. $t(n)=2.5n+4.5$

$textbf{(a)}$
$$
t(n)=t(n-1)+dfrac{t(n-1)}{10}
$$

Every year the ticket price is increasing 10 percent of the price in the previous year.

The general ticket price in nth year in the terms of price into the starting year is given by

$$
t(n)=50cdot (dfrac{110}{100})^{n}
$$

$textbf{(b)}$ The price of the admission in the 6th year can be found out by putting 6 at the place of $n$ into the general ticket price shown in previous part.

$$
begin{align*}
t(n)&=50cdot (dfrac{110}{100})^{n}\
t(6)&=50cdot (dfrac{110}{100})^{6}\
t(6)&=50cdot (1.771)\
t(6)&=88.578approx $88.58
end{align*}
$$

$textbf{(a)}$ Every year the ticket price is increasing 10 percent of the price in the previous year.

$textbf{(b)}$ $$88.58$

$textbf{(a)}$ The actual scatterplot will look like something like in the scatterplot attached below. As the height increases the magnitude of the residuals get smaller and smaller and converges to the line of best fit.

$textbf{(b)}$ The given residual plot tells that the predictions made for taller swimmers are more accurate than the prediction made for shorter swimmers. This can be explained by the magnitudes of the residuals. The far the point is from the best fit line, the more will be the magnitude of residual and hence more will be the error in the prediction.

$textbf{(a)}$ The actual scatterplot will look like something like in the scatterplot attached inside. As the height increases the magnitude of the residuals get smaller and smaller and converges to the line of best fit.

$textbf{(b)}$ The given residual plot tells that the predictions made for taller swimmers are more accurate than the prediction made for shorter swimmers. This can be explained by the magnitudes of the residuals. The far the point is from the best fit line, the more will be the magnitude of residual and hence more will be the error in the prediction.

The height of the sand dune can be found out using Pythagoras property. By pythagoras property in a Right angled triangle, we have a relation

$$
mathrm{hypotenuse}^2= mathrm{base}^2 + mathrm{height}^2
$$

In the given figure, we have
hypotenuse=137 feet
and base=105 feet

Now, applying pythagoras theorem

$$
137^2=105^2 + mathrm{height}^2
$$

This implies,

$$
mathrm{height}=sqrt{137^2 -105^2}
$$

$$
=88 >feet
$$

For the given sequence the first term is $-8$ and the common difference (d) can be found by taking difference of any two consecutive terms in the order of lower term gets subtracted from higher term number which gives $d=-2-(-8)=6$.

$textbf{(a)}$

Now, for calculating the $10^text{th}$ term one way is that we keep adding common difference to the previously known terms until we reach $10^text{th}$ term. From teh sequence we have $t(4)=10$ now, we can proceed in the way explained above.

$$
begin{align*}
t(5)&=10+6=16\
t(6)&=16+6=22\
t(7)&=22+6=28\
t(8)&=28+6=34\
t(9)&=34+6=40\
t(10)&=40+6=46
end{align*}
$$

We have got $t(10)=46$ but this method is not efficient and will not be helpful for finding the term of the sequence with very high term number. To avoid this, we can find the the explicit equation to represent any term $t(n)$, which is our method 2 and explained below.

$bullet$ The general term of an arithmetic progression is given by

$$
t(n)=a+(n-1)d
$$

where, $t(n)$ represents the $n^text{th}$ term of the sequence , $a$ represents the first term of the sequence, $d$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.

We have $a=-8$ and $d=6$, so we can put these value to obtain the explicit equation.

$$
begin{align*}
t(n)&=(-8)+(n-1)(6)\
&boxed{t(n)=-8+(n-1)6}
end{align*}
$$

Now, we can find the $t(10)$ by putting $n=10$ into the explicit equation.

$$
begin{align*}
t(10)&=-8+(10-1)6\
t(10)&=-8+54\
t(10)&=46
end{align*}
$$

$textbf{(b)}$ We have already written the explicit equation while explaining the part (a) of the question.

$textbf{(c)}$ The recursive equation is when we try to write any next term by using previous terms. As for an arithmetic sequence each next term can be obtained by adding $d$ to its previous term. Therefore, we can represent this in form of recursive equation as shown below.

$$
begin{align*}
t(n+1)&=t(n)+6 tag{put $t(n)=-8+(n-1)6$}\
t(n+1)&=-8+(n-1)6+6\
t(n+1)&=-8+6n-6+6\
&boxed{t(n+1)=-8+6n}
end{align*}
$$

$textbf{(a)}$ $t(n)=-8+(n-1)6$

$textbf{(b)}$ $t(n+1)=t(n)+6=-8+6n$

$textbf{(a)}$ Yes, Alejandro’s sequence has same recursive equation $t(n+1)=t(n)+6$ which is same as the recursive equation of sequence from the problem 5-72.

$textbf{(b)}$He got the different sequence because he started with 0 where in the correct sequence(previous problem) the first term is -8. For writing the correct sequence Alejandro should know the value of starting term. If he has access to $t(1)$ from the correct sequence then he can write the exact sequence by using recursive equation.

In other words, we can say that recursive equation is not unique unless we fixed a term of it.

$textbf{(a)}$ Yes, Alejandro’s sequence has same recursive equation $t(n+1)=t(n)+6$ which is same as the recursive equation of sequence from the problem 5-72.

$textbf{(b)}$He got the different sequence because he started with 0 where in the correct sequence(previous problem) the first term is -8. For writing the correct sequence Alejandro should know the value of starting term. If he has access to $t(1)$ from the correct sequence then he can write the exact sequence by using recursive equation.

$textbf{(a)}$ Given that $t(1) = 3$ and general term is given by $t(n+1)=t(n)^2-1$. So each next term can be found out by putting previous term value into the general term equation like shown below.

$$
begin{align*}
t(1) &= 3\
t(2)&=t(1)^2-1=3^2-1=9-1=8\
t(3)&=t(2)^2-1=8^2-1=64-1=63\
t(4)&=t(3)^2-1=63^2-1=3969-1=3968
end{align*}
$$

$textbf{(b)}$ Avery’s sequence is neither Arithmetic nor Geometric. It is some other king of the sequence as we cannot represent the terms of these sequence by general term equations of Arithmetic sequence which is given by $t(n)=t(0)+(n-1)d$ nor by Geometric sequences whose general term is given by $t(n)=acdot r^{n-1}$.

$textbf{(c)}$ Collin could determine the $10^{th}$ term by the method shown above. For this one needs to find the $9^{th}$ term first which needs $8^{th}$ and so on. At some point one will needing only first term which is already given in question.

$textbf{(a)}$ $(3,8,63,3968)$

$textbf{(b)}$ Avery’s sequence is neither Arithmetic nor Geometric.

$textbf{(c)}$ For this Collin needs to find the $9^{th}$ term first, which needs $8^{th}$ and so on. At some point he will be needing only first term which is already given in question.

$textbf{a.}$

Since you are subtracting $2$ from (or adding $-2$ to) the previous term, then the sequence is arithmetic. To find an explicit equation, use the formula:

$$
t(n)=mn+t(0)
$$

which is similar to the slope-intercept form of a line, $y=mx+b$.

The sequence generator is $m=-2$. Given that $t(2)=19$, we can use this to find $t(0)$:

$$
begin{align*}
19&=-2(2)+t(0)\
19&=-4+t(0)\
23&=t(0)
end{align*}
$$

So, the explicit equation is:

$$
color{#c34632}t(n)=-2n+23
$$

$textbf{b.}$

From the equation, we know that the sequence generator is $m=8$. We must state a known term, say $t(1)=6(1)+8=14$ to write a recursive rule so a possible answer is:

$$
color{#c34632}t(1)=14,t(n+1)=t(n)+8
$$

a. $t(n)=-2n+23$

b. $t(1)=14,t(n+1)=t(n)+8$

textbf{(a)} Given that for a Fibonacci sequence $t(1)=1$ and $t(2)=1$ and its recursive equation is given by $$t(n+1)=t(n)+t(n-1)$$
From the recursive equation we can see that a term of a Fibonacci sequence can be determined by sum of its previous two terms. So, we can find out the $t(3)$ as shown below.
begin{align*}
t(3)&=t(2)+t(1)\
t(3)&=1+1\
t(3)&=2
end{align*}
Now, we can find out the $t(4)$ as shown below.
begin{align*}
t(4)&=t(3)+t(2)\
t(4)&=2+1\
t(4)&=3
end{align*}
Similarly, we can find the each and every term by proceeding in the same manner. The table of value are attached below.\
begin{center}
begin{tabular}{|p{2cm}|p{2cm}|}
hline
n&t(n)\
hline
1&1\
hline
2&1\
hline
3&2\
hline
4&3\
hline
5&5\
hline
6&8\
hline
7&13\
hline
8&21\
hline
9&34\
hline
10&55\
hline
end{tabular}
end{center}

$textbf{(b}$ Fibonacci sequence is neither arithmetic nor geometric sequence as it does not fit in their general terms

$textbf{(c)}$ In order to calculate the value of $100^text{th}$ term of Fibonacci sequence we need values of $99^text{th}$ and $98^text{th}$ terms of Fibonacci sequence. Now, for the value of $99^text{th}$ term we need values of $98^text{th}$ and $97^text{th}$ terms of Fibonacci sequence. Which goes on like this until we reach the known value of a term of Fibonacci series.

In other way, we need to go from bottom to top, we have to start from term 1 and then find term 2 using recursive equation.we keep doing it until we reach $100^text{th}$ term.

$textbf{(a)}$ From the recursive equation we can see that a term of a Fibonacci sequence can be determined by sum of its previous two terms.The table of values is attached inside.

$textbf{(b)}$ Fibonacci sequence is neither arithmetic nor geometric sequence.

$textbf{(c)}$ In order to calculate the value of $100^text{th}$ term of Fibonacci sequence we need values of $99^text{th}$ and $98^text{th}$ terms of Fibonacci sequence. Now, for the value of $99^text{th}$ term we need values of $98^text{th}$ and $97^text{th}$ terms of Fibonacci sequence. Which goes on like this until we reach the known value of a term of Fibonacci series.

$textbf{(a)}$ We can find the first 4 terms by putting value of $n$ $0, 1, 2, 3$ one by one as shown below.

$$
begin{align*}
t(0)&=4.5(0)-8=-8\
t(1)&=4.5(1)-8=-3.5\
t(2)&=4.5(2)-8=1\
t(3)&=4.5(3)-8=5.5
end{align*}
$$

Therefore, first 4 terms for the given explicit equation are $(-8, -3.5, 1, 5.5)$

$textbf{(b)}$ To find the $15^text{th}$ term of the sequence we need to put $n$ equal to 14 in the explicit equation. This is because we have started from the $t(0)$ as the first term.

$$
begin{align*}
t(14)&=4.5(14)-8=55
end{align*}
$$

$textbf{(c)}$ For writing the explicit equation we need to write the general term the sequence with the help of its previous term. which can be proceeded in the way shown below.
lets assume that there is linear relationship between $t(n+1)$ and $t(n)$, so we can write it as

$$
t(n+1)=acdot t(n)+c
$$

Now we can put the known values of term in this equation to make two equations in two variables so that we can able to solve for the $a$ and $c$

We have $t(0)=-8$ and $t(1)=-3.5$

$$
begin{align*}
t(0+1)&=acdot t(0)+c tag{by putting $n=0$}\
t(1)&=acdot t(0)+c tag{put known values}\
-3.5&=acdot -8+c\
-3.5&=-8a+c
end{align*}
$$

We have $t(1)=-3.5$ and $t(2)=1$

$$
begin{align*}
t(1+1)&=acdot t(1)+c tag{by putting $n=1$}\
t(2)&=acdot t(1)+c tag{put known values}\
1&=acdot -3.5+c\
1&=-3.5a+c
end{align*}
$$

Now , we have 2 unknown variables and 2 equations, we can solve these for the $a$ and $c$.

$$
begin{align}
-3.5&=-8a+c\
1&=-3.5a+c
end{align}
$$

Subtract the eq(1) from the eq(2)

$$
begin{align*}
1-(-3.5)&=(-3.5-(-8))a+c-c\
1+3.5&=(-3.5+8)a\
4.5&=4.5atag{divide each side by 4.5}\
a&=1
end{align*}
$$

Use the value $a=1$ into the eq(2) and solve for $c$

$$
begin{align*}
1&=-3.5(1)+c\
1&=-3.5+ctag{add 3.5 to the each side}\
1+3.5&=-3.5+3.5+c\
4.5&=c
end{align*}
$$

Now, put $a=1$ and $c=4.5$, so the final recursive equation becomes

$$
begin{align*}
t(n+1)&=acdot t(n)+c\
t(n+1)&=1cdot t(n)+4.5\
t(n+1)&=t(n)+4.5
end{align*}
$$

$bullet$ The more convenient way to write the recursive equation is by observing that each next term can be obtained by adding 4.5 into the previous term. therefore, we can simply write the recursive equation as

$$
t(n)=t(n-1)+4.5
$$

which is also equivalent to

$$
boxed{t(n+1)=t(n)+4.5}
$$

$textbf{(a)}$ First 4 terms for the given explicit equation are $(-8, -3.5, 1, 5.5)$

$textbf{(b)}$ To find the $15^text{th}$ term of the sequence we need to put $n$ equal to 14 in the explicit equation. This is because we have started from the $t(0)$ as the first term.

$textbf{(c)}$ $t(n+1)=t(n)+4.5$

Since the graph of an arithmetic sequence is a line when the terms are connected, then we can find an equation for the arithmetic sequence using the slope-intercept form of a line:

$$
y=mx+b
$$

or in this case,

$$
t(n)=mn+t(0)
$$

The slope (or sequence generator) is the difference of two conseuctive terms:

$$
m=8-5=11-8=14-11=17-14=3
$$

Since $t(1)=5$ and the sequence generator is $3$, then:

$$
begin{align*}
t(1)&=t(0)+3\
5&=t(0)+3\
2&=t(0)
end{align*}
$$

So, the explicit equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=3n+2
$$

From the explicit equation, we know that the sequence generator is $m=3$. We must state a known term, say $t(1)=3(1)+2=5$ to write a recursive rule so a possible answer is:

$$
color{#c34632}t(1)=5,t(n+1)=t(n)+3
$$

explicit: $t(n)=3n+2$

recursive: $t(1)=5,t(n+1)=t(n)+3$

$textbf{a.}$

Use the Distributive Property: $a(b+c)=ab+ac$

$$
(4x+5)(4x-5)= (4x+5)4x-(4x+5)5
$$

Use the Distributive Property again:

$$
= 16x^2+20x-20x-25
$$

Simplify:

$$
= color{#c34632}16x^2-25
$$

$textbf{b.}$

The expression also means:

$$
(4x+5)^2= (4x+5)(4x+5)
$$

Use the Distributive Property: $a(b+c)=ab+ac$

$$
= (4x+5)4x+(4x+5)5
$$

Use the Distributive Property again:

$$
= 16x^2+20x+20x+25
$$

Simplify:

$$
= color{#c34632}16x^2+40x+25color{white}tag{1}
$$

a. $16x^2-25$

b. $16x^2+40x+25$

b.

It can be seen from the table above, Lona will have 264 stamps in a year.

c.

$t(mathrm{n})=120+12mathrm{n}$,
where n is the number of months

d.

$$
begin{align*}
500&=120+12mathrm{n} \
380&=12mathrm{n} tag{text{subtracting 120 from each side}}\
Rightarrow mathrm{n}&=dfrac{380}{12}\
end{align*}
$$

Now, $dfrac{380}{12}$ months implies 31 months and 20 days.

a. Table above

b. 264 stamps

c. $t(mathrm{n})=120+12mathrm{n}$

d. No, If we are talking about exactly 12 tickets for each month then on 32th month she will have 4 extra tickets. Her book will be filled completely in 31 months and 20 days.

The value of $r$ can vary from $-1$ to $+1$ Where, $-1$ denotes the perfect negative linear relationship between the variables and $+1$ represents perfect positive linear relationship between the variables.

In the given situation the $r$ value is $-0.45$, which shows there is somewhat negative relationship in dietary fibre and the blood cholesterol, which is not perfectly linear. Hence, Scientist’s finding shows that increase in dietary fibre results in decrease in blood cholesterol. Similarly, Decrease in Dietary fibre in the meals signifies more blood cholesterol.

Given that the company has sold 10,000 scientific calculators so far and continues to sell 1500 per month. If $m$ represent the month and let $y_{sc}$ represents the number of scientific calculators sold then this situation can be represented in the form of equation as shown below.

$$
begin{align*}
y_{sc}&=1500cdot m+10000
end{align*}
$$

Now, also given that It has also sold 18,000 graphing calculators and continues to sell 1300 of this model each month . let $y_{gc}$ represents the number of scientific calculators sold then this situation can be represented in the form of equation as shown below.

$$
begin{align*}
y_{gc}&=1300cdot m+18000
end{align*}
$$

$textbf{(a)}$ Given that the number of scientific calculators sold is equal to the number of graphing calculators sold.

$$
Rightarrow
$$

$$
begin{align*}
y_{sc}&=y_{gc}\
1500cdot m+10000&=1300cdot m+18000
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
1500cdot m+10000&=1300cdot m+18000 tag{subtract 10000 from each side}\
1500cdot m+10000-10000&=1300cdot m+18000-10000\
1500cdot m&=1300cdot m+8000tag{subtract $1300cdot m$ from each side}\
1500cdot m-1300cdot m&=1300cdot m-1300cdot m+8000\
200m&=8000tag{divide each side by 200}\
dfrac{200m}{200}&=dfrac{8000}{200}\
m&=40
end{align*}
$$

Therefore, after 40 months the sale of both types of calculators will become equal.

$textbf{(a)}$ $1500cdot m+10000=1300cdot m+18000$

$textbf{(b)}$ After 40 months the sale of both types of calculators will become equal.

$bullet$ Sequence$textbf{ A}$ grows linearly while the sequences $textbf{ B}$ and $textbf{C}$ grow exponentially where sequence $textbf{B}$ grows more rapidly than sequence $textbf{C}$.

Sequence$textbf{ A}$ grows linearly while the sequences $textbf{ B}$ and $textbf{C}$ grow exponentially where sequence $textbf{B}$ grows more rapidly than sequence $textbf{C}$.

$textbf{(a)}$ If $n$ represents the number of years , and $t(n)$ represents the amount of money in my savings account then, I will choose sequence B as account. Because this will give the highest rate of increase of amount with the increasing years. But if I want to have the account for a period less than 3 years then Sequence A will be more suitable.

$$
begin{align*}&text{For less than 3 years – Sequence A}\
&text{For more than 3 years – Sequence B}
end{align*}
$$

$textbf{(b)}$ Sequence B will be more suitable if amount is to be kept for many years.

textbf{(c)}\
begin{tabular}{|p{1cm}|p{1cm}|}
hline
multicolumn{2}{|c|}{Sequence A} \
hline
$n$ & $t(n)$\
hline
5&135\
hline
6&162\
hline
7&189\
hline
end{tabular}\\\
begin{tabular}{|p{1cm}|p{1cm}|}
hline
multicolumn{2}{|c|}{Sequence C} \
hline
$n$ & $t(n)$\
hline
5&96\
hline
6&192\
hline
7&384\
hline
end{tabular}
\\\
textbf{(d)} Due to extension the answer will not change. Because, after 3 years the Sequence gives the highest amount of returns than any other sequence.

$textbf{(a)}$ For less than 3 years, Sequence A is more suitable and for more than 3 years, Sequence B is more suitable.

$textbf{(b)}$ Sequence B will be more suitable if amount is to be kept for many years.

$textbf{(c)}$ See the tables Inside.

$textbf{(d)}$ Due to extension the answer will not change. Because, after 3 years the Sequence gives the highest amount of returns than any other sequence.

$textbf{(a)}$ Given sequence is $12,144,1728, cdots$

$$
begin{align*}
t(1)&=12\
t(2)&=144=12times 12=12cdot t(1)\
t(3)&=1728=12times 144=12cdot t(2)
end{align*}
$$

$Rightarrow$ The above sequence is Geometric with common ratio $=12$ and the growth is exponential for this sequence.

$textbf{(b)}$ $0,5,10,15,20,25,cdots$

$$
begin{align*}
t(1)&=0\
t(2)&=5=0+5=t(1)+5\
t(3)&=10=5+5=t(2)+5\
t(4)&=15=10+5=t(3)+5\
t(5)&=20=15+5=t(4)+5
end{align*}
$$

$Rightarrow$ The above sequence is Arithmetic with common difference $=5$ and the growth is linear for this sequence.

$textbf{(c)}$ $0,4,16,36,64,cdots$

For the above given sequence we cannot find neither common ratio nor common difference hence the given sequence is neither Geometric nor Arithmetic.

$textbf{(d)}$ $1.5, 2.25, 3.375, 5.0625,cdots$

$$
begin{align*}
t(1)&=1.5\
t(2)&=2.25=1.5times 1.5=1.5cdot t(1)\
t(3)&=3.375=1.5times 2.25=1.5cdot t(2)\
t(4)&=5.0625=1.5times 3.375=1.5cdot t(3)
end{align*}
$$

$Rightarrow$ The above sequence is Geometric with common ratio $=1.5$ and the growth is exponential for this sequence.

$textbf{(a)}$ Geometric sequence; Exponential growth

$textbf{(b)}$ Arithmetic sequence; Linear growth

$textbf{(c)}$ Neither

$textbf{(d)}$ Geometric sequence; Exponential growth

$textbf{(a)}$ The given recursive equation is $t(n+1)=-2cdot t(n)$ . We can find the first five terms of sequence by putting value of $n$ starting from 1 to 5.

$$
begin{align*}
t(1)&=-3\
t(2)&=-2cdot t(1)=-2cdot -3=6\
t(3)&=-2cdot t(2)=-2cdot 6=-12\
t(4)&=-2cdot t(3)=-2cdot -12=24\
t(5)&=-2cdot t(4)=-2cdot 24=-48
end{align*}
$$

Therefore, the first five terms for the sequence are $(-3, 6, -12, 24, -48)$

$textbf{(b)}$ The given recursive equation is $t(n+1)=t(n)-5$ . We can find the first five terms of sequence by putting value of $n$ starting from 1 to 5.

$$
begin{align*}
t(1)&=8\
t(2)&=t(1)-5=8-5=3\
t(3)&=t(2)-5=3-5=-2\
t(4)&=t(3)-5=-2-5=-7\
t(5)&=t(4)-5=-7-5=-12
end{align*}
$$

Therefore, the first five terms for the sequence are $(8, 3, -2, -7, -12)$

$textbf{(c)}$ The given recursive equation is $t(n+1)=(t(n))^{-1}$. We can rewrite it as
$$
t(n+1)=dfrac{1}{t(n)}
$$
Now, we can find the first five terms of sequence by putting value of $n$ starting from 1 to 5.

$$
begin{align*}
t(1)&=2\
t(2)&=dfrac{1}{t(1)}=dfrac{1}{2}\
t(3)&=dfrac{1}{t(2)}=dfrac{1}{dfrac{1}{2}}=2\
t(4)&=dfrac{1}{t(3)}=dfrac{1}{2}\
t(5)&=dfrac{1}{t(4)}=dfrac{1}{dfrac{1}{2}}=2
end{align*}
$$

Therefore, the first five terms for the sequence are $(2, dfrac{1}{2}, 2, dfrac{1}{2}, 2)$

$textbf{(a)}$ $(-3, 6, -12, 24, -48)$

$textbf{(b)}$ $(8, 3, -2, -7, -12)$

$textbf{(c)}$ $(2, dfrac{1}{2}, 2, dfrac{1}{2}, 2)$

The indicated angles $4x-5text{textdegree}$ and $2x+9text{textdegree}$ are the pair of alternate interior angles. Therefore, these should be equal. We can solve for the $x$ by equating these angles and solving the resultant equation.

$$
begin{align*}
4x-5&=2x+9tag{add 5 to the each side}\
4x-5+5&=2x+9+5\
4x&=2x+14tag{subtract $2x$ from each side}\
4x-2x&=2x-2x+14\
2x&=14tag{divide each side by 2}\
dfrac{2x}{2}&=dfrac{14}{2}\
x&=7
end{align*}
$$

As $x$ is a part of angle therefore, it should be mentioned as 7$text{textdegree}$.

$$
x=7text{textdegree}
$$

In the hint it given that the formula for the calculation of the area of a trapezoid is
$$
A=dfrac{1}{2}(b_1+b_2)h
$$

where, $b_1$ and $b_2$ are the lengths of the bases(parallel sides) and $h$ is perpendicular height.
Now, given that the area of the trapezoid is 56 $text{cm}^2$ and also $b_1=10$ cm and $b_2=6$ cm. By putting these known values into the equation of area we can solve for $h$.

$$
begin{align*}
A&=dfrac{1}{2}(b_1+b_2)h\
56&=dfrac{1}{2}(10+6)h\
56&=dfrac{16}{2}h\
56&=8htag{divide each side by 8}\
dfrac{56}{8}&=dfrac{8h}{8}\
7&=h\
&boxed{h=7 text{ cm}}
end{align*}
$$

$$
h=7 text{ cm}
$$

$$
{color{#4257b2}text{a)}}
$$

$$
begin{align*}
left(2m^3right)left(4m^2right)& {=}quad : 2m^3cdot :4m^2tag{Remove parentheses}\
&=8m^3m^2 tag{Multiply the numbers} \
&=8m^{3+2} \
&={color{#c34632}8m^5}
end{align*}
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad :a^bcdot :a^c=a^{b+c}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{b)}}
$$

$$
begin{align*}
frac{6y^5}{3y^2}& {=}quad : frac{2y^5}{y^2}tag{Divide the numbers}\
&=2y^{5-2} \
&={color{#c34632}2y^3}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:x^{a-b}
$$

$$
text{ }
$$

$$
{color{#4257b2}text{c)}}
$$

$$
begin{align*}
frac{-4y^2}{6y^7}& {=}quad : -frac{4y^2}{6y^7}\
&=-frac{2y^2}{3y^7} tag{Cancel the common factor 2}\
&frac{y^2}{y^7}=frac{1}{y^{7-2}}=frac{1}{y^5}tag{Simplify}\
&={color{#c34632}-frac{2}{3y^5}}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:the:fraction:rule}:quad frac{-a}{b}=-frac{a}{b}
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad frac{x^a}{x^b}:=:frac{1}{x^{b-a}}
$$

$$
{color{#4257b2}text{d)}}
$$

$$
begin{align*}
left(-2x^2right)^3& {=}quad : -left(2x^2right)^3\
&=-2^3left(x^2right)^3 \
&=-2^3x^{2cdot :3}\
&={color{#c34632}-8x^6}
end{align*}
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(-aright)^n=-a^n,:mathrm{if:}nmathrm{:is:odd}
$$

$$
color{#c34632} text{ } mathrm{Apply:exponent:rule}:quad left(acdot :bright)^n=a^nb^n
$$

$$
color{#c34632} text{ }mathrm{Apply:exponent:rule}:quad left(a^bright)^c=a^{bc}
$$

$$
text{ }
$$

$$
color{#4257b2}text{ a) } 8m^5
$$

$$
color{#4257b2} text{ b) } 2y^3
$$

$$
color{#4257b2} text{ c) } -frac{2}{3y^5}
$$

$$
color{#4257b2} text{ d) }-8x^6
$$

$textbf{(a)}$ Equation of a line of a given slope $m$ and which passes through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Now, given that the line has slope $=2$ and passing through (10, 17) hence the equation of line can be written as shown below.

$$
begin{align*}
y-17&=2cdot(x-10)\
y&=2x-3
end{align*}
$$

$textbf{(b)}$ Slope of a line passing through two given points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

Now, given that line passes through (1, -4) and (-2 , 5)

$Rightarrow$
$$
text{slope}=dfrac{5-(-4)}{-2-1}=-3
$$

Therefore, the line with the slope -3 and passing through $(1,-4)$ is

$$
begin{align*}
y-(-4)&=-3cdot(x-1)\
y+4&=-3x+3\
y&=-3x-1
end{align*}
$$

$textbf{(c)}$ We can choose any two points from the given table and can solve the question exactly like the case $textbf{(b)}$ as line passes through all the given points in the table

Let the two points are $(3,0)$ and $(6,2)$.

$Rightarrow$
$$
text{slope}=dfrac{2-0}{6-3}=dfrac{2}{3}
$$

Now, the equation of the line with slope $=dfrac{2}{3}$ and passing through $(3,0)$ is

$$
begin{align*}
y-(0)&=dfrac{2}{3}cdot(x-3)\
y&=dfrac{2}{3}x-2\
end{align*}
$$

$textbf{(d)}$ From the graph we can see that the line passes through points $(0,9)$ and $4,19$, hence this also can be solved similar to that of part $textbf{(b)}$.

Slope of line passing through points $(0,9)$ and $4,19$

$$
text{slope}=dfrac{19-9}{4-0}=dfrac{5}{2}
$$

Now, the equation of the line with slope $=dfrac{5}{2}$ and passing through $(0,9)$ is

$$
begin{align*}
y-(9)&=dfrac{5}{2}cdot(x-0)\
y&=dfrac{5}{2}x+9\
end{align*}
$$

$textbf{(a)}$ $y=2x-3$

$textbf{(b)}$ $y=-3x-1$

$textbf{(c)}$ $y=dfrac{2}{3}x-2$

$textbf{(d)}$ $y=dfrac{5}{2}x+9$

$textbf{(a)}$ The number of $pi$Phone sold in the week $0^{text{th}}$ is 100 and sale is increasing with a rate of 15 % per week.

The number of $pi$Phone sold in the first week are given by

$$
=100+100times dfrac{15}{100}=100times 1.15 =115
$$

The number of $pi$Phone sold in the Second week are

$$
=115+115times dfrac{15}{100}=115times(1.15)=100times 1.15^2 =132
$$

Here, the multiplier is 1.15 Therefore, number of $pi$Phone sold in $n^{text{th}}$ week can be represented as

$$
text{no. in } n^{text{th } } text{ week}=100times (1.15)^{n}
$$

$$
Rightarrow
$$

Number of $pi$Phone sold during $4^{text{th}}$ week $=100times (1.15)^4=174.9$

which can be approximated to 175 units.

Number of $pi$Phone sold during $10^{text{th}}$ week $=100times (1.15)^{10}=404.5$

which can be approximated to 405 units.

$textbf{(b)}$ It will make a Geometric sequence as its general term can be represented as the general term of a Geometric sequence which is given by $t(0)times (r)^{n-1}$

$textbf{(c)}$ If the number of $pi$Phones sold during week 51 is known, then the number of $pi$Phones sold in the week 52 can be found out by multiplying the week $51^{text{th}}$ data by 1.15

The recursive equation can be written as $t(n+1)=t(n)times (1.15)$

$textbf{(d)}$
$$
begin{align*}
t(0)&=100\
t(1)&=100+100times dfrac{15}{100}=100times (1.15)=115\
t(2)&=115+115times dfrac{15}{100}=115times (1.15)=100times (1.15)^2\
t(3)&=100times (1.15)^2+(100times (1.15)^2)times dfrac{15}{100}=100times (1.15)^2(1+0.15)=100times (1.15)^3\
& boxed{ t(n)=100times (1.15)^n}
end{align*}
$$

Yes, result matches with the part $textbf{(a)}$.

$textbf{(e)}$
$$
begin{align*}
t(52)&=100times (1.15)^52\
&=100times (1433.1369)\
&=143313.7
end{align*}
$$

Therefore, the number of $pi$Phones sold during week 52 are approximately 143314 units.

$textbf{(a)}$ 175 and 405 units

$textbf{(b)}$ Geometric Sequence

$textbf{(c)}$ By multiplying with 1.15

$textbf{(d)}$ $t(n)=100times (1.15)^n$, Yes

$textbf{(e)}$ 143314 units

$textbf{a.}$

Since the graph of a geometric sequence is an exponential graph when the terms are connected, then we can use the exponential equation:

$$
y=acdot b^x
$$

or in this case,

$$
t(n)= t(0)cdot b^n
$$

Let $n$ be the number of weeks so that $t(n)$ is the number of orders. The sequence generator is the multiplier $b$. Since the sales increase by 7% each week, then:

$$
b=100%+7%=107%=1.07
$$

Since $t(0)=10000$ and the sequence generator is $1.07$, the the explicit equation is:

$$
color{#c34632}t(n)=10000(1.07)^n
$$

From the explicit equation, we know that the sequence generator is $1.07$. We must state a known term, say $t(0)=10000$ (which is given) to write a recursive rule so a possible answer is:

$$
color{#c34632}t(0)=10000,t(n+1)= 1.07cdot t(n)
$$

$textbf{b.}$

Change the multiplier from 1.07 to 1.17 to find the new equations. The explicit equation will be:

$$
color{#c34632}t(n)=10000(1.17)^n
$$

and the recursive rule is:

$$
color{#c34632}t(0)=10000,t(n+1)= 1.17cdot t(n)
$$

a. explicit: $t(n)=10000(1.07)^n$ ; recursive: $t(0)=10000,t(n+1)= 1.07cdot t(n)$

b. explicit: $t(n)=10000(1.17)^n$ ; recursive: $t(0)=10000,t(n+1)= 1.17cdot t(n)$

Given that the 10,000 iTalks were made and sold initially and then after each week sales decreases by 3%.

$textbf{(a)}$
$$
begin{align*}
text{week(0) } t(0)&=10000\
text{week(1) } t(1)&=10000-dfrac{3}{100}cdot 10000=10000-300=9700\
text{week(2) } t(2)&=9700-dfrac{3}{100}cdot 9700=9700-291=9409\
text{week(3) } t(3)&=9409-dfrac{3}{100}cdot 9409=9409-282.27=9126.73\
text{week(4) } t(4)&=9126.73-dfrac{3}{100}cdot 9126.73=9126.73-273.80=8852.92
end{align*}
$$

Finding the sales after $10^text{th}$ week ($t(10)$) by above procedure will take longer, but if someone has to find the $t(10)$ by this method only, the idea is clear. Instead we can find explicit and recursive equation first. Then we can calculate the $t(10)$ very easily.

$textbf{(b)}bullet$ Finding the explicit equation
$$
begin{align*}
t(0)&=10000\
t(1)&=10000-dfrac{3}{100}cdot 10000=10000(dfrac{100-3}{100})=10000(0.97)\
t(2)&=10000 (0.97)-dfrac{3}{100}cdot 10000(0.97)=10000(0.97)(dfrac{100-3}{100})=10000(0.97)^2\
t(3)&=10000 (0.97)^2-dfrac{3}{100}cdot 10000(0.97)^2=10000(0.97)^2(dfrac{100-3}{100})=10000(0.97)^3\
&boxed{t(n)=10000(0.97)^n}
end{align*}
$$

$bullet$ Finding the recursive equation

$$
begin{align*}
t(0)&=10000\
t(1)&=10000(0.97)=t(0)(0.97)\
t(2)&=10000(0.97)(0.97)=t(1)(0.97)\
&boxed{t(n+1)=t(n)(0.97)}
end{align*}
$$

$textbf{(c)}$ We can verify our $4^text{th}$ week sale by calculating it with explicit equation.

$$
begin{align*}
t(n)&=10000(0.97)^n\
t(4)&=10000(0.97)^4\
t(4)&=10000(0.88529281)\
t(4)&=8852.92
end{align*}
$$

We can see that the result matches with the previous calculation. Hence we can say that our explicit equation is correct. similarly we can find the sale of the $10^text{th}$ week that is $t(10)$.

$$
begin{align*}
t(n)&=10000(0.97)^n\
t(10)&=10000(0.97)^10\
t(10)&=10000(0.7374241)\
t(10)&=7374.24
end{align*}
$$

$textbf{(a)}$ The number of iTalks sold during $4^text{th}$ week was approximately 8853 and the number of iTalks sold during $4^text{th}$ week was approximately 7374.

$textbf{(b)}$ $t(n)=10000(0.97)^n$ and $t(n+1)=t(n)(0.97)$

$textbf{(c)}$ Yes, the result from the part a matches after solving it via explicit equation. hence we can say that the explicit equation is valid.

$textbf{(a)}$ The multiplier for the problem 5-93

part (a) $rightarrow$ 1.07

part (b) $rightarrow$ 1.17

The multiplier for the problem 5-94 $rightarrow$ 0.97

$textbf{(b)}$ The given sequence is $(8, 8, 8, 8,cdots)$
The multiplier can be determined by taking the ratio of any two consecutive terms with higher term number in numerator and lower term number in denominator.

$Rightarrow$
$$
text{multiplier}=dfrac{8}{8}=1
$$

$textbf{(c)}$ $bullet$ When the multiplier is less than 1 but greater than zero, the sequence will have each of its terms smaller than its previous term. This will result in decreasing or decaying sequence.

$bullet$ When the multiplier is greater than 1, the sequence will have each of its terms greater than its previous term. This will result in increasing or growing sequence.

$textbf{(a)}$ The multiplier for the problem 5-93

part (a) $rightarrow$ 1.07

part (b) $rightarrow$ 1.17

The multiplier for the problem 5-94 $rightarrow$ 0.97

$textbf{(b)}$ $text{multiplier}=dfrac{8}{8}=1$

$textbf{(c)}$When the multiplier is less than 1 but greater than zero, the sequence will have each of its terms smaller than its previous term. This will result in decreasing or decaying sequence.

When the multiplier is greater than 1, the sequence will have each of its terms greater than its previous term. This will result in increasing or growing sequence.

#### (a)

In order to find how many weeks it will take for the weekly sales of one Talk per week, we will solve the following equation for $n$ and get the following:

$$
1=10000(0.97)^{n}
$$

$$
(0.97)^n=0.0001
$$

$$
napprox302
$$

So, it will take $302$ weeks.

#### (b)

On the following picture, there is a graph for the sales of $pi$Phones for the first year.

#### (c)

From the previous picture, we can notice that there is no $x$-intercept, but there is an $y$-intercept in the point $(0,100)$.

#### (d)

Here, we actually need to find intercept of those graphs.

From the previous picture, we can see that the point of intercept is $(63.869,752860.474)$.

So, the conclusion is that it will take about $64$ weeks to exceed the sales in the second equation.

#### (e)

On the following picture, there are graphed both functions in part d) and their intersection.

#### (f)

Using the calculator, we get that the answer is $302.3828$, so, our answer from the part a) is very close to this which we got using the calculator.

a) $napprox302$; c) $(0,100)$; d) about $64$ weeks; f) $n=302.3828$

$textbf{(a)}$ Lets find out the multiplier by taking the ratio of $t(1)=2000$ and $t(0)=1600$ which is given by
$$
text{multiplier}=dfrac{2000}{1600}=1.25
$$

Now, we can write its explicit equation as
$$
t(n)=t(0)cdot (1.25)^n=1600(1.25)^n
$$

Checking validity of the equation by finding the values of each term using the equation and then verifying it with table.

$$
begin{align*}
t(n)&=1600(1.25)^n\
t(0)&=1600(1.25)^0tag{first term}\
t(0)&=1600(1)\
t(0)&=1600\
t(1)&=1600(1.25)^1tag{second term}\
t(1)&=1600(1.25)\
t(1)&=2000\
t(2)&=1600(1.25)^2tag{third term}\
t(2)&=1600(1.5625)\
t(2)&=2500\
t(3)&=1600(1.25)^3tag{fourth term}\
t(3)&=1600(1.9531)\
t(3)&=3125\
t(4)&=1600(1.25)^4tag{fifth term}\
t(4)&=1600(2.4414)\
t(4)&=3906
end{align*}
$$

We can see that , each and every term matches with the given table, hence our equation $boxed{t(n)=1600(1.25)^n}$ is valid.

Now, we can write the recursive equation as
$$
boxed{t(n+1)=t(n)cdot (1.25)}
$$

$textbf{(b)}$ Lets find out the multiplier by taking the ratio of $t(1)=3125$ and $t(0)=3906.25$ which is given by
$$
text{multiplier}=dfrac{3125}{3906.25}=0.8
$$

Now, we can write its explicit equation as
$$
t(n)=t(0)cdot (0.8)^n=3906.25(0.8)^n
$$

Checking validity of the equation by finding the values of each term using the equation and then verifying it with table.

$$
begin{align*}
t(n)&=3906.25(0.8)^n\
t(0)&=3906.25(0.8)^0tag{first term}\
t(0)&=3906.25(1)\
t(0)&=3906.25\
t(1)&=3906.25(0.8)^1tag{second term}\
t(1)&=3906.25(0.8)\
t(1)&=3125\
t(2)&=3906.25(0.8)^2tag{third term}\
t(2)&=3906.25(0.64)\
t(2)&=2500\
t(3)&=3906.25(0.8)^3tag{fourth term}\
t(3)&=3906.25(0.512)\
t(3)&=2000\
t(4)&=3906.25(0.8)^4tag{fifth term}\
t(4)&=3906.25(0.4096)\
t(4)&=1600
end{align*}
$$

We can see that , each and every term matches with the given table, hence our equation $boxed{t(n)=3906.25(0.8)^n}$ is valid.

Now, we can write the recursive equation as
$$
boxed{t(n+1)=t(n)cdot (0.8)}
$$

$textbf{(c)}$ Lets find out the multiplier by taking the ratio of $t(1)=72$ and $t(0)=50$ which is given by
$$
text{multiplier}=dfrac{72}{50}=1.44
$$

Now, we can write its explicit equation as
$$
t(n)=t(0)cdot (1.44)^n=50(1.44)^n
$$

Checking validity of the equation by finding the values of each term using the equation and then verifying it with table.

$$
begin{align*}
t(n)&=50(1.44)^n\
t(0)&=50(1.44)^0tag{first term}\
t(0)&=50(1)\
t(0)&=50\
t(1)&=50(1.44)^1tag{second term}\
t(1)&=50(1.44)\
t(1)&=72\
t(2)&=50(1.44)^2tag{third term}\
t(2)&=50(2.0736)\
t(2)&=103.68\
t(3)&=50(1.44)^3tag{fourth term}\
t(3)&=50(2.985984)\
t(3)&=149.2992\
end{align*}
$$

We can see that , each and every term matches with the given table, hence our equation $boxed{t(n)=50(1.44)^n}$ is valid.

Now, we can write the recursive equation as
$$
boxed{t(n+1)=t(n)cdot (1.44)}
$$

$textbf{(d)}$ Lets assume that the explicit equation for this sequence is given by
$$
t(n)=t(0)(c)^n
$$

Now, we need to find out the values of $t(0)$ and $c$ by putting the values of known terms into this assumed equation.

$$
begin{align*}
t(n)&=t(0)(c)^n\
t(1)&=t(0)(c)^1tag{putting $t(1)=50$}\
50&=t(0)(c)\
t(3)&=t(0)(c)^3tag{putting $t(3)=72$}\
72&=t(0)(c)^3\
end{align*}
$$

Now, we have 2 equations as shown below

$$
begin{align}
50&=t(0)(c)\
72&=t(0)(c)^3
end{align}
$$

Divide the eq(2) by eq(1), which will result in

$$
begin{align*}
dfrac{72}{50}&=dfrac{t(0)}{t(0)}cdot dfrac{c^3}{c}\
1.44&=c^2tag{take $sqrt{}$ on the both side}\
sqrt{1.44}&=sqrt{c^2}\
1.2&=c
end{align*}
$$

putting $c=1.2$ in eq (1) , we can solve for $t(0)$ as shown below.

$$
begin{align*}
50&=t(0)(1.2)tag{divide each side by 1.2}\
dfrac{50}{1.2}&=dfrac{t(0)(1.2)}{1.2}\
41.66&=t(0)
end{align*}
$$

Now, putting $t(0)=41.66$ and $c=1.2$ in the unknown explicit equation, we can write our final explicit equation as

$$
boxed{t(n)=41.66(1.2)^n}
$$

Lets test our explicit equation for $t(5)=103.68$ and $t(7)=149.2992$

$$
begin{align*}
t(n)&=41.66(1.2)^n\
t(5)&=41.66(1.2)^5 tag{testing for t(5)}\
103.68&=41.66(2.48832)\
103.68&=103.68 tag{true always}\
t(5)&=41.66(1.2)^5 tag{testing for t(7)}\
149.2992&=41.66(3.5831808)\
149.2992&=149.2992 tag{true always}
end{align*}
$$

We can see that , our equation is valid hence we can use it to write the missing values in the table.

$$
begin{align*}
t(n)&=41.66(1.2)^n\
t(2)&=41.66(1.2)^2\
t(2)&=59.9904\
t(4)&=41.66(1.2)^4\
t(4)&=86.3861\
t(6)&=41.66(1.2)^6\
t(6)&=124.39609
end{align*}
$$

$textbf{(e)}$ The tables in part (a) is reversed order of the table in the part (d) and vice versa.
The multipliers in parts (a) and (b) are multiplicative inverse of each other. This makes sense because one table is written in increasing manner and other is written in decreasing manner. but the values are the same from the ends.

$textbf{(f)}$ we assumed the explicit equation first, then used the value given in the table to solve for the unknowns in the assumed equation.

The odd terms of the table in part (d) are the terms of the table in part (c).

$textbf{(g)}$ From the explicit equation, which we proved valid for part (d), we calculated the $t(2)$ was equal to 59.9904. which is calculated in solution of the part (d).

$textbf{(a)}$ $t(n)=1600(1.25)^n$

$textbf{(b)}$ $t(n)=3906.25(0.8)^n$

$textbf{(c)}$ $t(n)=50(1.44)^n$

$textbf{(d)}$ $t(n)=41.66(1.2)^n$

$textbf{(e)}$ The tables in part (a) is reversed order of the table in the part (d) and vice versa.
The multipliers in parts (a) and (b) are multiplicative inverse of each other. This makes sense because one table is written in increasing manner and other is written in decreasing manner. but the values are the same from the ends.

$textbf{(f)}$ we assumed the explicit equation first, then used the value given in the table to solve for the unknowns in the assumed equation.

The odd terms of the table in part (d) are the terms of the table in part (c).

$textbf{(g)}$ From the explicit equation, which we proved valid for part (d), we calculated the $t(2)$ was equal to 59.9904. which is calculated in solution of the part (d).

$textbf{(c)}$ There are 80 squares remained in case 1 after the first discount.

There are 70 squares remained in case 1 after the first discount.

$textbf{(d)}$ There are 56 squares remained in case 1 after the second discount.

There are 56 squares remained in case 2 after the second discount.

$textbf{(e)}$ This result makes sense as the final result here, can be obtained by mathematical multiplication of terms one after another.

For case 1 its multiplication by 0.2 followed by 0.3
For case 2 its multiplication by 0.3 followed by 0.2
Due to associative property of multiplication the result remained same.

$textbf{(c)}$ case 1- 80 squares
case 2- 70 squares

$textbf{(d)}$ case 1- 56 squares
case 1- 56 squares

$textbf{(e)}$ This result makes sense as the final result here, can be obtained by mathematical multiplication of terms one after another.

For case 1 its multiplication by 0.2 followed by 0.3
For case 2 its multiplication by 0.3 followed by 0.2
Due to associative property of multiplication the result remained same.

$textbf{(a)}$ If $x$ represent the cost of shirt and sales tax is 5% then
sales tax is represented on the cost as

$$
text{tax}= xcdot dfrac{5}{100}
$$

Now, total cost after adding the sales tax will become

$$
text{total cost}= x+xcdot dfrac{5}{100}=xcdot dfrac{105}{100}=1.05x
$$

$textbf{(b)}$ If $x$ represent the cost of shirt and discount is 20% then
discount is represented on the cost as

$$
text{discount}= xcdot dfrac{20}{100}
$$

Now, total cost after the discount becomes

$$
text{total cost}= x-xcdot dfrac{20}{100}=xcdot dfrac{80}{100}=0.8x
$$

$textbf{(c)}$ Lets, find out the cost Trixie has to pay if tax is added first then discount is given later

Cost after tax addition is $1.05x$ and after 20% discount is given at this cost the
final value to pay becomes

$$
begin{align*}
&1.05x-1.05x cdot dfrac{20}{100}\
&=1.05x(1-0.2)\
&=1.05xtimes 0.8\
&=0.84x
end{align*}
$$

Also, lets find out the cost Trixie has to pay if discount is added first then tax is added later

Cost after discount is $0.8x$ and after 5% taxis added at this cost the then final value to pay becomes

$$
begin{align*}
&0.8x+0.8x cdot dfrac{5}{100}\
&=0.8x(1+0.05)\
&=0.8xtimes 1.05\
&=0.84x
end{align*}
$$

Therefore, Trixie has to pay the same amount for any of the cases.

$textbf{(a)}$ Tax$= xcdot dfrac{5}{100}$ and cost with tax $=1.05x$

$textbf{(b)}$ Discount$=xcdot dfrac{20}{100}$ and cost after discount is $0.8x$

$textbf{(c)}$ No, she didn’t get charged extra money.

#### (a)

Required equation for this situation will be the following:

$$
y=6(3)^{x-1}
$$

#### (b)

We will substitute $30$ million in the previous formula for $y$ and solve it for $x$ and get the following:

$$
30 000 000=6cdot(3)^{x-1}
$$

$$
3^{x-1}=5000000
$$

$$
x-1approx14
$$

$$
xapprox15
$$

So, it passed at least $15$ months.

#### (c)

Equation for this situation would be the following:

$$
y=6(0.3)^{x-1}
$$

We will substitute $100$ for $y$ and solve the equation for $x$ and get the following:

$$
100=6(0.3)^{x-1}
$$

$$
(0.3)^{x-1}=16.67
$$

$$
x-1=-2.34
$$

$$
x=-1.34
$$

Because we got negative number, and in this situation it is not possible, so, the conclusion is that never be less than $100$ rabbits.

a) $y=6(3)^{x-1}$; b) $xapprox15$; c) $y=6(0.3)^{x-1}$

For converting any percent increase or decrease into the multiplier we need to assume any starting value lest say $x$. Now we can find the multiplier any of the given case as shown below.

$textbf{(a)}$ 3% increase

$$
begin{align*}
&x+xcdot dfrac{3}{100}tag{take $x$ common}\
&=x(1+dfrac{3}{100})\
&=x(dfrac{100+3}{100})\
&=x(dfrac{103}{100})\
&=x(1.03)
end{align*}
$$

With 3% increase the starting term $x$ changes to $1.03x$, hence the multiplier is $boxed{1.03}$

$textbf{(b)}$ 25% decrease

$$
begin{align*}
&x-xcdot dfrac{25}{100}tag{take $x$ common}\
&=x(1-dfrac{25}{100})\
&=x(dfrac{100-25}{100})\
&=x(dfrac{75}{100})\
&=x(0.75)
end{align*}
$$

With 25% decrease the starting term $x$ changes to $0.75x$, hence the multiplier is $boxed{0.75}$

$textbf{(c)}$ 13% decrease

$$
begin{align*}
&x-xcdot dfrac{13}{100}tag{take $x$ common}\
&=x(1-dfrac{13}{100})\
&=x(dfrac{100-13}{100})\
&=x(dfrac{87}{100})\
&=x(0.87)
end{align*}
$$

With 13% decrease the starting term $x$ changes to $0.87x$, hence the multiplier is $boxed{0.87}$

$textbf{(d)}$ 2.08% increase

$$
begin{align*}
&x+xcdot dfrac{2.08}{100}tag{take $x$ common}\
&=x(1+dfrac{2.08}{100})\
&=x(dfrac{100+2.08}{100})\
&=x(dfrac{102.08}{100})\
&=x(1.0208)
end{align*}
$$

With 2.08% increase the starting term $x$ changes to $1.0208x$, hence the multiplier is $boxed{1.0208}$

$textbf{(a)}$ $1.03$

$textbf{(b)}$ $0.75$

$textbf{(c)}$ $0.87$

$textbf{(d)}$ $1.0208$

$$
textbf{(a)}
$$

#1
For the given sequence each term can be represented as by $t(n+1)=t(n)+d$, hence its an arithmetic sequence with $t(0)=16$ and the common difference $d=-3$

#2
Its neither Arithmetic nor Geometric sequence. this is because the sequence is decreasing for some terms then starts increasing which is not possible with any of arithmetic or geometric sequence as either they will keep on increasing or keep on decreasing.

#3
Given sequence is the Geometric with common ratio$=dfrac{t(n+1)}{t(n)}=dfrac{8}{4}=2$

$$
textbf{(b)}
$$

#1
General term of an arithmetic sequence is given by $t(n)=t(0)+(n-1)d$

$Rightarrow$
$$
t(n)=16+(n-1)cdot -3
$$

#3
General term of an geometric sequence is given by $t(n)=ar^{n-1}$

$Rightarrow$
$$
t(n)=16cdot 2^{n-1}
$$

$textbf{(a)}$ #1 Arithmetic; #2 None; #3 Geometric

$textbf{(b)}$
$$
t(n)=16+(n-1)cdot -3
$$

$$
t(n)=16cdot 2^{n-1}
$$

Two parallel lines should have the same slopes. Therefore, a line parallel to given line with the equation $y=-dfrac{1}{3}x+5$ should have the slope equal to $-dfrac{1}{3}$.

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Therefore, the equation of a line of slope $-dfrac{1}{3}$ passing through a point $(9,-1)$ is given by

$$
begin{align*}
y-(-1)&=-dfrac{1}{3}cdot (x-9)\
y+1&=-dfrac{1}{3}cdot x -dfrac{1}{3}cdot (-9)\
y+1&=-dfrac{1}{3}cdot x+3\
y&=-dfrac{1}{3}cdot x+3-1\
&boxed{y=-dfrac{1}{3}cdot x+2}
end{align*}
$$

$$
y=-dfrac{1}{3}cdot x+2
$$

$textbf{(a)}$
$$
begin{align*}
8-(2x+1)&=3tag{use distributive property}\
8-2x-1&=3\
7-2x&=3tag{subtract 7 from each side}\
7-7-2x&=3-7\
-2x&=-4tag{divide each side by -2}\
dfrac{-2x}{-2}&=dfrac{-4}{-2}\
&boxed{x=2}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
sqrt{x}+4&=9tag{subtract 4 from each side}\
sqrt{x}+4-4&=9-4\
sqrt{x}&=5tag{squaring both side}\
(sqrt{x})^2&=5^2\
x&=25\
&boxed{x=25}
end{align*}
$$

$textbf{(a)}$ $x=2$

$textbf{(b)}$ $x=25$

$textbf{(a)}$ For describing the linear association between the given data, we need to calculate the correlation coefficient between the length of the organelle and the diameter of the cell, with the help of the given data.The value of correlation coefficient is represented by $r$ and vary between $-1$ to $+1$ where $r=-1$ signifies a perfect negative linear relationship between the given parameters and $r=+1$ signifies a perfect positive linear relationship between the given parameters and $r=0$ means there is no association at all. Positive relationship means with the increase in the value of one parameter the other will also increase and a negative relationship means with the increase in the value of one parameter the other will decrease.

We can also make scatterplot for the given data and determine the line of best fit. A line of best fit with lower magnitude of residual signifies a better linear association between the parameters.

$textbf{(b)}$ The electronically calculated value of correlation coefficient $r$ and the scatterplot along with the line of best fit is attached below.

The value of $r$ is $-0.928$ which is very close to $-1$ which signifies that there is strong negative linear association between Length of Organelle and the Diameter of cell which means with the increase in the length of the organelle the diameter of the cell will decrease.

We can also observe the scatterplot and the line of best fit on it. The equation of line of best fit is $y=-1.604x+49.497$ and we can see that the residuals are relatively small hence, it represents a good linear association.

$textbf{(a)}$ For describing association between any to parameters we need to calculate the correlation coefficient. we can also draw the scatterplot and draw the line of best fit on it. For more detailed explanation see inside.

$textbf{(b)}$ The value of correlation coefficient is calculated for the given case which is $r=-0.9281$ and the scatterplot along with the line of best fit is attached inside.

Theoretically, Mantias can never leave the hall. because he is allowed to move exactly halfway to the door each time. Given that he started with the distance 100 meters. let $t(n)$ represents his distance from the door after $n^text{th}$ time walking and waiting 1 minute each for every walk which means $n$ minutes of only waiting time(excluding walking time)

$$
Rightarrow
$$

$$
begin{align*}
t(0)&=100\
t(1)&=dfrac{100}{2}=50\
t(2)&=dfrac{50}{2}=25\
t(3)&=dfrac{25}{2}=12.5\
t(4)&=dfrac{12.5}{2}=6.25\
t(5)&=dfrac{6.25}{2}=3.125\
t(6)&=dfrac{3.125}{2}=1.5625\
t(7)&=dfrac{1.5625}{2}=0.78125
end{align*}
$$

The distance after 7 times walking and 7 minutes waiting between him and the door becomes $0.78125$. The distance will keep getting half every time but will never become 0 as it will constitute a small value after every time of walking. This means theoretically there will always be a minute distance between him and the door. And hence, he will only reach the door after infinite time.

Other way to prove that is, we can represent his distance from the door in the form of general term which is given by

$$
t(n)=100cdot(dfrac{1}{2})^n
$$

We can see that $t(n)$ will become 0 only when value of $n$ tends to $infty$

Given that the Zeke ran 4 miles in 45 minutes. So we can find his speed per minute by dividing 4 miles by 45 minutes.

$Rightarrow$
$$
text{speed}=dfrac{4}{45} dfrac{text{miles}}{text{minute}}
$$

Now, time taken by him to cover 10 miles can be found out by dividing the distance by his speed.

$$
text{time}=dfrac{text{distance}}{text{speed}}=dfrac{10}{(4/45)}=dfrac{45cdot 10}{4}=112.5 text{ minutes}
$$

For his unit rate in miles per hour we need to covert given time in hours first. We know that 60 minutes is equal to 1 hour. This implies that 45 minutes is $dfrac{45}{60}=dfrac{3}{4}$ hours. So , we can say that Zeke ran 4 miles in $dfrac{3}{4}$ hours.So we can find his speed in miles per hour by dividing 4 miles by $dfrac{3}{4}$ hours.

$Rightarrow$
$$
text{speed}=dfrac{4}{frac{3}{4}}=dfrac{16}{3}=5.333
dfrac{text{miles}}{text{hour}}
$$

112.5 minutes and $5.333
dfrac{text{miles}}{text{hour}}$

$textbf{(a)}$ Given that all the sequences are Arithmetic, hence we can write the $t(n)$ in the general form as $t(n)=t(1)+(n-1)d$ where $d$ is the common difference of the sequence which can be found out by taking difference of any two consecutive term in the way that the term with lower term number gets subtracted from term of higher numbered term.

$bullet$ Sequence 1.

$$
d=6-2=4
$$

$$
boxed{t(n)=2+(n-1)4}
$$

$$
begin{align*}
t(3)&=2+(3-1)4tag{term 3rd}\
t(3)&=2+(2)4\
t(3)&=10\
t(4)&=2+(4-1)4tag{term $4^text{th}$}\
t(4)&=2+(3)4\
t(4)&=14\
t(5)&=2+(5-1)4tag{term $5^text{th}$}\
t(5)&=2+(4)4\
t(5)&=18\
t(6)&=2+(6-1)4tag{term $6^text{th}$}\
t(6)&=2+(5)4\
t(6)&=22\
end{align*}
$$

$bullet$ Sequence 2.

$$
d=12-24=-12
$$

$$
boxed{t(n)=24+(n-1)(-12)}
$$

$$
begin{align*}
t(3)&=24+(3-1)(-12)tag{term 3rd}\
t(3)&=24+(2)(-12)\
t(3)&=0\
t(4)&=24+(4-1)(-12)tag{term $4^text{th}$}\
t(4)&=24+(3)(-12)\
t(4)&=-12\
t(5)&=24+(5-1)(-12)tag{term $5^text{th}$}\
t(5)&=24+(4)(-12)\
t(5)&=-24\
t(6)&=24+(6-1)(-12)tag{term $6^text{th}$}\
t(6)&=24+(5)(-12)\
t(6)&=-36\
end{align*}
$$

$bullet$ Sequence 3.

$$
d=5-1=4
$$

$$
boxed{t(n)=1+(n-1)4}
$$

$$
begin{align*}
t(3)&=1+(3-1)4=9tag{term 3rd}\
t(4)&=1+(4-1)4=13tag{term $4^text{th}$}\
t(5)&=1+(5-1)4=17tag{term $5^text{th}$}\
t(6)&=1+(6-1)4=21tag{term $6^text{th}$}\
end{align*}
$$

$textbf{(b)}$ Yes, next terms for the given sequences will be different if the sequence were geometric.

The general term for a geometric sequence is given by $t(n)=t(1)r^{n-1}$ where $r$ represents the common ratio which can be found by dividing two consecutive terms such that a term with higher term number is divided by the term with lower term number.

$bullet$ sequence 1.

$$
r=dfrac{t(2)}{t(1)}=dfrac{6}{2}=3
$$

$$
boxed{t(n)=2(3^{n-1})}
$$

$$
begin{align*}
t(3)&=2(3^{3-1})=2(9)=18tag{term 3rd}\
t(4)&=2(3^{4-1})=2(27)=54tag{term $4^text{th}$}\
t(5)&=2(3^{5-1})=2(81)=162tag{term $5^text{th}$}\
t(6)&=2(3^{6-1})=2(243)=486tag{term $6^text{th}$}\
end{align*}
$$

$bullet$ sequence 2.

$$
r=dfrac{t(2)}{t(1)}=dfrac{12}{24}=dfrac{1}{2}
$$

$$
boxed{t(n)=24cdot(dfrac{1}{2})^{n-1}}
$$

$$
begin{align*}
t(3)&=24cdot(dfrac{1}{2})^{3-1}=24(dfrac{1}{4})=6tag{term 3rd}\
t(4)&=24cdot(dfrac{1}{2})^{4-1}=24(dfrac{1}{8})=3tag{term $4^text{th}$}\
t(5)&=24cdot(dfrac{1}{2})^{5-1}=24(dfrac{1}{16})=1.5tag{term $5^text{th}$}\
t(6)&=24cdot(dfrac{1}{2})^{65-1}=24(dfrac{1}{32})=0.75tag{term $6^text{th}$}\
end{align*}
$$

$bullet$ sequence 3.

$$
r=dfrac{t(2)}{t(1)}=dfrac{5}{1}=5
$$

$$
boxed{t(n)=1(5^{n-1})}
$$

$$
begin{align*}
t(3)&=1(5^{3-1})=25tag{term 3rd}\
t(4)&=1(5^{4-1})=125tag{term $4^text{th}$}\
t(5)&=1(5^{5-1})=625tag{term $5^text{th}$}\
t(6)&=1(5^{6-1})=3125tag{term $6^text{th}$}
end{align*}
$$

$textbf{(c)}$

$bullet$ sequence 1.

$$
t(n)=1+5^{n-1}
$$

$bullet$ sequence 2.

$$
t(n)=25-13^{n-1}
$$

$bullet$ sequence 3.

$$
t(n)=6^{n-1}-2^{n-1}+1
$$

General term of an Arithmetic sequence is given by $t(n)=t(1)+(n-1)d$ , so for part (a) we can try to write the sequences in this form and then we can solve for the value of next term by putting different values of $n$ in it.

General term of an Geometric sequence is given by $t(n)=t(1)cdot r^{n-1}$ , so for part (b) we can try to write the sequences in this form and then we can solve for the value of next term by putting different values of $n$ in it.

The complete explanation of each part is included inside.

$textbf{(a)}$
$$
begin{align*}
(x+2)(x+3)&=x^2-10tag{use distributive property}\
x(x+3)+2(x+3)&=x^2-10tag{solve brackets}\
xcdot x+xcdot 3+2cdot x+2cdot 3&=x^2-10\
x^2+3x+2x+6&=x^2-10tag{subtract $x^2$ from each side}\
x^2-x^2+5x+6&=x^2-x^2-10\
5x+6&=-10tag{subtract 6 from each side}\
5x+6-6&=-10-6\
5x&=-16tag{divide each side by 5}\
dfrac{5x}{5}&=dfrac{-16}{5}\
&boxed{x=dfrac{-16}{5}}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
dfrac{1}{2}x+dfrac{1}{3}x-7&=dfrac{5}{6}xtag{multiply each term by 6 (LCM) }\
dfrac{1}{2}xcdot 6+dfrac{1}{3}xcdot 6-7cdot 6&=dfrac{5}{6}xcdot 6\
3x+2x-42&=5x\
5x-42&=5xtag{subtract $5x$ from each side}\
5x-5x-42&=5x-5x\
-42&=0tag{not true}
end{align*}
$$

Therefore, equation $dfrac{1}{2}x+dfrac{1}{3}x-7=dfrac{5}{6}x$ has no solution.

$textbf{(c)}$
$$
begin{align*}
dfrac{x+1}{3}&=dfrac{x}{2}tag{multiply each side by 6 (LCM)}\
dfrac{x+1}{3}cdot 6&=dfrac{x}{2}cdot 6\
dfrac{6}{3}(x+1)&=dfrac{6}{2}(x)\
2(x+1)&=3(x)\
2x+2&=3xtag{subtract 2x from each side}\
2x-2x+2&=3x-2x\
2&=x\
&boxed{x=2}
end{align*}
$$

$textbf{(d)}$
$$
begin{align*}
9^x&=(dfrac{1}{3})^{x+3}tag{rewrite 9 as $3^2$}\
(3^2)^x&=(dfrac{1}{3})^{x+3}tag{use $(a^m)^n=a^{mcdot n}$}\
3^{2x}&=dfrac{1^{x+3}}{3^{x+3}}\
3^{2x}&=dfrac{1}{3^{x+3}}tag{use $dfrac{1}{a^m}=a^{-m}$}\
3^{2x}&=3^{-1cdot (x+3)}tag{equate the powers, as base is same}\
Rightarrow\
2x&=-x-3tag{add $x$ to the each side}\
2x+x&=-x+x-3\
3x&=-3tag{divide each side by 3}\
dfrac{3x}{3}&=dfrac{-3}{3}\
&boxed{x=-1}
end{align*}
$$

$textbf{(a)}$ $x=dfrac{-16}{5}$

$textbf{(b)}$ No solution

$textbf{(c)}$ $x=2$

$textbf{(d)}$ $x=-1$

Slope of a line which passes through a pair or points $(x_1,y_1)$ and $(x_2,y_2)$ is given by

$$
text{slope}=dfrac{y_2-y_1}{x_2-x_1}
$$

The equation of a line of given slope $m$ passing through a point $(x_1,y_1)$ is given by

$$
y-y_1=mcdot (x-x_1)
$$

Now, we have given points as $(0,-3)$ and $(-9,-9)$, therefore slope of line passing through these point is
$$
begin{align*}
text{slope}&= dfrac{-9-(-3)}{-9-0}\\
&=dfrac{-9+3}{-9}\
&=dfrac{2}{3}
end{align*}
$$

Now, the equation of a line of slope $dfrac{2}{3}$ passing through a point $(0,-3)$ is given by

$$
begin{align*}
y-(-3)&=dfrac{2}{3}cdot (x-0)\
y+3&=dfrac{2}{3}cdot x\
y&=dfrac{2}{3}cdot x-3
end{align*}
$$

$$
y=dfrac{2}{3}cdot x-3
$$

$textbf{(a)}$
$$
begin{align*}
f(x)&=dfrac{1}{x+2}\
f(-3)&=dfrac{1}{(-3)+2}\
f(-3)&=dfrac{1}{-1}\
f(-3)&=-1\
&boxed{f(-3)=-1}
end{align*}
$$

$textbf{(b)}$
$$
begin{align*}
f(x)&=dfrac{1}{x+2}\
f(-1.5)&=dfrac{1}{(-1.5)+2}\
f(-1.5)&=dfrac{1}{0.5}\
f(-1.5)&=dfrac{1}{5/10}\
f(-1.5)&=dfrac{1cdot 10}{5}\
f(-1.5)&=2\
&boxed{f(-1.5)=2}
end{align*}
$$

$textbf{(c)}$
$$
begin{align*}
f(x)&=dfrac{1}{x+2}\
f(-2)&=dfrac{1}{(-2)+2}\
f(-3)&=dfrac{1}{0}tag{$dfrac{a}{0}$ is not defined}
end{align*}
$$

Any number divide by 0 is not defined,therefore, $f(-2)$ is not defined.

$textbf{(d)}$
$$
begin{align*}
f(x)&=dfrac{1}{x+2}\
(5)&=dfrac{1}{x+2}tag{multiply each side by $x+2$}\
5cdot (x+2)&=dfrac{1}{x+2}cdot (x+2)\
5(x+2)&=1tag{solve bracket}\
5x+10&=1tag{subtract 10 from each side}\
5x+10-10&=1-10\
5x&=-9tag{divide each side by 5}\
dfrac{5x}{5}&=dfrac{-9}{5}\
x&=dfrac{-9}{5}\
boxed{x=dfrac{-9}{5}}
end{align*}
$$

$textbf{(a)}$ $-1$

$textbf{(b)}$ $2$

$textbf{(c)}$ Not defined

$textbf{(d)}$ $dfrac{-9}{5}$

$textbf{(a)}$ Area of a square with given side length $l$ units is given by

$$
text{Area}=l^2 text{ } text{ units}^2
$$

We are given a square with side length 7 cm, therefore the area of this square is

$$
begin{align*}
text{Area}&=7^2=49 text{cm}^2
end{align*}
$$

$textbf{(b)}$ Area of any triangle is given by the formula shown below

$$
text{Area}=dfrac{1}{2}times text{base}times text{ perpendicular height}
$$

For the given triangle we have base length $=10$ inches and perpendicular height $=4$ inches.
Hence, the area of the given triangle is

$$
text{Area}=dfrac{1}{2}times 10 times 4=20 text{ inches}^2
$$

$textbf{(c)}$ The area of the given trapezium can be calculated by dividing it into parts. These parts consists 3 parts which are

$bullet$A triangle with base 6 units and perpendicular height 6 units

$Rightarrow$
$$
text{Area}_1=dfrac{1}{2}times 6 times 6=18 text{units}^2
$$

$bullet$A rectangle with length 8 units and breath 6 units

$Rightarrow$
$$
text{Area}_2= 8 times 6=48 text{units}^2
$$

$bullet$A triangle with base 2 units and perpendicular height 6 units

$Rightarrow$
$$
text{Area}_3=dfrac{1}{2}times 2 times 6=6 text{units}^2
$$

Therefore the total area of trapezium is summation of all the above three calculated areas.
Total area$= 18+48+6=72$units$^2$

$textbf{(a)}$ 49 cm$^2$

$textbf{(b)}$ 20 inches$^2$

$textbf{(c)}$ 72 units$^2$

textbf{(a)} Given sequence is $-5, -1, 3, 7, cdots$, we can see that each next term of this sequence can be obtained by adding $4$ to its previous term. Therefore, the given sequence is an arithmetic sequence and $d=4$ is its common difference.\
As we know that the general term $t(n)$ of an arithmetic sequence is given by $$t(n)=t(1)+(n-1)d$$
So, we can write its explicit equation as shown below.
$$boxed{t(n)=-5+(n-1)4}$$
The table of values and the graph is attached below.
begin{center}
begin{tabular}{|p{1.5cm}|p{1.5cm}|}
hline
n&t(n)\
hline
1&-5\
hline
2&-1\
hline
3&3\
hline
4&7\
hline
5&11\
hline
6&15\
hline
7&19\
hline
end{tabular}
end{center}

$textbf{(b)}$ We have explicit equation for the given sequence as $t(n)=-5+(n-1)4$, so we can put $t(n)=400$ in this equation and solve for $n$. If the value of $n$ comes to be a positive integer then 400 is a term of the sequence otherwise it is not.

$$
begin{align*}
t(n)&=-5+(n-1)4\
400&=-5+(n-1)4tag{add 5 to the each side}\
400+5&=-5+5+(n-1)4\
405&=(n-1)4tag{divide each side by 4}\
dfrac{405}{4}&=dfrac{(n-1)4}{4}\
101.25&=n-1tag{add 1 to the each side}\
101.25+1&=n-1+1\
102.25&=n
end{align*}
$$

We can see, for the term 400 we did not got the term number an integer value. Therefore, 400 cannot be a term of the given sequence.

textbf{(c)} The given function is $f(x)=4x-9$. We can find the values of $f(x)$ by putting different values of $x$ in it. The difference between $f(x)$ and $t(x)$ is that $f(x)$ is a continuous function and it can takes all real values as input and can give all real values as output. whereas the $t(x)$ is a discrete function which can takes only non negative integers as input and can give only discrete values as output.\
begin{align*}
f(x)&=4x-9\
f(-5)&=4(-5)-9\
f(-5)&=-20-9\
f(-5)&=-29\
f(-0.5)&=4(-0.5)-9\
f(-0.5)&=-2-9\
f(-0.5)&=-11\
f(0)&=4(0)-9\
f(0)&=-9\
f(1)&=4(1)-9\
f(1)&=4-9\
f(1)&=-5\
f(2.5)&=4(2..5)-9\
f(2.5)&=10-9\
f(2.5)&=1
end{align*}
So this can be represented as table, see the table attached below.\
begin{center}
begin{tabular}{|p{2cm}|p{3cm}|}
hline
x&f(x)\
hline
-5&-29\
hline
-0.5&-11\
hline
0&-9\
hline
1&-5\
hline
2.5&1\
hline
end{tabular}
end{center}

$textbf{(d)}$ Yes, the $f(x)=4x-9$ can take value 400 corresponding to $x=102.25$. We can say this as unlike $t(n)=-5+(n-1)4=4n-9$ this is a continuous function and can take all real values as input.

$textbf{(a)}$ The different representation of the sequence is attached inside.

$textbf{(b)}$ No

$textbf{(c)}$ This is a continuous function and its table values and graph is attached inside.

$textbf{(d)}$ Yes, The function $f(x)=4x-9$ can take value 400 corresponding to $x=102.25$

$textbf{(a)}$ Yes, $f(x)=2cdot 3^x$ is a function as for every given input $x$ there is single output $f(x)$ which means there is one to one relation. Now, $f(x)$ can take all real values as input which , means domain of $f(x)$ is all real numbers and range of $f(x)$ is $(0,+infty)$ and it is continuous function.

But $t(n)=2cdot 3^n$ is a discrete function.These can only take natural numbers as its input because of the reason that each of its term is given a term number and a term number can never be negative or fractional.The output corresponding to only natural number as inputs will be discrete only and will not able to cover all real values of numbers as output. Hence, sequences are discrete function.

$textbf{(b)}$ We can equate the $t(n)$ equal to 1400 for the equation and try to find the value of $n$ corresponding to it. If this value belongs to integers then we can say it is possible for $t(n)$ to be equal to 1400 otherwise it is not.
This restriction is due to the reason that $t(n)$ can take only natural numbers in its domain.
$$
begin{align*}
t(n)&=2cdot 3^n\
1400&=2cdot 3^ntag{divide each side by 2}\
700&=3^ntag{taking log to the base 3 on both side}\
log_3{700}&=log_3{3^n}\
log_3{700}&=nlog_3{3}\
log_3{700}&=n(1)\
5.963&=n
end{align*}
$$

We can see that the input corresponding to $t(n)=1400$ is not integral. hence, $t(n)$ cannot take the value 1400.

$textbf{(c)}$ Yes, we can solve just like above to obtain that $f(x)$ takes value 1400 corresponding to the value $x=5.93$. This is allowed because $x$ can take all real values as $f(x)$ has its domain equal to all real numbers.

$textbf{(d)}$ Functions and sequences are similar as these both are functions. But the difference between both is that function can take all real values as input and are continuous unless it becomes not defined at some point. whereas the sequences can take only natural number as input and can give only discrete value of output.
$f(x)=2cdot 3^x$ is a function but $t(n)=2cdot 3^n$ is a discrete function or sequence (more precisely).

$textbf{(a)}$ Yes, $f(x)=2cdot 3^x$ is a function and $t(n)=2cdot 3^n$ is a discrete function(sequence).

$textbf{(b)}$ No, it can only take natural number as input.

$textbf{(c)}$ Yes

$textbf{(d)}$ Functions and sequences are similar as these both are functions. But the difference between both is that function can take all real values as input and are continuous unless it becomes not defined at some point. whereas the sequences can take only natural number as input and can give only discrete values of output.

Sequences vs. Functions
Yes, sequences are also a function. But these are special type of functions. These can only take natural numbers as its input because of the reason that each of its term is given a term number and a term number can never be negative or fractional.

The output corresponding to only natural number as inputs will be discrete only and will not able to cover all real values of numbers as output. Hence, sequences are discrete function whereas a simple function is continuous and can take all real values as input unless it becomes not defined for some inputs.

A function is represented by $f(x)$ and takes $x$ as input whereas a sequence is represented by $t(n)$ which takes $n$ as input.

$textbf{(a)}$
$$
begin{align*}
a_n&=dfrac{3}{2}n-1\
a_1&=dfrac{3}{2}(1)-1=dfrac{3}{2}-1=dfrac{3-2}{2}=dfrac{1}{2}\
a_2&=dfrac{3}{2}(2)-1=dfrac{2}{2}(3)-1=3-1=2\
a_3&=dfrac{3}{2}(3)-1=dfrac{9}{2}-1=dfrac{9-2}{2}=dfrac{7}{2}\
a_4&=dfrac{3}{2}(4)-1=dfrac{4}{2}(3)-1=2cdot3-1=5\
a_5&=dfrac{3}{2}(5)-1=dfrac{15}{2}-1=dfrac{15-2}{2}=dfrac{13}{2}
end{align*}
$$

Therefore, first five term are $(dfrac{1}{2}, 2, dfrac{7}{2}, 5, dfrac{13}{2}, cdots)$

$textbf{(b)}$
$$
begin{align*}
a_n&=dfrac{1}{4}(-8)^n\
a_1&=dfrac{1}{4}(-8)^1=dfrac{1}{4}(-8)=dfrac{-8}{4}=-2\
a_2&=dfrac{1}{4}(-8)^2=dfrac{1}{4}(64)=dfrac{64}{4}=16\
a_3&=dfrac{1}{4}(-8)^3=dfrac{1}{4}(-512)=dfrac{-512}{4}=-128\
end{align*}
$$

Therefore, first three term are $(-2, 16, -128, cdots)$

$textbf{(c)}$ In the given sequence each next term can be obtained by multiplying the previous term by $-1$. Therefore, the common ratio for this is $r=dfrac{a_{n+1}}{a_n}=-1$ and its general term can be written by using the general term of geometric sequence which is given by
$$
a_n=a_1(r)^{n-1}
$$

where $a_1$ is the first term and $r$ is the common ratio of the sequence. we have $r=-1$ and $a=-4$ therefore, the equation becomes

$$
boxed{a_n=-4(-1)^{n-1}}
$$

$textbf{(d)}$ Given sequence is $(7, 13, 19, 25, cdots)$
We can see the common difference for this sequence is $d=13-7=6$, which means each next term of this sequence can be obtained by adding 6 to the previous term.

We know that the general term of an arithmetic progression is given by

$$
a_n=a_1+(n-1)d
$$

where, $a_n$ represents the $n^text{th}$ term of the sequence , $a_1$ represents the first term of the sequence, $d$ represents the common difference (sequence generator) of the sequence and $n$ represents the term number.

We can put the known value $a_1=7$ and $d=6$ in the general equation, the final equation becomes

$$
boxed{a_n=7+(n-1)6}
$$

$textbf{(a)}$ The first five term are $(dfrac{1}{2}, 2, dfrac{7}{2}, 5, dfrac{13}{2}, cdots)$

$textbf{(b)}$ The first three term are $(-2, 16, -128, cdots)$

$textbf{(c)}$ $a_n=-4(-1)^{n-1}$

$textbf{(d)}$ $a_n=7+(n-1)6$

A sequence has its domain as the set of natural numbers or a subset of the natural numbers but function can have domain as real numbers or a subset of real number.

We are given that the output of $p(r)=2cdot 5^r$ is 78,000. On solving for the value of $r$ we get

$$
begin{align*}
78000&=2cdot 5^r\
39000&=5^r\
r&=6.569
end{align*}
$$

The given condition is possible for a value of $r=6.569$ which is not a natural number. Therefore, p(r) has real numbers in its domain and hence $p(r)$ cannot be a sequence.

$p(r )$ is a function.

$textbf{a.}$

By Equal Values Method, we can treat each side of the equation as two separate equations:

$$
begin{align*}
y&=200(0.5)^x\
y&=3.125
end{align*}
$$

Graph both equations on Y1 and Y2, adjust the window if necessary, then use the $textbf{intersect}$ feature to find the point of intersection of the graphs. The $x$-coordinate of the point of intersection is the solution of the equation. In this case,

$$
color{#c34632}x=6
$$

$textbf{b.}$

By Equal Values Method, we can treat each side of the equation as two separate equations:

$$
begin{align*}
y&=318\
y&=6cdot 3^x
end{align*}
$$

Graph both equations on Y1 and Y2, adjust the window if necessary, then use the $textbf{intersect}$ feature to find the point of intersection of the graphs. The $x$-coordinate of the point of intersection is the solution of the equation. In this case,

$$
color{#c34632}xapprox 3.61
$$

a. $x=6$

b. $xapprox 3.61$

Domain of any function is the set of values that the input variable can take so that function does not becomes not defined.

$textbf{(a)}$ For the given function $f(x)=3x-5$ the input variable $x$ can take value from all real numbers. therefore, domain of this function is $(-infty,+infty)$

$textbf{(b)}$ For the given sequence $t(n)=3n-5$ the input variable $n$ can take only positive integral values. Therefore, domain of this sequence is all natural number $(1, 2, 3, cdots)$

$textbf{(c)}$ For the given function $f(x)=dfrac{5}{x}$ the input variable $x$ can take value from all real numbers except $x=0$ which makes the $f(x)$ not defined as $dfrac{5}{0}$ is not defined. Therefore, domain of this function is $(-infty,+infty)-(0)$

$textbf{(d)}$ For the given sequence $t(n)=dfrac{5}{n}$ the input variable $n$ can take only positive integral values. Therefore, domain of this sequence is all natural number $(1, 2, 3, cdots)$. The $n=0$ is already is not in its domain so we need not to worry about it.

$textbf{(a)}$ $(-infty,+infty)$

$textbf{(b)}$ The domain of this sequence is all natural number $(1, 2, 3, cdots)$

$textbf{(c)}$ $(-infty,+infty)-(0)$

$textbf{(d)}$ The domain of this sequence is all natural number $(1, 2, 3, cdots)$

$textbf{(a)}$ Given sequence has an explicit equation as $t(n) = 5cdot 2^n$

For this sequence to have a term with the value of 200, we should get the value of $n$ corresponding to it as a natural number.

$$
begin{align*}
200&= 5cdot 2^n\
40&=2^n\
n&=dfrac{log 40}{log 2}\
n&=5.322
end{align*}
$$

The value of $n=5.322$ doesn’t belong to a natural number for a value of $t(n)=200$ hence it is not possible for the sequence to have a term as 200.

$textbf{(b)}$ Function is given with the equation $f(x) =5cdot 2^x$. A function can have real numbers in its domain. Hence, value of $x=5.322$ belongs to real number for which $f(x)$ gives output 200. Therefore, it is possible for $f(x)$ to have an output 200.

$textbf{(a)}$ No, It is not possible for the sequence to have a term as 200.

$textbf{(b)}$ Yes, 200 can be the output of $f(x)$

Given that $a_1=5$ and $a_n =a_{n-1}+6$ Also, $a_2=11$ and $a_3=17$
Now, putting the values of $n$ into the general term of the sequence we can easily find $a_4$ and $a_5$ as shown below.

$textbf{(a)}a_4 =a_3+ 6 = 17 + 6 =23$

$textbf{(b)}a_5 =a_4+ 6 = 23 + 6 =29$

$textbf{(c)}$ First five terms of the sequence are $(5,11,17,23,29)$

$textbf{(a)}$ 23

$textbf{(b)}$ 29

$textbf{(c)}$ First five terms of the sequence are $(5,11,17,23,29)$

Given that Geoffrey ran $x$ meters of the race, let Ricardo ran $y$ meters. If they both ran 775 meters in total, then this can be represented as an equation shown below

$$
x+y=775
$$

Therefore the distance ran by Ricardo can be found out by rearranging the above equation to write it in term of x
$Rightarrow$
$$
y=775-x
$$

Distance ran by Ricardo can be represented by the equation $y=775-x$

If $x$ is the figure number and $y$ is the number of tiles, then we have two points:

$$
(2,6)text{ and }(5,15)
$$

We notice that the $y$-value is 3 times the $x$-value in both points. Since the pattern grows linearly, then the rule is:

$$
color{#c34632}y=3x
$$

$y=3x$ where $x$ is the figure number and $y$ is the number of tiles.