a. Every month $3 is added to the previous amount:

$50, $53, $56, $59, $62, $65, $68, $71, $74, $77, $80, $83

Thus we note that in the 12th month $83 is added to the college fund.

\
\

b. Since the amount starts at 50 at$n=1$and every month the amount increases by 3:$ $a(n)=50+3(n-1)=50+3n-3=47+3n$$

a. $83

b.$a(n)=47+3n$

a. Both are correct because
$$
50+3(n-1)=50+3n-3=47+3n
$$

The first form is easier to use for calculations, while the second form is easier to use to determine what had been originally given.

b. Replace $n$ with 12:

$$
t(12)=47+3(12)=47+36=83
$$

c.The fifth bithday is then $5cdot 12=60$ deposits:

$$
t(60)=47+3(60)=47+180=227
$$

b)

When the manufacturer wants to calculate

the raw material amount required to make

this toy, in this situation he should

find the sum of its diameters

If it consists of 12 dolls, the series will be:

5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16

The sum is:

S = (5 + 16) $times$ 6 = 126 cm

c)

t(n) = 8 + 3(n – 1)

The smallest doll diameter is 8,

And it increases by 3

So, the series is:

8, 11, 14, 17, 20, 23, 26, 29, 32, 35,

38, 41, 44, 47, 50, 53, 56, 59, 62, 65

The line length is the sum of the series:

S = (8 + 65) $times$ 10 = 730 mm

a. Since the amount starts at 1 at $n=1$ and every step the amount increases by 2:

$$
t(n)=1+2(n-1)=1+2n-2=-1+2n
$$

The first term means that during the first second Flo travels 1 m and every second the distance traveled increases by 2 m.

b. The series represents then the total distance traveled during the 6 seconds.

c. The terms are shown by a number of squares equal to the value of the term. The total distance is then represented by all the squares next to each other.

d. You can approximate the area of the squares by approximated it by a trapezium and calculating the area of this trapezium.

a. $t(n)=-1+2n$

b. Total distance traveled during the 6 seconds.

c. The total distance is then represented by all the squares next to each other.

d. Approximate area by trapezium

The area of a rectangle is the product of the length and the width, whlie the area of a trapezium is the product of the sum of both bases and the height divided by 2:

a.
$$
5cdot 7 = 35
$$

b.
$$
dfrac{(4+10)cdot 16}{2}=320
$$

c.
$$
dfrac{(228+48)cdot 6}{2}=828
$$

d.
$$
dfrac{(14+20)cdot 32}{2}=544
$$

The area of a trapezium is the product of the sum of both bases and the height divided by 2 and the sum of a arithmetic series is the product of the sum of the first and last term and the number of terms divided by 2:

a. The number of terms is $dfrac{72-18}{4.5}=12$ and the series is $18,22.5,27,31.5,…,72$.

$$
SUM=dfrac{(18+72)cdot 12}{2}=540
$$

b. The common difference is the sum multiplied by 2 divided by the difference last/first term:

$$
dfrac{336cdot 2}{21-6}=44.8
$$

c. It is not possible to determine the series because you need either the first term or the common difference.

d. The first term is than 9 and the last 179: 9,19,29,…,179.

$$
SUM=dfrac{(9+179)cdot 18}{2}=1692
$$

a. Every 15 minutes the number of people that enter increase by 250 and at noon 16 quarters have passed

$$
300,550,800,1050,1300,1550,1800,2050,2300,2550,
$$

$$
2800,3050,3300,3550,3800,4050
$$

Thus we note that the last quarter before noon 4050 people have entered.

b. The number of people start at 300 with $n=1$ and increase by 250 every quarter:

$$
t(n)=300+250(n-1)=300+250n-250=50+250n
$$

a. 4,050 people

b. $t(n)=50+250n$

a)

On Angla’s second birthday, n = 24

So,

t(24) = 50 + 5 $times$ (24 -1) = $165

b)

The formula of $text{n}^{text{th}}$ term of this sequence is:

t(n) = 50 + 5(n – 1)

c)

Deposit on the $text{12}^{text{th}}$ is:

t(12) = 50 + 5 $times$ (12 – 1) = $105\

The fund after 12 months is:\

S = (50 + 105)$times $dfrac{12}{2}$ = $930$

a) $165

b) t(n) = 50 + 5(n – 1)

c) $930

(a.)\
We determine the number of ways in which different four-person teams can be made by using combination because it is given that the order in which they bowl does not matter.\
Then, the number of combinations of $n$ object taken $r$ at times is determined by the following formula as stated:\
begin{align*}
^{n}C_{r} &= dfrac {n!}{(n-r)! r !}\
intertext {Put the values in the formula, we get}\
^{6}C_{4} &= dfrac {6!}{(6 – 4)! 4!} \
&= dfrac {6cdot5cdot4!}{2! cdot 4!}\
&= dfrac{6cdot5}{2cdot1}\
^{6}C_{4} &= 15\
intertext {Thus, there are 15 different four-person teams can be made.}\
end{align*}

(b.)\
We determine the number of bowling lineups of four player by using permutations, because it is given that the order of bowling lineups does matter.\
Then, the number of permutation of $n$ object taken $r$ at times is determined by the following formula as stated:\
begin{align*}
^{n}P_{r} &= dfrac {n!}{(n-r)!}\
intertext {Put the values in the formula, we get}\
^{6}P_{4} &= dfrac {6!}{(6 – 4)!}\
&= dfrac {6cdot5cdot4cdot3cdot2!}{2!}\
&= dfrac{6cdot5cdot4cdot3}{1}\
^{6}P_{4} &= 360\
intertext {Thus, there are 360 different bowling lineups of four players can be made.}\
end{align*}

$a.) 15$

$b.) 360$

The common difference is the difference of the two given terms divided by the number of terms in between them:

$$
dfrac{-29-13}{17-3}=dfrac{-42}{14}=-3
$$

We can then continue from the third term to obtain the eigth term:

$$
13,10,7,4,1,-2
$$

Thus the eigth term is $-2$.

$$
-2
$$

Now find the value of $x$.

$(a.) 46.70$

$(b.) 8.19$

Put the values in equation (1) to get the equation of sine.

$$
begin{align*}
text {Y}&= 1 times text {Sin} (1 (text x – dfrac{pi}{4})) + 2\
text {Y} &= text {Sin} (text x – dfrac {pi}{4}) + 2\
end{align*}
$$

(b.)

As per the given graph.

We can say that the graph represents a Cosine function.

The equation of sine is stated below:

$$
begin{align*}
text {Y} &= text{A Cos(B (x – C)) + D}tag{1}\
end{align*}
$$

Put the values in equation (1) to get the equation of Cosine.

$$
begin{align*}
text {Y}&= 1.5 times text {Cos} (1 (text x)) + 0.5\
text {Y} &= 1.5 text {Cos x} + 0.5\
end{align*}
$$

(c.)

As per the given graph.

We can say that the graph represents a Cosine function.

The equation of sine is stated below:

$$
begin{align*}
text {Y} &= text{A Cos(B (x – C)) + D}tag{1}\
end{align*}
$$

Put the values in equation (1) to get the equation of Cosine.

$$
begin{align*}
text {Y} &= 1 times text {Cos} (1 (text x + dfrac{pi}{3})) + 2\
text {Y} &= text {Cos} (text x + dfrac{pi}{3}) + 2\
end{align*}
$$

(d.)

As per the given graph.

We can say that the graph represents a sine function.

The equation of sine is stated below:

$$
begin{align*}
text {Y} &= text{A Sin(B (x – C)) + D}tag{1}\
end{align*}
$$

Put the values in equation (1) to get the equation of sine.

$$
begin{align*}
text {Y}&= 3 times text {Sin} (1 (text x – dfrac{2pi}{3})) – 1\
text {Y} &= 3text {Sin} (text x – dfrac {2pi}{3}) – 1\
end{align*}
$$

$(a.) Y = text {Sin} (text x – dfrac {pi}{4}) + 2\$$(b.) Y = 1.5 Cos x + 0.5
(c.) Y = Cos ( x + $dfrac{pi}{3}$) + 2
(d.) Y = 3Sin ( x – $dfrac {2pi}{3}$) – 1
$

The area of a trapezium is the product of the sum of both bases and the height divided by 2 and the sum of a arithmetic series is the product of the sum of the first and last term and the number of terms divided by 2:

$$
dfrac{(350+4050)16}{2}=35,200
$$

Determine the diameter of the 1th doll:

$$
t(1)=13+3(1-1)=13+0=13
$$

Determine the diameter of the 20th doll:

$$
t(20)=13+3(20-1)=13+57=70
$$

The sum of a arithmetic series is the product of the sum of the first and last term and the number of terms divided by 2:

$$
dfrac{(70+13)20}{2}=830mm=83cm
$$

Thus we note that the dolls will fit on the windowsill.

Determine the 1st term:

$$
1
$$

Determine the 10th term:

$$
46
$$

The sum of a arithmetic series is the product of the sum of the first and last term and the number of terms divided by 2:

$$
dfrac{(1+46)10}{2}=235
$$

$$
235
$$

The $9^{text{th}}$ term is 18

So,

18 = first term + common difference(9 – 1)

First term = 18 – 8 $times$ common difference

And the $5^{text{th}}$ term is 6

6 = first term + common difference (5 – 1)

So,

6 = 18 – 8 $times$ common difference + 4 $times$ common difference

So,

Common difference = 3

First term = 18 – 8 $times$ 3 = – 6

So, the series is:

-6, -3, 0, 3, 6, 9, 12, 15, 18

First term is: 3

Common difference is: 8

Check the term 435,

if n is integer it will be a term of the series

435 = 3 + 8 $times$ (n – 1)

n = 55

So,

435 is the $55^{text{th}}$ term of this series

Yes, the $55^{text{th}}$ term

begin{tabular}{ |p{1.5cm}|p{0.75cm}|p{1cm}|p{0.225cm}|p{1cm}|p{0.5cm}|p{1cm}|p{0.75cm}|p{0.5cm}|p{0.75cm}| }
hline
$xtext{(angle)}$ & $-90^circ$ & $-45^circ$ & $0^circ$ & $45^circ$ & $90^circ$ & $135^circ$ & $180^circ$ & $cdots$ & $270^circ$ \
hline
$xtext{(height)}$ & $-30$ & $-15 sqrt{2}$ & $0$ & $15 sqrt{2}$ & $30$ & $15 sqrt{2}$ & $0$ & $cdots$ & $-30$ \
hline

end{tabular}

a)

$$
y=30 sin x
$$

b)

The maximum distance above the center that the top of the seat attains during the ride occurs when $x=90^circ$ and it is equal to $y=30 sin 90^circ=30$

The maximum distance below the center that the top of the seat attains during the ride occurs when $x=-90^circ$ and it is equal to $left|30 sin -90^circright|=30$

c)

The equation that fits the County Fair Ferris wheel ride is

$$
y=30 cos x
$$

a)

Graph

b)

The maximum distance above the center that the top of the seat attains during the is $30$

The maximum distance below the center that the top of the seat attains during the ride is $30$

c)

$$
y=30 cos x
$$

The graph of $y=mid x+5mid$ and $y=mid xmid-5$ are equivalent when $xleq-5$

a. As long as $n$ is even the stacking will always result in a rectangle because the stacks have a common difference.

b. The sum is the area of the rectangle which is the product of the length and the width of the rectangle:
$$
8cdot 13=104
$$

c. The first column contains the first and the 16th term. The height of this column is 13 which is the sum of the first and the 16th term.

d. The first element of the series is 2 and every consective element is increased by $dfrac{11-2}{16}=dfrac{9}{16}$:

$$
t(n)=2+dfrac{9}{16}(n-1)=2+dfrac{9}{16}n-dfrac{9}{16}=dfrac{23}{16}+dfrac{9}{16}n
$$

The sum of the series is the sum of the first and last term multiplied by the number of terms divided by 2:

$$
dfrac{(2+11)16}{2}=104
$$

a. As long as $n$ is even the stacking will always result in a rectangle

b. 104

c. The first column contains the first and the 16th term. The height of this column is 13 which is the sum of the first and the 16th term

d. 104

For example, the series $1+2+3$

You can then use Antonio’s method by dividing the middle block into two even pieces and placing the second piece on the first.

Then the area of the rectangle would be
$$
4cdot 1.5=6=1+2+3
$$

Yes, because the sum of a series is the sum of the first and last term multiplied by the number of terms divided by two.

a. The number of terms is the difference between the last and first term divided by the common difference increased by 1:
$$
dfrac{74-11}{7}+1=dfrac{63}{7}+1=9+1=10
$$

The sum is then:
$$
dfrac{(74+11)10}{2}=425
$$

b.
$$
dfrac{(8+151)14}{2}=1113
$$

c. The first term is 4 and the 25th term is $t(25)=4+6(24)=148$

$$
dfrac{(4+148)25}{2}=1900
$$

a. $425$

b. 1113

c. 1900

Horizontal dimension is the number of terms = 12 $times$ 18 = 216

Value of the last term is:

t(216) = 50 + 3 $times$ (216 – 1) = 695

Vertical dimension = 50 + 695 = 745

Total money for Angla on her $18^{text{th}}$ birthday is the sum of series:

S = $dfrac{745 times 216}{2}$ = $80460

$80460

a. The series is:

$$
6+12+18+24+30+36+42+48+54+60
$$

The total amount is then the sum of the first and last payment multiplied by the number of payments divided by 2:

$$
dfrac{(6+60)10}{2}=330
$$

b. Sequence, every term is the previous term increased by 55:

$$
190,245,300,355,410,465,520,575
$$

c. Series, determine the allowance after one year:

$$
t(52)=1.5+0.25(52-51)=14.25
$$

The total amount is then the sum of the first and last payment multiplied by the number of payments divided by 2:

$$
dfrac{(1.5+14.25)52}{2}=409.5
$$

Thus he has enough money.

a. 330 cells

b. $575

c. Yes

Since, ticket sales is constantly increasing,

And every sales is added to the sum,

So, this is an arithmetic series,

Last term = 1800000 + 20000 $times$ (14 – 1) = $2060000\

The sum of series is:\

S = (1800000 + 2060000)$times $$dfrac{14}{2}$ = $27020000

Since, the jackpot started with $29000000,

So, the jackpot when the tickets are drown is expected to be:

29000000 + 27020000 = $56020000

So, the other state will have larger jackpot

The sum of an arithmetic series is then the sum of the first and last payment multiplied by the number of terms divided by 2:

$$
dfrac{(t(1)+t(n))n}{2}
$$

$$
dfrac{(t(1)+t(n))n}{2}
$$

The sum of an arithmetic series is then the sum of the first and last payment multiplied by the number of terms divided by 2:

$$
dfrac{(t(1)+t(n))n}{2}
$$

$$
dfrac{(t(1)+t(n))n}{2}
$$

b)

For odd numbers:

149 = 1 + 2 $times$ (n – 1)

n = 75

S = (1 + 149) $times dfrac{75}{2}$ = 5625

For even numbers:

150 = 2 + 2 $times$ (n – 1)

n = 75

S = (2 + 150) $times dfrac{75}{2}$ = 5700

a. The first term is 21 and every consecutive term is the previous term decreased by 4:

$$
t(n)=21-4(n-1)=21-4n+4=25-4n
$$

b. The number of terms is the difference between the last and the first term divided by the common difference increased by 1 (we increase by 1, because else we only count the number of increments and thus forgot one term):

$$
dfrac{-99-21}{-4}+1=31
$$

c. The sum of an arithmetic series is then the sum of the first and last payment multiplied by the number of terms divided by 2:

$$
dfrac{(21+(-99))31}{2}=-1,209
$$

a. $t(n)=25-4n$

b. 31 terms

c. $-1,209$

a.

Using one of the Laws of Cosine

$$
c^2=a^2+b^2-2ab cos C
$$

$$
8^2=10^2+12^2-2(10)(12) cos C
$$

$$
64=100+144-240 cos C
$$

$$
64=244-240 cos C
$$

$$
64textcolor{#c34632}{-244}=244textcolor{#c34632}{-244}-240 cos C
$$

$$
-180=-240 cos C
$$

$$
dfrac{-180}{textcolor{#c34632}{-240}}=dfrac{-240}{textcolor{#c34632}{-240}} cos C
$$

$$
0.75=cos C
$$

$$
cos^{-1}(0.75)=C
$$

$$
boxed{41.41text{textdegree} approx C}
$$

b.

Using Trig Ratios

$$
tan theta = dfrac{opposite}{adjascent}
$$

$$
tan C= dfrac{7}{13}
$$

$$
C=tan^{-1}left( dfrac{7}{13}right)
$$

$$
boxed{Capprox 28.30text{textdegree}}
$$

$_{n}P_{r} = dfrac{n!}{(n-r)}= n(n-1)(n-2)..(n-r+1)$ for $n$ items chosen $r$ at a time

in this case

$n$ represents the number of murals

$r$ represents the number of times

$$
_{4}P_{4}=dfrac{4!}{(4-4)!}= dfrac{4!}{0!}=dfrac{4cdot3cdot2cdot1}{1} = dfrac{24}{1} =boxed{24}
$$

a.

$x=-2+sqrt3$,

$x=-2-sqrt3$

factored polynomial for

$x=-2+sqrt3$ is $x+2-sqrt3$ , and

$x=-2-sqrt3$ is $x+2+sqrt3$

$boxed{(x+2-sqrt3)(x+2+sqrt3)}$

standard polynomial

$(x+2-sqrt3)(x+2+sqrt3)$

$(x+2)^2-sqrt3^2$

$(x+2)^2-3$

$x^2+2x+2x+4-3$

$$
boxed{x^2+4x+1}
$$

b.

$x=-2+i$

$x=-2-i$

factored polynomial for

$x=-2+i$ is $x+2-i$ , and

$x=-2-i$ is $x+2+i$

$boxed{(x+2-i)(x+2+i)}$

standard polynomial

$(x+2-i)(x+2+i)$

$(x+2)^2-i^2$

$left[Rule: i^2=-1right]$

$(x+2)^2-(-1)$

$(x+2)^2+1$

$x^2+2x+2x+4+1$

$$
boxed{x^2+4x+5}
$$

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

To solve the problem, we use this formula
$$
boldsymbol{x^{circ}=frac{pi}{180}}textbf{ radians}
$$

$$
begin{align*}
&60^{circ}cdot frac{pi}{180}text{ radians}&&boxed{text{Given proportion}}\
&frac{60cdot pi}{180}text{ radians}&&boxed{text{Multiply fractions}}\
&frac{pi}{3}text{ radians}&&boxed{text{Cancel the common factor: }60}\\
&boxed{{color{#c34632}frac{pi}{3}text{ radians}} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ }mathrm{Multiply:fractions}:quad :acdot frac{b}{c}=frac{a:cdot :b}{c} }
$$

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

To solve the problem, we use this formula
$$
boldsymbol{x^{circ}=frac{pi}{180}}textbf{ radians}
$$

$$
begin{align*}
&75^{circ}cdot frac{pi}{180}text{ radians}&&boxed{text{Given proportion}}\
&frac{75cdot pi}{180}text{ radians}&&boxed{text{Multiply fractions}}\
&frac{5pi}{12}text{ radians}&&boxed{text{Cancel the common factor: }15}\\
&boxed{{color{#c34632}frac{5pi}{12}text{ radians}} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ } mathrm{Multiply:fractions}:quad :acdot frac{b}{c}=frac{a:cdot :b}{c}}
$$

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

To solve the problem, we use this formula
$$
boldsymbol{x^{circ}=frac{pi}{180}}textbf{ radians}
$$

$$
begin{align*}
&210^{circ}cdot frac{pi}{180}text{ radians}&&boxed{text{Given proportion}}\
&frac{210cdot pi}{180}text{ radians}&&boxed{text{Multiply fractions}}\
&frac{7pi}{6}text{ radians}&&boxed{text{Cancel the common factor: }30}\\
&boxed{{color{#c34632}frac{7pi}{6}text{ radians}} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ } mathrm{Multiply:fractions}:quad :acdot frac{b}{c}=frac{a:cdot :b}{c}}
$$

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

To solve the problem, we use this formula
$$
boldsymbol{x^{circ}=frac{pi}{180}}textbf{ radians}
$$

$$
begin{align*}
&225^{circ}cdot frac{pi}{180}text{ radians}&&boxed{text{Given proportion}}\
&frac{225cdot pi}{180}text{ radians}&&boxed{text{Multiply fractions}}\
&frac{5pi}{4}text{ radians}&&boxed{text{Cancel the common factor: }45}\\
&boxed{{color{#c34632}frac{5pi}{4}text{ radians}} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
boxed{ color{#c34632} text{ } mathrm{Multiply:fractions}:quad :acdot frac{b}{c}=frac{a:cdot :b}{c}}
$$

$$
color{#4257b2} text{ a) } frac{pi}{3}text{ radians}
$$

$$
color{#4257b2} text{ b) }frac{5pi}{12}text{ radians}
$$

$$
color{#4257b2} text{ c) }frac{7pi}{6}text{ radians}
$$

$$
color{#4257b2} text{ d) } frac{5pi}{4}text{ radians}
$$

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a. 3 terms:
$$
dfrac{(1+3)3}{2}=6
$$

10 terms:
$$
dfrac{(1+10)10}{2}=55
$$

b.
$$
dfrac{(1+n)n}{2}
$$

c.
$$
dfrac{(1+100)100}{2}=5050
$$

a. 6, 55

b. $dfrac{(1+n)n}{2}$

c. 5,050

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a.
$$
dfrac{(2+2n)n}{2}=(1+n)n
$$

b.
$$
dfrac{(1+2n-1)n}{2}=dfrac{n^2}{2}=n^2
$$

c.
$$
Even: (1+100)100=10,100
$$

$$
Odd: 100^2=10,000
$$

a. $(1+n)n$

b. $n^2$

c. 10,100 and 10,000

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a. The new series is arithmetic since the common difference is 3 and the series contains $n$ terms.

b.
$$
text{First series}:dfrac{(1+n)n}{2}
$$

$$
text{Second series}:dfrac{(2+2n)n}{2}
$$

Add the two expression to obtain the sum of the new series:

$$
dfrac{(1+n)n+(2+2n)n}{2}=dfrac{(1+n+2+2n)n}{2}=dfrac{(3+3n)n}{3}
$$

c. You could also determine the sum of the new series directly:

$$
dfrac{(3+3n)n}{2}
$$

d. We note that the equations obtained in (b) and (c) are identical.

a. Yes, $n$

b. $dfrac{(3+3n)n}{3}$

c. $dfrac{(3+3n)n}{3}$

d. Identical

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a. Since every term in the sum is multiplied by 2, the sum itself will also be multiplied by 2 as we note is obvious in the formulas as well:

$$
text{First series}:dfrac{(1+n)n}{2}
$$

$$
text{Second series}:dfrac{(2+2n)n}{2}=(1+n)n
$$

b. We note that ever term is decreased by 1 and thus in total the sum is decreased by $n$; as we note in the formulas as well:

$$
text{First series}:dfrac{(2+2n)n}{2}=(1+n)n=n^2+n
$$

$$
text{Second series}:dfrac{(1+2n-1)n}{2}=dfrac{2n^2}{2}=n^2
$$

a. Sum is multiplied by 2

b. Sum is decreased by $n$

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a. We note that the terms contain: 1, 3, 5, 7, 9, etc. terms and thus the nth terms is $2n-1$. (See previous exercise).

b.
$$
text{3 terms}:dfrac{(1+5)3}{2}=9
$$

$$
text{5 terms}:dfrac{(1+9)5}{2}=25
$$

The sum could be representated by adding the blocks together and wethen note that they form a square with length $n$ (when $n$ terms were added)

c.
$$
n text{terms}:dfrac{(1+2n-1)n}{2}=dfrac{2n^2}{2}=n^2
$$

Thus the sum is a square with length $n$.

d. The sum of the first $(n+1)$ odd numbers is then: $(n+1)^2$.

a. $2n-1$

b. 9, 25

c. $n^2$, square with length $n$

d. $(n+1)^2$

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

Let us add the second and third series: $3+7+11+15+…+(4n-1)$

$$
text{Second series}:dfrac{(2+2n)n}{2}=(1+n)n=n^2+n
$$

$$
text{Third series}:dfrac{(1+2n-1)n}{2}=dfrac{2n^2}{2}=n^2
$$

Add the two expression to obtain the sum of the new series:

$$
n^2+n+n^2=2n^2+n
$$

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

Solution to this example is given below

The first term is $5$,
and the difference between the terms is $boldsymbol{5}$
Because $5+5$ = $10$.

The number of terms is the
difference between the last and first term
divided by the common difference
and incerased by 1.

$$
begin{align*}
t&=frac{400-5}{5}+1&&boxed{text{Use the formula}}\
t&=frac{395}{5}+1&&boxed{text{Subtract the numbers: }400-5=395}\
t&=79+1&&boxed{text{Divide the numbers: } frac{395}{5}=79}\\
t&=color{#c34632}{80}&&boxed{text{Simplify}}\\
end{align*}
$$

The sum of an arithmetic series is
the sum of the first and last term,
multiplied by the number of terms
and divided by 2

$$
begin{align*}
S(n)&=frac{(5+400)n}{2}&&boxed{text{Use the formula}}\
S(80)&=frac{5+400)color{#c34632}{80}}{2}&&boxed{text{Substitute }80 text{ for }n}\
S(80)&=frac{405cdot80}{2}&&boxed{text{Calculate within parentheses: }5+400=405}\
S(80)&=frac{32,400}{2}&&boxed{text{Multiply the numbers }}\
S(80)&=16,200&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}S(80)=16,200} }&&boxed{text{Final solution}}\
end{align*}
$$

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

Solution to this example is given below

The first term is $3$,
and the difference between the terms is $boldsymbol{5}$
Because $3+5$ = $8$.

The number of terms is the
difference between the last and first term
divided by the common difference
and incerased by 1.

$$
begin{align*}
t&=frac{398-3}{5}+1&&boxed{text{Use the formula}}\
t&=frac{395}{5}+1&&boxed{text{Subtract the numbers: }398-3=395}\
t&=79+1&&boxed{text{Divide the numbers: } frac{395}{5}=79}\\
t&=color{#c34632}{80}&&boxed{text{Simplify}}\\
end{align*}
$$

The sum of an arithmetic series is
the sum of the first and last term,
multiplied by the number of terms
and divided by 2

$$
begin{align*}
S(n)&=frac{(3+398)n}{2}&&boxed{text{Use the formula}}\
S(80)&=frac{3+398)color{#c34632}{80}}{2}&&boxed{text{Substitute }80 text{ for }n}\
S(80)&=frac{401cdot80}{2}&&boxed{text{Calculate within parentheses: }3+398=401}\
S(80)&=frac{32,080}{2}&&boxed{text{Multiply the numbers }}\
S(80)&=16,040&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}S(80)=16,040} }&&boxed{text{Final solution}}\
end{align*}
$$

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

Solution to this example is given below

The first term is $80$,
and the difference between the terms is $boldsymbol{-6}$
Because $80-6$ = $74$.

The number of terms is the
difference between the last and first term
divided by the common difference
and incerased by 1.

$$
begin{align*}
t&=frac{14-80}{-6}+1&&boxed{text{Use the formula}}\
t&=frac{-66}{-6}+1&&boxed{text{Subtract the numbers: }14-80=-66}\
t&=11+1&&boxed{text{Divide the numbers: } frac{-66}{-6}=11}\\
t&=color{#c34632}{12}&&boxed{text{Simplify}}\\
end{align*}
$$

The sum of an arithmetic series is
the sum of the first and last term,
multiplied by the number of terms
and divided by 2

$$
begin{align*}
S(n)&=frac{(80+14)n}{2}&&boxed{text{Use the formula}}\
S(12)&=frac{80+14)color{#c34632}{12}}{2}&&boxed{text{Substitute }12 text{ for }n}\
S(12)&=frac{94cdot12}{2}&&boxed{text{Calculate within parentheses: }80+14=94}\
S(12)&=frac{1,128}{2}&&boxed{text{Multiply the numbers }}\
S(12)&=564&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}S(12)=564} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
color{#4257b2} text{ a) }S(80)=16,200
$$

$$
color{#4257b2} text{ b) }S(80)=16,040
$$

$$
color{#4257b2} text{ c) }S(12)=564
$$

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

The series is:
$$
11+22+33+44+55+66+77+88+99
$$

The sum is then:

$$
dfrac{(11+99)9}{2}=495
$$

$$
11+22+33+44+55+66+77+88+99
$$

Sum: 495

The sum of an arithmetic series is the sum of the first and last term multiplied, by the number of terms, divided by 2.

a. We note that the second term is the first term decreased by 2 and the next term is then increased by 7, then decreased by 2 again and then increased by 7 again and so on.

b. Yes, since the sequence does not have a common difference, the sequence is not arithmetic, however we do note that this sum is the sum of two arithmetic sequences and thus we can determine the sum.

c.
$$
text{First series}:dfrac{(5+400)80}{2}=16,200
$$

$$
text{Second series}:dfrac{(3+398)80}{2}=16,040
$$

Add the two sums to obtain the sum of the original series:

$$
16,200+16,040=32,240
$$

$$
32,240
$$

a. Since the order does not matters, we need to use a combination:

$$
C_{10}^{12}=dfrac{12!}{(12-10)!10!}=dfrac{12cdot 11}{2cdot 1}=66
$$

b. Since the order does not matters, we need to use a combination:

$$
C_{7}^{9}=dfrac{9!}{(9-7)!7!}=dfrac{9cdot 8}{2cdot 1}=36
$$

a-

$x^2=(35)^2+(43)^2-2 cdot 35 cdot 43 cos 50$          (The law of cosins)

$x^2=1225+1849-2 cdot 35 cdot 43 cdot 0.6428$

$x^2=1225+1849-1934.83$

$x^2=1139.17$

$$
x approx 33.75 mathrm { ~ft}
$$

b-

$dfrac {sin 25text{textdegree}}{x}=dfrac {sin 41text{textdegree}}{15}$          (The law of sins)

$dfrac {0.4226}{x}=dfrac {0.6561}{15}$

$x=0.4226 times dfrac {15}{0.6561}$

$$
x approx 9.66 mathrm { ~ft}
$$

a-          $x approx 33.75 mathrm { ~ft}$

b-          $x approx 9.66 mathrm { ~ft}$

$y=sintheta$

$$
y=cosleft(theta-dfrac{pi}{2}right)
$$

$theta=angle AOB$

$y=sintheta=BX$

In the unit circle $y=sintheta$ represents the $y$-coordinate of the point in which one the side of the angle $theta$ which is not $Ox$ intersects the circle:

$theta-dfrac{pi}{2}=angle AOC$

$$
angle COB=90text{textdegree}
$$

$thetadfrac{pi}{2}$ is the angle having the side which is not $OX$ situated $90text{textdegree}$ before $theta$,therefore in the previous Quadrant.

$$
angle BOXcongangle OCY
$$

The angles $BOX$ and $OCY$ are complementary with the same angle $COY$, therefore they are congruent:

$$
triangle BOXcongtriangle YCO
$$

$$
OY=BX
$$

$$
cosleft(theta-dfrac{pi}{2}right)=cos AOC=OY=BX=sintheta
$$

It is given that,
The average cost of a movie ticket is 9.50 dollars, i.e,
$$begin{aligned}
a_{1} &= 9.50\
end{aligned}$$
As per the statement, the cost is increasing 4% per annum, which means:
$$begin{aligned}
q &= 1+ dfrac{4}{100}\\
q &= 1.04\
end{aligned}$$

(a.) The average cost of a movie could be calculated by the given equation is state below:
$$begin{aligned}
a_{n} &= a_{0}.r^{n-1} tag 1\\
end{aligned}$$
Now to determine the value of $n$ ,so that
$$begin{aligned}
a_{n} &= 2 . a_{0}\\
a_{n} &= 2 times 9.50\\
a_{n} &= 19\
end{aligned}$$

Put the value in equation (1),
$$begin{aligned}
19 &= 9.50 (1.04)^{n-1}\\
dfrac {19}{9.50} &= (1.04)^{n-1}\\
2 &= (1.04)^{n}\\
log {2} &= (n-1) log (1.04)\\
n-1 &= dfrac {log(2)}{log(1.04)}\\
n-1 &= 17.67\\
n &= 18.67\\
n &approx 19\
end{aligned}$$

(b.) To determine the average rate of change of the movie ticket cost over the next 10 years.
$$begin{aligned}
&=dfrac {a_{11}-a_{1}}{11-1}\\
&=dfrac {9.50 (1.04)^{10}-9.50}{10}\\
&= dfrac{14.062 – 9.50}{10}\\
&approx 0.46\
end{aligned}$$
Thus, during the next 10 years, the ticket’s cost will be increased by $ $0.46$ per year.

(a.) 19 years.
(b.) $0.46 per year.

a.He adds the first to the last term, then the second to the next to last term and so on, unitl you have added the last term to the first term.

b. The sum of the series is then 100 times 101 divided by 2:

$$
dfrac{100cdot 101}{2}=5,050
$$

c. The sum of the series is also the sum of the first and last term, multiplied by the number of terms divided by 2; we note that this is the same as has been done in (a) and (b).

d. The sum is the sum of the first and last term, multiplied by the number of terms divided by 2:

$$
dfrac{(2+59)20}{2}=610
$$

a. Note that the sum contains five terms and the common difference is 4:
$$
sum_{n=1}^{5}(15+4(n-1))=sum_{n=1}^{5}(15+4n-4)=sum_{n=1}^{5}(11+4n)
$$

b.
$$
1^2+2^2+3^3+4^2+5^2+6^2=1+4+9+16+25+36=91
$$

a. $sum_{n=1}^{5}(11+4n)$

b. $1^2+2^2+3^3+4^2+5^2+6^2=91$

a. Note that the sum contains 80 terms and the common difference is 5:
$$
sum_{n=1}^{80}(5n)
$$

Note that the sum contains 80 terms and the common difference is 5:
$$
sum_{n=1}^{80}(3+5(n-1))=sum_{n=1}^{80}(3+5n-5)=sum_{n=1}^{80}(-2+5n)
$$

b. Add the two expression following the sigma sign:

$$
sum_{n=1}^{80}(5n-2+5n)=sum_{n=1}^{80}(-2+10n)
$$

a. $sum_{n=1}^{80}(5n)$ and $sum_{n=1}^{80}(-2+5n)$

b. $sum_{n=1}^{80}(-2+10n)$

(a.) Given arithmetic series are:
$$begin{aligned}
sum ^{10} _{t=1} (13t-5)\
end{aligned}$$

As the value of $n$ is given from 1 and 10
$$begin{aligned}
a_{1} &= 13times1-5\\
a_{1} &= 8\\
a_{10} &= 13times10-5\\
a_{10} &= 125\
end{aligned}$$

Now determine the sum $S_{10}$ using the formula:
$$begin{aligned}
S_{n} &= dfrac{n(a_{1}+a_{n})}{2}\\
S_{10} &= dfrac{10(8+125)}{2}\\
S_{10} &= dfrac {10(133)}{2}\\
S_{10} &= 665
end{aligned}$$

(b.) Identify the elements of the series:
$$begin{aligned}
a_{1} &= 105.5\\
a_{23} &= 6.5\
end{aligned}$$

Now determine the sum $S_{23}$ using the formula:
$$begin{aligned}
S_{n} &= dfrac{n(a_{1}+a_{n})}{2}\\
S_{23} &= dfrac{23(105.5+6.5)}{2}\\
S_{23} &= dfrac {23 (112)}{2}\\
S_{23} &= 1288
end{aligned}$$

(c.) The given series are:

4+10+16+22+……..+(6n-2)
$$begin{aligned}
a_{1} &=4\\
d &= 10-4\\
d &= 6\
end{aligned}$$

Now determine the number $N$ of elements in the sum:
$$begin{aligned}
a_{N} &= a_{1}+(N-1)d\\
6n-2 &= 4+(N-1) (6)\\
6n-2 &= 4+6N-6\\
6n-2+2 &= 6N\\
6n &= 6N\
end{aligned}$$

So, there are $n$ elements in the series. Now, determine $S_{n}$:
$$begin{aligned}
S_{n} &= dfrac{n(a_{1}+a_{n})}{2}\\
S_{n} &= dfrac{n(4+6n-2)}{2}\\
S_{n} &= dfrac {n (6n+2)}{2}\\
S_{n} &= dfrac {2n (3n+1)}{2}\\
S_{n} &= n (3n+1)\
end{aligned}$$

(d.) Given arithmetic series are:
$$begin{aligned}
sum ^{30} _{t=1} (-5t+10)\
end{aligned}$$

As the value of $n$ is given from 1 and 30
$$begin{aligned}
a_{1} &= -5times1+10\\
a_{1} &= 5\\
a_{30} &= -5times30+10\\
a_{30} &= -140\
end{aligned}$$

Now determine the sum $S_{30}$ using the formula:
$$begin{aligned}
S_{n} &= dfrac{n(a_{1}+a_{n})}{2}\\
S_{30} &= dfrac{30 (5 – 140)} {2}\\
S_{30} &= dfrac {30 (-135)} {2}\\
S_{30} &= -2025
end{aligned}$$

$(a.)$ $665$
$(b.)$ $1288$
$(c.)$ $n(3n+1)$
$(d.)$ $-2025$

The sum of a arithmetic series is the product of the sum of the first and last term and the number of terms divided by 2:

a.

$$
Sigma_{n=1}^{11}(47-13(n-1))=dfrac{(47-83)11}{2}=-198
$$

b.

$$
Sigma_{n=1}^{n}(3+7(n-1))=dfrac{(3+3+7(n-1))n}{2}=dfrac{(6+7n-7)n}{2}=dfrac{(-1+7n)n}{2}
$$

a. $Sigma_{n=1}^{11}(47-13(n-1))=-198$

b. $Sigma_{n=1}^{n}(3+7(n-1))=dfrac{(-1+7n)n}{2}$

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

Solution to this example is given below

The first term is $100$,
and the difference between the terms is $boldsymbol{1}$
Because $100+1$ = $101$.

The number of terms is the
difference between the last and first term
divided by the common difference
and incerased by 1.

$$
begin{align*}
t&=frac{1,000-100}{1}+1&&boxed{text{Use the formula}}\
t&=frac{900}{1}+1&&boxed{text{Subtract the numbers: }1,000-100=900}\
t&=900+1&&boxed{text{Divide the numbers: } frac{900}{1}=900}\\
t&=color{#c34632}{901}&&boxed{text{Simplify}}\\
end{align*}
$$

The sum of an arithmetic series is
the sum of the first and last term,
multiplied by the number of terms
and divided by 2

$$
begin{align*}
S(n)&=frac{(100+1,000)n}{2}&&boxed{text{Use the formula}}\
S(901)&=frac{100+1,000)color{#c34632}{901}}{2}&&boxed{text{Substitute }901 text{ for }n}\
S(901)&=frac{1100cdot901}{2}&&boxed{text{Calculate within parentheses: }100+1,000=1,100}\
S(901)&=frac{991,100}{2}&&boxed{text{Multiply the numbers }}\
S(901)&=495,550&&boxed{text{Simplify}}\\
&boxed{{color{#c34632}S(901)=495,550} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
color{#4257b2} text{ }S(901)=495,550
$$

It is given that there are 12 members in a PEP squad.

(a.) The total number of ways the five students from the 12 members can be lined up will be calculated by using the Permutation formula is stated below.

$$begin{aligned}
^{n}P_{r}&= dfrac{n!}{(n-r)}
end{aligned}$$
where,
$n$ represents the number of members.
$r$ represents the number of students.

$$begin{aligned}
^{12}P_{5}&= dfrac{12!}{(12 – 5)}\\
^{12}P_{5}&= dfrac{12!}{7!}\\
^{12}P_{5}&= dfrac{12.11.10.9.8.7!}{7!}\\
^{12}P_{5}&= 12.11.10.9.8\\
^{12}P_{5}&= 95040\\
end{aligned}$$
Hence, the total number of ways 5 students can be lined for a picture will be 95040.

(b.) To find the number of ways that the five students be selected from 12 members by using the Combination formula is stated below.
$$begin{aligned}
^{n}C_{r} &= dfrac{^{n}P_{r}}{(r!)}\\
^{n}C_{r} &= dfrac{n!}{(n-r)! r!}\
end{aligned}$$
where,
$n$ represents the number of members.
$r$ represents the number of students.

$$begin{aligned}
^{12}C_{5} &= dfrac{^{12}P_{5}}{5!}\\
^{12}C_{5} &= dfrac{95040}{5!}\\
^{12}C_{5} &= dfrac{95040}{5.4.3.2.1}\\
^{12}C_{5} &= dfrac{95040}{120}\\
^{12}C_{5} &= 792\\
end{aligned}$$
Hence, 792 students are chosen to visit the rival school before the big game.

$(a.)$ $95040$
$(b.)$ $792$

(a.) Find the value of $x$ by using the Law of Tangent is stated below.


As per the figure, we get,

$$begin{aligned}
tantheta &= dfrac{opp}{adj}\\
tan (24degree) &= dfrac {36}{x}\\
x &= dfrac {36}{tan 24degree}\\
x &= 80.85
end{aligned}$$

(b.) Find the value of $x$ by using the Law of Sine’s as stated below.


As per the figure, we get

Let’s say A = 23, a = 10, B = 70, b = $x$
$$begin{aligned}
dfrac{Sin A}{a} &= dfrac{Sin B}{b}\\
end{aligned}$$

Put the given values in the equation.

$$begin{aligned}
dfrac{Sin 23degree}{10} &= dfrac{Sin 70degree}{x}\\
end{aligned}$$
Cross multiply the equation to get the value of $x$.
$$begin{aligned}
Sin (23degree).x &= 10. Sin(70degree)\\
x &= dfrac{10.Sin 70degree}{sin 23degree}\\
x &= dfrac{9.3969}{0.390}\\
x &approx 24.05\
end{aligned}$$

(c.) Find the value of $x$ by using the Law of Cosine as stated below.


As per the figure, we get

Let’s say A = $x$, a = 3, b = 10, c = 11
$$begin{aligned}
a^{2} &= b^{2} + c^{2} – 2.b.c.CosA
end{aligned}$$

Put the given values in the equation.

$$begin{aligned}
(3)^{2} &= (10)^{2} + (11)^{2}+2.10.11.Cos(x)\\
9 &= 100+121+220.Cos(x)\\
220.Cos(x) &= 221-9\\
Cosx &= dfrac {212}{220}\\
Cosx &= 0.9636\\
x &= Cos^{-1}(0.9636)\\
x &approx 15.50
end{aligned}$$

$(a.)$ $80.85$
$(b.)$ $24.05$
$(c.)$ $15.50$

According to the graph.

Find the amplitude |a|. So,
$$begin{aligned}
text{ Amplitude} &= 4\
end{aligned}$$

Now, find the period (b).

The period of the function can be calculated using
$$begin{aligned}
dfrac{2pi}{|b|}\
end{aligned}$$

Replace $b$ with 1 in the formula for the period.
$$begin{aligned}
dfrac{2pi}{|1|}\
end{aligned}$$
The absolute value is the distance between a number and zero. The distance between 0 and 1 is 1.

Divide 2 pi by 1.

$$begin{aligned}
text{ Period} &= 2pi\
end{aligned}$$

Vertical shifted up by 2 units.
Phase Shift = 0

As per the values.

So, the given graph is that of a Sine function, having the same period 2 pi, vertically stretched by a factor of 4 and shifted up by 2 units.

The equation is given below:

$$begin{aligned}
y &= 4cdot sin x + 2\
end{aligned}$$

$y = 4sin x + 2$

a. If a function contains a factor $(x-a)^2$ then $x=a$ is a double root of the function.

b. If a function contains a factor $(x-a)^2$ then $x=a$ is a double root of the function.

(a.) Given:
$$begin{aligned}
cx – a &= b\\
end{aligned}$$
Adding $a$ on both sides, to solve the equation,
$$begin{aligned}
cx – a+a &= b+a\\
cx &= a+b\\
end{aligned}$$
Divide L.H.S and R.H.S by $c$, to get the value of $x$.
$$begin{aligned}
dfrac{cx}{c} &= dfrac{a+b}{c}\\
x &= dfrac {a+b}{c}\
end{aligned}$$

(b.) Given:
$$begin{aligned}
dfrac{x}{a} – b^{2}&= c\\
end{aligned}$$
Adding $b^{2}$ on both sides, to solve the equation,
$$begin{aligned}
dfrac{x}{a} – b^{2} + b^{2} &= c + b^{2}\\
dfrac{x}{a} &= c + b^{2}\\
end{aligned}$$
Multiply L.H.S and R.H.S by $a$, to get the value of $x$.
$$begin{aligned}
(a)dfrac{x}{a} &= a(b^{2} + c)\\
x &=ab^{2} + ac\
end{aligned}$$

(c.) Given:
$$begin{aligned}
(x-a)(x-b) &= 0\\
x-a &= 0, x-b = 0\\
x-a+a &= 0+a, x-b+b =0+b\\
x &= a, x = b
end{aligned}$$

(d.) Given:
$$begin{aligned}
ax^{3} – acx^{2} &= 0\\
ax^{2} (x-c) &=0\\
ax^{2} &=0 and x-c=0\\
x^{2} &= dfrac{0}{a} and x = c\\
sqrt {x^{2}} &= 0 and x = c\\
x &=0 and x = c

end{aligned}$$

(e.) Given:
$$begin{aligned}
dfrac {x}{a+b} &= dfrac {1}{c}\\
end{aligned}$$
Multiplying $(a+b)$ on both side i.e L.H.S and R.H.S
$$begin{aligned}
(a+b)dfrac {x}{a+b} &= (a+b) dfrac {1}{c}\\
x &= dfrac {a+b}{c}\\
end{aligned}$$

(f.) Given:
$$begin{aligned}
dfrac {1}{x^{3}} + a&= b\\
end{aligned}$$
Subtracting $(a)$ on both side i.e L.H.S and R.H.S
$$begin{aligned}
dfrac {1}{x^{3}} + a – a &= b-a\\
dfrac {1}{x^{3}} &= b-a\\
end{aligned}$$
Multiplying $(x^{3})$ on both side i.e L.H.S and R.H.S
$$begin{aligned}
(x^{3})dfrac {1}{x^{3}} &= (x^{3}) (b-a)\\
1 &= (x^{3}) (b-a)\\
end{aligned}$$
Divide $(b-a)$ on both side i.e L.H.S and R.H.S
$$begin{aligned}
dfrac {1}{(b-a)} &= (x^{3})dfrac {(b-a)} {(b-a)}\\
dfrac {1}{(b-a)} &= x^{3}\\
sqrt[3]{dfrac {1}{(b-a)}} &= x\\

end{aligned}$$

$(a.)$ $dfrac{a+b}{c}$

$(b.)$ $ab^{2} + ac$
$(c.)$ $x = a, x = b$
$(d.)$ $x = 0 , x = c$
$(e.)$ $dfrac{a+b}{c}$

$(f.)$ $sqrt[3]{dfrac{1}{b-a}}$

(a.) Given series: 1 + 3 + 5 + … + (2n – 1)

Identify the elements of the arithmetic series:
$$begin{aligned}
a_{1} &=1\
d &= 3 – 1\
d &= 2\
end{aligned}$$

Geometrically, it can be seen that in order to add $a_{1}$ and $a_{2}$. Place the first square near the other 3 from $a_{2}$ and get a square of side 2, so
$$S_{2} = 2^{2}$$
$$S_{2} = 4$$
In the same way, to get $S_{3}$ add the square of side 2 which we got by adding $a_{1}$ and $a_{2}$ to $a_{2}$ and we get a square of side 3 which means
$$S_{3} = 3^{2}$$
$$S_{3} = 9$$
Therefore,
$$S_{n} = n^{2}$$

(b.) Now, check the formula which is true for $n = 1$
$$begin{aligned}
a_{1} &= 1^{2}\
1 &= 1\
end{aligned}$$

Check the formula which is true for $n = 5$
$$begin{aligned}
a_{1} +a_{2}+a_{3}+a_{4}+a_{5} &= 5^{2}\
1+3+5+7+9 &= 25\
25 &= 25\
end{aligned}$$

Check the formula which is true for $n = 8$
$$begin{aligned}
a_{1} +a_{2}+a_{3}+a_{4}+a_{5}+a_{6} +a_{7} +a_{8} &= 8^{2}\
1+3+5+7+9+11+13+15 &= 64\
64 &= 64\
25 &= 25\
end{aligned}$$

(c.) Now prove the formula is right either by using the formula for the sum of the first $n$ terms of an arithmetic series or by induction.
$$begin{aligned}
S_{n} &= dfrac {n(a_{1}+a_{n})}{2}\\
&= dfrac {n(1+(2n-1))}{2}\\
&= dfrac {n(1+2n-1)}{2}\\
&= dfrac {2n^{2}}{2}\\
&boxed {S_{n} = n^{2}}\
end{aligned}$$

(a.) We proved that the formula is correct for a few particular values of $n$. This does not allow us to assume that it is true for all values of $n$.
It is not reasonable to try to test all $n$ because there is an infinity of numbers and we cannot do this an infinite number of times.

(b.) If a property $p$ is true for one integer value $k$, it doesn’t mean it is also true for the next time $k+1$, but if we are able to prove that if $p(k)$ is true, then $p(k+1)$ is also true, then this is true for any two consecutive elements.

(c.) It is not enough to prove that $p(k)$ true implies $p(k+1)$ true.
We need to prove that for the starting point the property is also true. If we want to prove the property is true. for all the positive integers we must prove that $p(1)$ is true, which means that the property is true for $n=1$.

$a$. Two tiles.

$b$. It is true that the sum can be expressed algebraically because from the above calculation performed in above step (2) i.e. (the factorizing part) equated to ,
$$= k (k+1)+1(k+1)$$
$$=(k+1) (k+1)$$

$c$. Proving algerically,
$$1 + 3 + 5 + …. + (2n-1)= n^2$$
for all $n geq 1$

For $n = 1$.
$$(2(1)- 1)^2= (1)^2$$
$$(2-1)^2= 1$$
$$1=1rightarrow(text{True for n = 1})$$

For $n = k$.
$$begin{aligned}
1 + 3 + 5 + …. + (2(k)-1)&= k^2\
1 + 3 + 5 + …. +(2k-1)&=k^2rightarrow(1)
end{aligned}$$

For $n = k+1$.
$$begin{aligned}
1 + 3 + 5 + …. +(2k – 1)+ (2(k+1)-1)&= (k+1)^2rightarrow(2)\
end{aligned}$$
Solving left-hand side of equation (2):
$$begin{aligned}
&= 1 + 3 + 5 + …. +(2k – 1)+ (2(k+1)-1)\
&= 1 + 3 + 5 + …. +(2k – 1)+ (2k+2-1)\
&= 1 + 3 + 5 + …. +(2k – 1)+ (2k+1)\
&= k^2 + 2k + 1\
text{Factorizing,}\
&= k^2+ k+k+1\
&= k (k+1)+1(k+1)\
&= (k+1) (k+1) text{or} k^2
end{aligned}$$
Hence, the above series is true for every value of $n geq 1$.

$d$.
Conclusion: We conclude that the series is true for every value of $n geq 1$.

$e$. Student part.

$a$.
We will use the value $n =1$ in order to demonstrate the relationship is true at the first step.
Mathematical explanation is shown below:
$$n = dfrac{n(n+1)}{2}$$
for , $n = 1$
$$1 = dfrac{1(1+1)}{2}$$
$$1= dfrac{1(2)}{2}$$
$$1= dfrac{2}{2}$$
$$1=1rightarrow(TRUE, text{for} n = 1)$$

$b$.
For $n = k+1$.
$$begin{aligned}
1 + 2 + 3 + ………. n&= dfrac{n(n+1)}{2}\
text{Put , n = k+1.}\
1 + 2 + 3 + ………. k+1&= dfrac{k+1(k+1+1)}{2}\\
&boxed{1 + 2 + 3 + ………. k+1=dfrac{(k+1)(k+2)}{2}}\\

end{aligned}$$

$c$.
Since,
$$begin{aligned}
1 + 2 + 3 + ………. k&= dfrac{k(k+1)}{2}rightarrow(1)\
text{Also,}\
1 + 2 + 3 + ………. k + k+1&= dfrac{k(k+1)}{2}\\
1 + 2 + 3 + ……….k+k+1&=dfrac{(k+1)(k+2)}{2} rightarrow(2)\\
end{aligned}$$
Solving the left hand side of the above equation and from(1), we can say that
$$begin{aligned}
&= dfrac{k(k+1)}{2} +k + 1\\
&= dfrac{k^2+k+2k+2}{2}\\
&=dfrac{k^2+3k + 2}{2}\\
&= dfrac{k^2+2k + 1k + 2}{2}\\
&= dfrac{k(k+2)+1(k+2))}{2}\\
&= dfrac{(k+1)(k+2)}{2}rightarrow(text{True})\\
end{aligned}$$

$d$.
Therefore by mathematical induction we concluded that the result is true for every $n$ natural number greater than one.

(a.) Given relationship $p(n)$:

$$begin{aligned}
p(n): 2+4+6+……+ 2n = n(n+1)
end{aligned}$$

Prove $p(1)$ is true.
$$begin{aligned}
2 cdot 1 &= 1(1+1)\
2 &= 2\
end{aligned}$$
Thus, $p(1)$ is true.

Now, assume $p(k)$ is true and prove $p(k+1)$ is also true.

For $p(k)$
$$begin{aligned}
p(k): 2+4+6+…….2k &= k(k+1)\
end{aligned}$$

For $p(k+1)$
$$begin{aligned}
p(k+1): 2+4+6+…….2k+ 2(k+1) &= (k+1)(k+2)\
end{aligned}$$

Now start from $p(k)$, by adding $2(k+1)$ on each side:

$$begin{aligned}
2+4+6+…….2k &= k(k+1)\
2+4+6+…….2k + 2(k+1) &= k(k+1)+ 2(k+1)\
2+4+6+…….2k + 2(k+1) &= (k+1)+(k+2)\
end{aligned}$$

Thus, $p(k)$ is true and $p(k+1)$ is also true, which means that now both steps are accomplished and the property is proved.

(b.) (a.) Given relationship $p(n)$:

$p(n): 5+8+11+…….+ (3n+2) = dfrac{n(3n+7)}{2}$

Prove $p(1)$ is true.
$$begin{aligned}
3 cdot 1 + 2 &= dfrac{1(3 cdot 1+7)}{2}\\
3+2 &= dfrac{1(10)}{2}\\
5 &= 5\\
end{aligned}$$
Thus, $p(1)$ is true.

Now, assume $p(k)$ is true and prove $p(k+1)$ is also true.

For $p(k)$
$$begin{aligned}
p(k): 5+8+11+…….+ (3k+2) &= dfrac{k(3k+7)}{2}\
end{aligned}$$

For $p(k+1)$
$$begin{aligned}
p(k+1): 5+8+11+…….+ (3k+2) + (3(k+1)+2) &= dfrac{(k+1)(3(k+1)+7)}{2}\\
p(k+1): 5+8+11+…….+ (3k+2) + (3(k+1)+2) &= dfrac{(k+1)(3k+10)}{2}\\
end{aligned}$$

Now start from $p(k)$, by adding $(3k + 5)$ on each side:

$$begin{aligned}
p(k): 5+8+11+…….+ (3k+2) &= dfrac{k(3k+7)}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{k(3k+7k)}{2} + (3k+5)\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k^{2} + 7k + 2(3k+5)}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k^{2} + 7k + 6k+10}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k^{2} + 13k +10}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k^{2} + 3k + 10k +10}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k(k + 1) + 10(k +1)}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{3k(k + 1) + 10(k +1)}{2}\\
5+8+11+…….+ (3k+2)+(3k+5) &= dfrac{(k + 1)(3k+10)}{2}\\
end{aligned}$$

Thus, $p(k)$ is true and $p(k+1)$ is also true, which means that now both steps are accomplished and the property is proved.

(c.) (a.) Given relationship $p(n)$:

$$begin{aligned}
p(n): 1^{2}+2^{2}+3^{2}+…….+ n^{2} = dfrac{n(n+1)(2n+1)}{6}
end{aligned}$$

Prove $p(1)$ is true.
$$begin{aligned}
1^{2} &= dfrac{1(1+1)(2 cdot 1+1)}{6}\
1 &= dfrac{6}{6}\
1 &= 1\
end{aligned}$$
Thus, $p(1)$ is true.

Now, assume $p(k)$ is true and prove $p(k+1)$ is also true.

For $p(k)$
$$begin{aligned}
p(k): 1^{2}+2^{2}+3^{2}+…….+ k^{2} &= dfrac{k(k+1)(2k+1)}{6}\\
end{aligned}$$

For $p(k+1)$
$$begin{aligned}
p(k+1): 1^{2}+2^{2}+3^{2}+…….+ k^{2}+ (k+1)^{2} &= dfrac{(k+1)(k+2)(2(k+1)+1)}{6}\\
p(k+1): 1^{2}+2^{2}+3^{2}+…….+ k^{2}+ (k+1)^{2} &= dfrac{(k+1)(k+2)(2k+3)}{6}\\
end{aligned}$$

Now start from $p(k)$, by adding $(k+1)^{2}$ on each side:

$$begin{aligned}
p(k): 1^{2}+2^{2}+3^{2}+…….+ k^{2} &= dfrac{k(k+1)(2k+1)}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{k(k+1)(2k+1)}{6} + (k+1)^{2}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{k(k+1)(2k+1) + 6(k+1)^{2}}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{(k+1)(2k^{2}+k + 6k+6)}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{(k+1)(2k^{2}+7k+6)}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{(k+1)(2k^{2}+4k+3k+6)}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{(k+1)(2k(k+2)+3(k+2)}{6}\\
1^{2}+2^{2}+3^{2}+…….+ k^{2} + (k+1)^{2} &= dfrac{(k+1)(k+2)(2k+3)}{6}\\
end{aligned}$$

Thus, $p(k)$ is true and $p(k+1)$ is also true, which means that now both steps are accomplished and the property is proved.

Given relationship $p(n)$:

$$begin{aligned}
p(n):1 + 2+ 3+……+ n = dfrac {n(n+1)}{2}
end{aligned}$$

Prove $p(1)$ is true.
$$begin{aligned}
1 &= dfrac {1(1+1)}{2}\\
1 &= dfrac {2}{2}\\
1 &= 1
end{aligned}$$
Thus, $p(1)$ is true.

Now, assume $p(k)$ is true and prove $p(k+1)$ is also true.

For $p(k)$
$$begin{aligned}
p(k):1 + 2+ 3+……+ k = dfrac {k(k+1)}{2}\\
end{aligned}$$

For $p(k+1)$
$$begin{aligned}
p(k+1): 1 + 2+ 3+……+ k+ (k+1)= dfrac {(k+1)(k+2)}{2}\\
end{aligned}$$

Now start from $p(k)$, by adding $(k+1)$ on each side:

$$begin{aligned}
1 + 2+ 3+……+ k = dfrac {k(k+1)}{2}\\
1 + 2+ 3+……+ k+ (k+1)= dfrac {k(k+1)}{2}+(k+1)\\
1 + 2+ 3+……+ k+ (k+1)= dfrac {k(k+1)+2(k+1)}{2}\\
1 + 2+ 3+……+ k+ (k+1)= dfrac {(k+1)(k+2)}{2}\\
end{aligned}$$

Thus, $p(k)$ is true and $p(k+1)$ is also true, which means that now both steps are accomplished and the property is proved.

To demonstrate,
$$dfrac{3k^2 + 7k}{2}+ 3k + 5 = dfrac{3k^2+13k+10}{2}rightarrow(1)$$

Solving the left- hand side of equation (1) :
$$begin{aligned}
&=dfrac{3k^2+7k + 2(3k + 5)}{2}\\
&= dfrac{3k^2+7k+ 6k + 10}{2}\\
&= dfrac{3k^2+13k+10}{2}rightarrow(text{Equivalent to right-hand side})\\
end{aligned}$$

Given series,
$$21, 26,31, 36, 41,………..$$

Determining whether the given series is $AP$ or $GP$:
Since, the difference between first term and second term $(i.e. 26-21 = 5)$ and the difference between the second term and third term $(i.e. 31-26 = 5)$ is equal. Thus, the given series is an Arithmetic Progression.

So, we can conclude from the given series,
$$begin{aligned}
text{First term}, a&= 21\
text{Common Difference}, d&= 5\
end{aligned}$$

Determining $a_{11}$ using the formula stated below:
$$begin{aligned}
a_n&= a + (n-1)d\
a_{11}&= 21+ (11-1)5\
a_{11}&= 21+ 10(5)\
a_{11}&=21+ 50\
&boxed{a{11}=71}
end{aligned}$$

Now, in order to determine $S_{11}$ use the formula stated below:
$$begin{aligned}
S_n&= dfrac{n}{2}[a + a_{n}]\
text{Substituting the values,}\
S_{11}&= dfrac{11}{2}[21+ 71]\\
S_{11}&= 5.5 (92)\\
&boxed{S_{11}=506}
end{aligned}$$

Given series,
$$2 + 4 + 6+ ……$$

From the given arithmetic series, we can conclude that,
$$begin{aligned}
text{First term}, a &= 2\
text{Common difference}, d&= 2\
end{aligned}$$

From the given series,
$sum 2n, n= 1 text{to} 10$.

On putting $n = 10$. The sum calculates as follows:
$$2(10)= 20$$

$$20$$

To find the sum of all multiples of 7 between 0 and 500.

Multiples of 7 between 0 and 500 =

$$begin{aligned}
7,14,21,……….,497 text { i.e,}\\
S_{n} &= 0+ 7cdot1 + 7cdot 2 + 7 cdot 3 + 7 cdot 4+…………+7 cdot n\
end{aligned}$$
This forms an Arithmetic progression.

The first form, $a = 7$

Common difference, $d = 14 – 7 = 7$

$a_{n} = 497$

Now, The formula of $n^{th}$ term,

$$begin{aligned}
a_{n} &= a + (n – 1) cdot d\\
497 &= 7 + (n – 1) cdot 7\\
497 – 7 &= (n-1) cdot 7\\
(n-1) &= dfrac{490}{7}\\
n &= 70+1\\
&boxed {text {n} =71}\
end{aligned}$$

Now, formula of $n$ terms is stated below:
$$begin{aligned}
S_{n} &= dfrac{n}{2}(a+a_{n})\\
S_{71} &= dfrac{71}{2}(7 + 497)\\
S_{71} &= dfrac{71}{2}(504)\\
S_{71} &= 252cdot 71\\
&boxed {S_{71} = 17892}\\
end{aligned}$$
Hence the sum of all multiples of 7 between 0 and 500 is 17892

$$17892$$

a. At $x=0$ we obtain $y=sin{dfrac{pi}{2}}=1$ and thus $(0,1)$ has to lie on the graph. Then only graph 2 is possible.

b. At $x=0$ we obtain $y=sin{0}=0$ and thus $(0,0)$ has to lie on the graph. At $x=pi$ we note that $y=sin{2pi}=0$ and thus $(pi,0)$ also has to lie on the graph. Then only graph 4 is possible.

c. At $x=0$ we obtain $y=sin{0}=0$ and thus $(0,0)$ has to lie on the graph. At $x=pi$ we note that $y=2sin{dfrac{pi}{2}}=2(1)=2$ and thus $(pi,2)$ also has to lie on the graph.Then only graph 5 is possible.

d. At $x=0$ we note that $y=sin{0}-3=-3$ and thus $(0,-3)$ also has to lie on the graph. Then only graph 3 is possible.

e. At $x=0$ we note that $y=-sin{left(-dfrac{pi}{4}right)}=dfrac{sqrt{2}}{2}approx 0.7$ and thus $(0,0.7)$ also has to lie on the graph. Then only graph 1 is possible.

$textbf{a.)}$

$$
begin{align*}
logleft(8^{2/3}right)&=left(2/3right)log 8 tag{Using log property $log_{a}b^c=clog_{a}b$}\ \
&=dfrac{2}{3} log 8 tag{Simplifying}
end{align*}
$$

$textbf{b.)}$

$$
begin{align*}
-2log(5)&=log 5^{(-2)} tag{Using log property $clog_{a}b=log_{a}b^c$}\ \
&=log_{}left(frac{1}{25}right) tag{Simplifying}
end{align*}
$$

$textbf{c.)}$

$$
begin{align*}
log(na)^{ba}&=left(abright)log (na) tag{Using log property $log_{a}b^c=clog_{a}b$}\ \
&=(ab)[log n+log a] tag{Using log property $log ab=log a+log b$}\ \
&=ablog n+ablog a tag{Simplifying}\
end{align*}
$$

$$
logleft(8^{2/3}right)=dfrac{2}{3} log 8
$$

(a.) Given equation :
$$begin{aligned}
h &= -4.9 t^{2} + 49t +11.27\\
end{aligned}$$
The height of the platform is the height at $t = 0$.

Substitute $t = 0$, in the given equation,
$$begin{aligned}
h &= -4.9 (0)^{2} + 49 (0) +11.27\\
&boxed {h = 11.27 text {m}}\
end{aligned}$$

(b.) Write the equation in vertex form:

$$begin{aligned}
h &= -4.9 t^{2} + 49t +11.27\
end{aligned}$$
On comparing the given equation form to the standard parabolic form,

$$begin{aligned}
y &=ax^{2} + bx +c\
end{aligned}$$

Then, the axis of symmetry is:

$$begin{aligned}
x &= -dfrac {b}{2a}\\
x &= -dfrac {49}{2 (-4.9)}\\
&boxed {x =5}\
end{aligned}$$
The maximum height is then obtained at $t = 5$.

Now, put the value of &t=5$ in the given equation.

$$begin{aligned}
h &= -4.9 (5)^{2} + 49 (5) +11.27\\
h &= -4.9 cdot 25 + 245 + 11.27\
&boxed {h = 133.77 text { m}}\
end{aligned}$$

(c.) Determine the roots using the Quadratic equation.

$$begin{aligned}
t &= dfrac{- b pmsqrt{ b^{2} – 4 cdot a cdot c}}{2a}\\
t &=dfrac{- 49 pmsqrt{ (49)^{2} – 4 cdot (-4.9) cdot (11.27)}}{2cdot (-4.9)}\\
t &= dfrac{- 49 pmsqrt{2621.892}}{-9.8}\\
t &= dfrac{- 49 pm 51.20}{-9.8}\\
&boxed {t = -0.22 text{ or } 10.22}\
end{aligned}$$
Since only the positive times make sense, we know that it took $10.22$ s to hit the ground.

$(a.)$ 11.27 m
$(b.)$ 133.77 m
$(c.)$ 10.22 s

Given series,
$$9 + 14 + 19 + ….. + (4n + 5)= n (2n + 7)$$
Checking whether the series is true for $n geq 1$.

For , $n = 1$.
$$boxed{4 (1) + 5 = 1(2(1) + 7)}$$
Solving further,
$$begin{aligned}
4 + 5& = 2 + 7\
9 &= 9rightarrow(text{TRUE})\
end{aligned}$$

For any integer $n = k$, the series equivalents to,
$$9 + 14 + 19 + ….. + (4k + 5)=k (2k + 7)rightarrow(1)$$

For any integer $n = k+1$, the series equivalents to,
$$9 + 14 + 19 + ….. +(4k + 5)+(4(k+1) + 5)=k+1 (2(k+1) + 7)$$
$$9 + 14 + 19 + ….. +(4k + 5)+(4(k+1) + 5)=k+1 (2k+2 + 7)$$
$$9 + 14 + 19 + ….. +(4k + 5)+(4(k+1) + 5)=k+1 (2k+9)$$
Solving further,
$$begin{aligned}
&=9 + 14 + 19 + …..+(4k+5) + (4k+4 + 5)\
text{From (1),}\
&=k (2k+7)+ (4k+ 9)\
&= 2k^2 + 7k + 4k + 9\
&= 2k^2 + 11k+9\
text{Factorizing,}\
&= 2k^2+ 2k + 9k+9\
&= 2k(k + 1)+ 9 (k + 1)\
&= (2k+9) (k+1)rightarrow(text{TRUE})\
end{aligned}$$

$a$.
Kari proved that
$$3+6+9+…. + 3n = dfrac{3}{2}n(n +1)$$
It is true for all integer $n geq 2$.

$b$.
She verified the following result:
$3 + 6 + 9 + …….+3n = dfrac{3}{2}n (n +1)$
It is true for $n = 1$.

Given sequence:

5,-2,-9 ,- 16, …

The arithmetic sequence is stated as follows.
$$begin{aligned}
a_{n} &= a_{1} + (n-1)d\
end{aligned}$$
where,

n is the $n^{th}$ term : 50

$a_{1}$ is the first term: 5

d is the common difference between the terms i.e (-2 – 5) = – 7

(a.) Put the values in the formula.

$$begin{aligned}
a_{n} &= a_{1} + (n-1)d\
a_{50} &= 5 + (50-1)cdot (-7)\
a_{50} &= 5+49 cdot (-7)\
a_{50} &= 5 – 343\
&boxed {a_{50} = -338}\
end{aligned}$$

(b.) Now, determine the arithmetic sequence sum of the first 50 terms.

$$begin{aligned}
S_{n} &= n left(a_{1}+ dfrac {(n-1)d}{2}right)\\
end{aligned}$$

Put the value in the above formula.

$$begin{aligned}
S_{50} &= 50 left(5 + dfrac {(50-1) cdot (-7)}{2}right)\\
S_{50} &= 50 left(5 + dfrac {(49) cdot (-7)}{2}right)\\
S_{50} &= 50 left(5 + dfrac {-343}{2}right)\\
S_{50} &= 50 (-166.5)\\
&boxed {S_{50} = -8325}\\
end{aligned}$$

$$(a.) -338$$
$$(b.) -8325$$

Given series: an
1 + 4 + 7 + 10 + 13 + 16

The above-given series is an arithmetic sequence.

The arithmetic sequence is stated as follows.
$$begin{aligned}
a_{n} &= a_{1} + (n-1)d\
end{aligned}$$
where,

$a_{1}$ is the first term : 1

d is the common difference between the terms i.e (4 – 1) = 3

Put the values in the formula.

$$begin{aligned}
a_{n} &= a_{1} + (n-1)d\
a_{n} &= 1 + (n-1)cdot (3)\
a_{n} &= 1 + (n-1) cdot (3)\
a_{n} &= 1 + 3 text {n} -3\
&boxed {a_{n} = 3 text{n}-2}\
end{aligned}$$

The formula of Summation Notation is stated as follows:
$$begin{aligned}
displaystylesum_{k=1}^n a_{n}\
end{aligned}$$
Where,

k = 1 $rightarrow$ begin at smallest value
n $rightarrow$ end at greatest value
$a_{n}$ $rightarrow$ sequence formula

Put all the values in the above formula.
$$begin{aligned}
boxed {displaystylesum_{k=1}^6 (3 text{n}-2)}\
end{aligned}$$

Given:
$$begin{aligned}
tan(theta) &= dfrac{1}{2}\\
end{aligned}$$
$tan(theta)$ is positive in the third quadrant.

and

$$begin{aligned}
pi < theta <dfrac{3pi}{2}\
end{aligned}$$
means, $theta$ will be in the third quadrant.

Thus,

$$begin{aligned}
tan(theta) &= dfrac{y}{x} = dfrac{text{opp}}{text {adj}} = dfrac{1}{2}\\
end{aligned}$$
Determine the value of $sin(theta)$,

$$begin{aligned}
sin(theta) &= dfrac{text{opp}}{text {hyp}}\
end{aligned}$$

Finding the value of hypotheses, by using the Pythagorean theorem.
$$begin{aligned}
a^{2} + b^{2} &= c^{2}
end{aligned}$$
Put the values in above formula,
$$begin{aligned}
1^{2} + 2^{2} &= c^{2}\
1+ 4 &= c^{2}\
5 &= c^{2}\
&boxed {c = sqrt{5}} rightarrowtext{hyp.}\
end{aligned}$$

Now,
$$begin{aligned}
sin(theta) &= dfrac{text{opp}}{text {hyp}}\\
&boxed {sin(theta) = dfrac{-1}{sqrt{5}}}\\
end{aligned}$$

$$dfrac{-1}{sqrt{5}}$$

Given ,
$$tan theta = dfrac{1}{2}$$
Range $rightarrow pi leq theta leq dfrac{3pi}{2}$

Determining $sin theta$ :
Since we know that,
$tan theta = dfrac{text{Perperdicular}}{text{Base}}$
Thus, we conclude that
$$boxed{text{Perpendicular} rightarrow 1}$$
$$boxed{text{Base} rightarrow 2}$$

Applying Pythagorean theorem,
$$begin{aligned}
text{Hypotenuse}&= sqrt{(text{Base})^2+ (text{Perpendicular})^2}\
text{Substituting the values,}\
text{Hypotenuse}&=sqrt{(2)^2+(1)^2}\
text{Hypotenuse}&=sqrt{4+1}\
&boxed{text{Hypotenuse}=sqrt{5}}\
end{aligned}$$

Since,
$$sin theta = dfrac{text{perpendicular}}{text{Hypotenuse}}$$
Thus,
$$boxed{sintheta= dfrac{1}{sqrt{5}}}$$

$$dfrac{1}{sqrt{5}}$$

(b.) Given dimensions of the cylinder are:
$$begin{aligned}
text {h} &= 2.5 text {ft}\
text {h} &= 2.5 cdot 12\
text {h} &= 30 text { inches}\
text {Circumference} &= 2pi text {r}\
14 &= 2pi text {r}\
dfrac {14}{2 pi} &= text {r}\
dfrac {14}{2cdot 3.14} &= text {r}\
text {r} &approx 2.23 text { inches}\
end{aligned}$$

Now determines the volume of a cylinder.
$$begin{aligned}
text {V} &= pi r^{2} cdot h\
text {V} &= 3.14 cdot (2.23)^{2} cdot 30\
&boxed {text {V} = 468.68 text { cubic meter}}\
end{aligned}$$

It is given that only 75% of the cylindrical body can hold the dirt from cleaning.
Now determine the quantity of dirt.
$$begin{aligned}
75 text { Percent of volume} &= dfrac {75}{100} cdot (text {V})\\
&= 0.75 cdot (text {V})\
&= 0.75 cdot 468.68\
&boxed {355.51 text { cubic inches}}\
end{aligned}$$

$(a.)$ Prism or cylinder
$(b.)$ 355.51 Cubic inches.

a. They are both correct because
$$
y=10cdot 5^{n-1}=2cdot 5 cdot 5^{n-1}=2cdot 5^{n-1+1}=2cdot 5^n
$$

b. The advange of the equation of Katelyn is that it is easier to determine the graph, while the equation of Janelle makes it easier to determine the first and consecutive terms.

c.
$$
t(n)=18cdot 3^n=18cdot 3 cdot 3^{n-1}=54cdot 3^{n-1}
$$

a. Both are correct

b. See explanation

c. $t(n)=54cdot 3^{n-1}$

a. The first term is $$1,500,000$ and the growth factor is then $100%+5%=105%=1.05$. The equation then becomes:

$$
t(n)=1,500,000(1.05)^{n-1}
$$

b. Replace $n$ with 20:
$$
t(20)=1,500,000(1.05)^{19}=$3,790,425.29
$$

c.
$$
Sigma_{n=1}^{20} (1,500,000(1.05)^{n-1})
$$

d. The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. $t(n)=1,500,000(1.05)^{n-1}$

b. $3,790,425.29

c.$Sigma_{n=1}^{20} (1,500,000(1.05)^{n-1})$d.$a$dfrac{1-r^n}{1-r} with$a$the first term and$r$ the ratio.

We can not use addition or subtraction for solving the equations as it yield no useful result as shown below:\\
begin{tabular}{lllllllll}
& $s(6)$ & = & 2 & +10 & +50 & +250 & +1250 & +6250 \
+ & $s(6)$ & = & 6250 & +1250 & +250 & +50 & +10 & +2 \ hline
& $2s(6)$ & = & 6252 & +1260 &+ 300 & +300 & +1260 &+ 6252
end{tabular}
begin{align*}
2s(6)&= 2(6252 +1260 + 300 ) \
2s(6)&= 2(7812)\
s(6)&=7812
end{align*}
It can be observed that it doesn’t simplify the process instead makes it more complex than before\

textbf{(a)} Elimination method is used when we have sum of two or more equations in in order to extract a single equation\\
textbf{(b)} Using Luann’s method we have below result:\\
begin{align*}
s(6)&=2+10+50+250+1250+6250\
rcdot s(6)&=5(s(6))\
&=5(2+10+50+250+1250+6250)\
&=10+50+250+1250+6250+31250
intertext{Evaluating $5(s(6))-s(6)$:}
5(s(6))-s(6)&=(10+50+250+1250+6250+31250)-(2+10+50+250+1250+6250)\
4(s(6))&=10+50+250+1250+6250+31250-2-10-50-250-1250-6250\
4s(6)&=31250-2\
4s(6)&=31248\
s(6)&=textcolor{blue}{7812}
end{align*}

(b) $s(6)=7812$

Using Luann’s method we have below result:\
begin{align*}
s(n)&=2+10+50+dots+2cdot5^{n-1}\
rcdot s(n)&=5(s(n))\
&=5(2+10+50+dots+2cdot5^{n-1})\
&=10+50+250+dots+2cdot5^{n}
intertext{Evaluating $5(s(6))-s(6)$:}
5(s(n))-s(n)&=(10+50+250+dots+2cdot5^{n})-(2+10+50+dots+2cdot5^{n-1})\
4s(n)&=2cdot5^{n}-2\
4s(n)&=2(5^{n}-1)\
s(n)&=textcolor{blue}{dfrac{(5^{n}-1)}{2}}
end{align*}

$$
s(n)=dfrac{(5^{n}-1)}{2}
$$

a. The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

b. Sum of the first 37 terms:

$$
30dfrac{1-1.2^{37}}{1-1.2}=-150(1-1.2^{37})approx 127,434
$$

Sum of the first 99 terms:

$$
30dfrac{1-1.2^{99}}{1-1.2}=-150(1-1.2^{99})approx 1,035,224,667
$$

The sum of the 38th through the 99th term is then:

$$
1,035,224,667-127,434=1,035,097,233
$$

a. $adfrac{1-r^n}{1-r}$

b. $1,035,097,233$

The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

Option 1:

$$
51% times $50,000,000(1.04)^{20}=$ 55,873,640.15
$$

Option 2:

$$
$1,500,000dfrac{1-1.05^{20}}{1-1.05}approx $49,598,931
$$

Thus option 1 is better.

The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a.
$$
51% times $50,000,000=$25,500,000
$$

b.
$$
$25,500,000(1.04)^{20}=$ 55,873,640.15
$$

c. No, it is a sequence.

a. $25,500,000

b. $ 55,873,640.15

c. No

The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a.
$$
$1,500,000(1.05)^{2}=$ 1,653,750
$$

$$
$1,500,000(1.05)^{5}=$ 1,914,422
$$

b.

$$
$1,500,000dfrac{1-1.05^{n}}{1-1.05}
$$

c.
$$
$1,500,000dfrac{1-1.05^{20}}{1-1.05}approx $49,598,931
$$

a. $1,653,750; $1,914,422

b.$$1,500,000dfrac{1-1.05^{n}}{1-1.05}$

c. $49,598,931

The sum of a geometric serie with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

Option 1:

$$
51% times $50,000,000(1.04)^{20}=$ 55,873,640.15
$$

Option 2:

$$
$1,500,000dfrac{1-1.05^{20}}{1-1.05}approx $49,598,931
$$

Thus option 1 is better.

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a.
$$
t(n)=0.1(1+0.02)^{n-1}=0.1(1.02)^{n-1}
$$

b. Replace $n$ with 100:

$$
t(100)=0.1(1.02)^{99}=$0.71
$$

Replace $n$ with 200:

$$
t(200)=0.1(1.02)^{199}=$5.15
$$

c. 150 pieces:
$$
0.1dfrac{1-1.02^{150}}{1-1.02}=$92.50
$$

200 pieces:
$$
0.1dfrac{1-1.02^{200}}{1-1.02}=$257.42
$$

d.
$$
t(n)=0.1+0.01(n-1)=0.1n
$$

Replace $n$ with 100:

$$
t(100)=0.01(100)=$1
$$

Replace $n$ with 200:

$$
t(200)=0.1(200)=$2
$$

e. Replace $n$ with 150:

$$
t(150)=0.1(150)=$1.50
$$

The sum is then:

$$
dfrac{(0.1+1.5)150}{2}=$120
$$

Thus we note that Greg’s payment scheme results in more money.

f. Greg’s method gives:
$$
dfrac{(0.1+2)200}{2}=$210
$$

Thus we note that Scott’s payment scheme gives more money.

a. $t(n)=0.1(1.02)^{n-1}$

b. $$0.71$, $$5.15$

c. $$92.50$, $$257.42$

d. Greg’s payment scheme results in more money.

e. Scott’s payment scheme results in more money.

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a.
$$
t(n)=15-0.35n
$$

Replace $n$ with 2:

$$
t(2)=15-0.35(2)=14.3
$$

Replace $n$ with 7:

$$
t(7)=15-0.35(7)=12.55
$$

The last bucket was used in week 12:

$$
t(12)=15-0.35(12)=10.8
$$

b. We note that this series is arithmetic, because the common difference is 0.35 gallons.

c.
$$
dfrac{(14.65+10.8)12}{2}=152.70
$$

a. 14.3, 12.55, 10.8 gallons

b. Arithmetic

c. $152.70

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. Since every child has two parents: $1,2,4$

b. The sequence is geometric since the common ratio is 2.

c.

$$
t(n)=2^{n-1}
$$

Evaluate at $n=12$:

$$
t(12)=2^{11}=2,048
$$

Then the total number of ancestors in the first 12 generations are:

$$
1dfrac{1-2^{12}}{1-2}=4,095
$$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. Geometric since the common ratio is 2:

$$
5dfrac{1-2^{16}}{1-2}=327,675
$$

b. Arithmetic, since the common difference is 4:

$$
dfrac{(16-696)179}{2}=-60,860
$$

c. Arithmetic, since the common difference is 4:

$$
dfrac{(4+392)98}{2}=19,404
$$

d. Arithmetic, since the common difference is 11:

$$
dfrac{(11+3,432)312}{2}=537,108
$$

e. Geometric since the common ratio is $dfrac{1}{2}$:

$$
5dfrac{1-0.5^{12}}{1-0.5}=9.99755859375
$$

a. Geometric: $327,675$

b. Arithmetic: $-60,860$

c. Arithmetic, 19,404

d. Arithmetic, 537,108

e. Geometric, $9.99755859375$

We know sum of an arithmetic series can be given as:
$$S(n)=dfrac{(a+l)n}{2}$$
where $n$ is number of terms in series, $a$ is first term and $l$ is last term\\
According to given data:
begin{align*}
dfrac{(12+(-9+21n))n}{2}&=3429\
dfrac{(3+21n)n}{2}&=3429\
3n+21n^2&=6858\
3(7n^2+n-2286)&=0\
7n^2+n-2286&=0
intertext{Evaluating determinant:}
D&=(1)^2-4(7)(-2286)\
&=1+64008\
&=64009\
&=253^2
intertext{Finding $n$ using determinant method:}
n&=dfrac{-1pm sqrt{253^2}}{2(7)}\
&=dfrac{-1pm 253}{14}\
&=dfrac{252}{14} textrm{ or } dfrac{-254}{14}\
&=18 textrm{ or } -18.1428
end{align*}
As number of terms in a series is always non negative therefore textcolor{blue}{$n=18$}

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

A series is arithmetic if the terms have a common difference and geometric if the terms have a common ratio.

$$
adfrac{1-r^n}{1-r}
$$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. $3, 33, 333, 3333, 33333, 333333$

b.The sum of $n$ terms is the number with $n$ digits and each digit is a 3.

c.$Sigma_{k=1}^n(3cdot 10^{k-1})=333333333($n digits)

a. $3, 33, 333, 3333, 33333, 333333$

b.The sum of $n$ terms is the number with $n$ digits and each digit is a 3.

c.$Sigma_{k=1}^n(3cdot 10^{k-1})=333333333($n digits)

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. The sequence represents the number of graduates in each year while the series represents the total number of graduates of the school up to that year.

b.
$$
dfrac{(42+42+12(9))10}{2}=960
$$

c.
$$
dfrac{(42+42+12(n-1))n}{2}=(42+6(n-1))n=(36+6n)n=6n^2+36n
$$

a. See explanation

b. 960 students

c. $6n^2+36n$ invitations

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. The series is arithmetic since the common difference is $-7$. The number of terms is then the difference between the last and first term divided by the common difference, increased by 1:

$$
dfrac{-90-8}{-7}+1=14+1=15
$$

b.
$$
dfrac{(8-90)15}{2}=-615
$$

a. 15 terms

b. $-615$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

The series is arithmetic since the common difference is $5$. The sum is then:

$$
dfrac{(15+45)7}{2}=210
$$

As it is given.

Cold Scoop has 18 flavors, $n = 18$.

and

Amelia wants three scoops, $r = 3$.

(a.) As it is given that the order matter, then using the Permutation formula which is stated as follows:
$$begin{aligned}
^{n}P_{r} &=dfrac{n!}{(n- r)} = n(n-1)(n-2)…….(n-r+1)\
end{aligned}$$
for $n$ items chosen $r$ at a time.

$$begin{aligned}
^{18}P_{3} &=dfrac{18!}{(18-3)}\\
^{18}P_{3} &=dfrac{18!}{15!}\\
^{18}P_{3} &= dfrac{18 cdot 17 cdot 16 cdot 15…..cdot1}{15 cdot 14 cdot 13 cdot 12…..cdot1}\\
^{18}P_{3} &= 18 cdot 17 cdot 16 \\
&boxed {^{18}P_{3} = 4896}\
end{aligned}$$
$4896$ different cones can Amelia get if all other scoops are different.

(b.) As it is given that the order does not matter, then using the Combination formula which is stated as follows:
$$begin{aligned}
^{n}C_{r} &=dfrac{^{n}P_{r}}{(r)} \
end{aligned}$$

$$begin{aligned}
^{18}C_{3} &=dfrac{4896}{3!}\\
^{18}C_{3} &=dfrac{4896}{3 cdot 2 cdot 1}\\
^{18}C_{3} &=dfrac{4896}{6}\\
&boxed {^{18}C_{3} = 816}\
end{aligned}$$
$816$ ways can she order.

$$(a.) 4896$$
$$(b.) 816$$

(a.) It is given that
$$begin{aligned}
theta &= 34degree\
text{Hyp} &= 124
end{aligned}$$

In order to find the value of $x$. Using Law of Sine is stated as follows.


$$begin{aligned}
text{ sin}(theta) &= dfrac{text{opp}}{text{hyp}}\\
text{ sin}(34degree) &= dfrac {text{x}}{124}\\
text {x} &= text{ sin}(34degree) cdot 124\\
text {x} &= text{ sin}(0.559) cdot 124\\
text {x} &= 69.33\\
&boxed {text {x}approx 69.3}\\
end{aligned}$$

(b.) It is given that
$$begin{aligned}
text{A} &= 42degree\
text{a} &= text{x}\
text{b} &= 6\
text{c} &= 8\
end{aligned}$$


This is side-angle-side (SAS)

In order to find the value of $x$. Using Law of Cosine is stated as follows.

$$begin{aligned}
text {a}^{2} &= text{b}^{2} + c^{2} -2 cdottext {b} cdot text {c} cdotcos text {A}\
text {x}^{2} &= 6^{2} + 8^{2} -2 cdot 6 cdot 8cdotcos text {42degree}\
text {x}^{2} &= 100 – 96(0.743)\
text {x}^{2} &= 100-71.432\
text {x} &= sqrt{}28.658\
&boxed {text {x}approx 5.35}\\
end{aligned}$$

$$(a.) 69.3$$
$$(b.) 5.35$$

a. The graph is the graph of $y=sin(x)$ translated to the left by $dfrac{pi}{2}$ units and stretched vertically by factor 3.

b. The graph is the graph of $y=sin(x)$ stretched horizontally by factor $dfrac{1}{4}$ units and stretched vertically by factor $-2$.

According to the question, is he choose the first option whose explanation is written below.
$$begin{aligned}
text{ The amount Joe will get at the age of 55}, A&= $30,000\
text{ Rate of interest}, r&= 3%\
&=0.03\
end{aligned}$$
Also, the amount he will get next year can be expressed using the formula stated below.
$$begin{aligned}
A&=P (1 + r)\
end{aligned}$$
Thus, next year amount is given as follows.
$$begin{aligned}
&boxed{A= 30,000(1+0.03)}
end{aligned}$$

Since each year the sum of $$30,000$ is increasing at the rate of $3%$. So, this can be considered as geometric progression where the rate of the interest will be the common ratio.
The series can be defines as follows.
$30,000 + 30,000(1 + 0.03) + 30,000(1 + 0.03)^{2}+………. 30,000(1 + 0.03)^{n}$

From the above concluded series , we can say that.
$$begin{aligned}
text{First term}, a&= 30,000\
text{ Common ratio}, r&= 1.03\
text{Number of terms or years}, n&= 80-55 text{ years}\
&= 25\
end{aligned}$$

The sum is given by the formula stated below.
$$begin{aligned}
S_{n} &= dfrac{a(r^{n-1})}{r-1}\
end{aligned}$$
Since, $r > 1$.

Putting the above concluded values, the amount he will be receiving at the age of $80$ is given as follows.
$$begin{aligned}
S_{25}&= dfrac{30,000left[(1.03)^{25}-1
right]}{1.03-1}\\
&=dfrac{30,000(2.09-1)}{0.03}\\
&= dfrac{30,000(1.09)}{0.03}\\
&= dfrac{32,700}{0.03}\\
&=1,090,000\
end{aligned}$$
The amount Joe will get at the age of $80$ is $$1,090,000$ at the rate of $3%$ with the starting pension at $%30,000$ at the age of $55$.

According to the question, is he choose the second option whose explanation is written below.
$$begin{aligned}
text{ The amount Joe will get at the age of 65}, A&= $60,000\
text{ Rate of interest}, r&= 3%\
&=0.03\
end{aligned}$$
Also, the amount he will get next year can be expressed using the formula stated below.
$$begin{aligned}
A&=P (1 + r)\
end{aligned}$$
Thus, next year amount is given as follows.
$$begin{aligned}
&boxed{A= 60,000(1+0.03)}
end{aligned}$$

Since each year the sum of $$60,000$ is increasing at the rate of $3%$. So, this can be considered as geometric progression where the rate of the interest will be the common ratio.
The series can be defines as follows.
$60,000 + 60,000(1 + 0.03) + 60,000(1 + 0.03)^{2}+………. 60,000(1 + 0.03)^{n}$

From the above concluded series , we can say that.
$$begin{aligned}
text{First term}, a&= 60,000\
text{ Common ratio}, r&= 1.03\
text{Number of terms or years}, n&= 80-65 text{ years}\
&= 15\
end{aligned}$$

The sum is given by the formula stated below.
$$begin{aligned}
S_{n} &= dfrac{a(r^{n-1})}{r-1}\
end{aligned}$$
Since, $r > 1$.

Putting the above concluded values, the amount he will be receiving at the age of $80$ is given as follows.
$$begin{aligned}
S_{15}&= dfrac{60,000left[(1.03)^{25}-1
right]}{1.03-1}\\
&=dfrac{60,000(2.09-1)}{0.03}\\
&= dfrac{60,000(1.09)}{0.03}\\
&= dfrac{65,400}{0.03}\\
&=2,180,000\
end{aligned}$$
The amount Joe will get at the age of $80$ is $$2,180,000$ at the rate of $3%$ with the starting pension at $%60,000$ at the age of $65$.

So, according to the calculations performed above the second plan will pay out Joe the largest amount by age $80$ i.e. $$2,180,000$.

Let $x$ be the amount invested at the rate of $8%$, and
$y$ be the amount invested at $6%$.

According to the question, Ms. Fernandez invests $$50,000$ of her lottery winning into two different mutual funds. Mathematically it can be expressed as follows.
$$begin{aligned}
x+y&= 50,000\
&boxed{x= 50,000 – y}\
end{aligned}$$

According to the second condition, at the end of one year, both funds together are worth $$53,550$. This can be mathematically expresses as follows.
Let us assume, if $$1$ invested today.
So, after one year at the rate of the interest of $8%$ the amount of $$1$ will become,
$$begin{aligned}
text{ Amount}, A&= 1 times (1 + 0.08)\
&=$1.08\
end{aligned}$$
Similarly, the amount after one year at the rate of $6.5%$ is given as follows.
$$begin{aligned}
text{ Amount}, A&= 1 times (1 + 0.065)\
&=$1.065\
end{aligned} $$

Thus, the second condition can now be mathematically expressed as,
$$begin{aligned}
&boxed{1.08x + 1.065y = 53,550}\
end{aligned}$$

On putting, $x = 50,000 – y$ in second condition we get.
$$begin{aligned}
1.08(50,000-y) + 1.065y&= 53,550\
54,000- 1.08y + 1.065y &= 53,550\
-0.015y&= 53,550-54,000\
-0.015y&= -450\
y&= dfrac{-450}{-0.015}\
y&=30,000\
end{aligned}$$

Putting $y = 30,000$ in the first equation i.e. $x = 50,000 – y$, we get.
$$begin{aligned}
x&= 50,000-30,000\
x&= 20,000\
end{aligned}$$

Thus, the amount she get at $8%$ is $$20,000$. And the amount she get at $6.5%$ is $$30,000$.

$$20,000$ ; $$30,000$

From the given figure,
Concluding the given things,
$$begin{aligned}
QR&= 7 text{cm}\
PQ&= 8 text{cm}\
angle PQR&= 70^o\
angle RSP&= 110^o\
angle SRP &= 35^o\
angle SPR&= 35^o\
end{aligned}$$

In order to determine $PR$, we will use law of cosines in $triangle PQR$ :
$$begin{aligned}
PR^2&= PQ^2 + QR^2 – 2cdot PQ cdot QR cdot cosQ\
text{Substituting the values,}\
PR^2&=8^2 + 7^2 – 2 cdot 8 cdot 7 cdot cos 70^o\
PR^2&= 64 + 49 – 112 cdot 0.34202014\
PR^2&=74. 693744\
text{Taking square root.}
PR&= pm sqrt{74. 693744}\
&boxed{PRapprox 8.64}\
end{aligned}$$

$b$. Determining perimeter of quadrilateral $PQRS$:

In $triangle PRS$,
$$angle SRP = angle SPR$$
We know that sides opposite to equal angles are also equal.
Thus,
$$SR = PSrightarrow(1)$$
From condition (1), $triangle PRS$ is an isosceles triangle.

Determining $SR$ using the law of sines:
$$dfrac{SR}{ sin 35^o}= dfrac{PR}{ sin 110^o}$$
Substituting the values,
$$begin{aligned}
SR&= dfrac{PR sin 35^o}{ sin 110^o}\\
SR&= dfrac{8.64 cdot 0.57357644}{0.93969262}\\
&boxed{SRapprox 5.27}\\
end{aligned}$$

Thus, perimeter of quadrilateral is given as follows:
$$begin{aligned}
text{Perimeter}&=PQ + QR + 2SR\
text{Substituting the values,}\
&= 8 + 7 + 2(5.27)\
&= 25.54 text{units}\
end{aligned}$$

It is given that,
$$begin{aligned}
cosleft(dfrac{pi}{3}right)= dfrac{1}{2}
end{aligned}$$
In order to find the three other true statements using the given function an the angles that will be multiple of $dfrac{pi}{3}$, we will set some values for eg. $dfrac{2pi}{3}$, $dfrac{4pi}{3}$, and $dfrac{5pi}{3}$ which are the multiple of $dfrac{pi}{3}$.

Evaluating $cosleft(dfrac{2pi}{3}right)$,
$$begin{aligned}
cosleft(dfrac{2pi}{3}right) &= -cosleft(pi-dfrac{2pi}{3}right)\
end{aligned}$$
Since, $cos(x)$ is negative in second quadrant.
Thus,
$$begin{aligned}
cosleft(dfrac{2pi}{3}right) &= -cosleft(dfrac{3pi-2pi}{3}right)\\
&=-cosleft(dfrac{pi}{3}right)\\
&= -dfrac{1}{2}\\
end{aligned}$$

Evaluating $cosleft(dfrac{4pi}{3}right)$,
$$begin{aligned}
cosleft(dfrac{4pi}{3}right) &= -cosleft(dfrac{4pi}{3}-piright)\
end{aligned}$$
Since, $cos(x)$ is negative in third quadrant.
Thus,
$$begin{aligned}
cosleft(dfrac{4pi}{3}right) &= -cosleft(dfrac{4pi-3pi}{3}right)\\
&=-cosleft(dfrac{pi}{3}right)\\
&= -dfrac{1}{2}\\
end{aligned}$$

Evaluating $cosleft(dfrac{5pi}{3}right)$,
$$begin{aligned}
cosleft(dfrac{5pi}{3}right) &= cosleft(2pi-dfrac{5pi}{3}right)\
end{aligned}$$
Since, $cos(x)$ is positive in fourth quadrant.
Thus,
$$begin{aligned}
cosleft(dfrac{5pi}{3}right) &= cosleft(dfrac{6pi-5pi}{3}right)\\
&=cosleft(dfrac{pi}{3}right)\\
&= dfrac{1}{2}\\
end{aligned}$$

$a.$ $ax + bx$

In order to simplify further we will take $x$ common.

$$begin{aligned}
&=x(a+b)\
end{aligned}$$

$a.$ $ax + bx$
In order to simplify further we will take $x$ common.
$$begin{aligned}
&=x(a+b)\
end{aligned}$$

$c$ $dfrac{ax+a}{x^{2}+2x+1}$

using $(a+b)^{2}$ = $a^{2} + 2ab + b^{2}$

$$begin{aligned}
&=dfrac{a(x+1)}{(x+1)^{2}}\\
&= dfrac{a}{x+1}\\
end{aligned}$$

$d$ $dfrac{x^{2}-b^{2}}{ax+ab}$

using , $a^{2} – b^{2}$ = $(a+b) (a-b)$

$$begin{aligned}
&=dfrac{(x+b)(x-b)}{a(x+b)}\\
&= dfrac{x-b}{a}\\
end{aligned}$$

According to the question , the pattern determined by Julian is stated as follows.
$$begin{aligned}
1 + 3& = 4 &= 2^2\
4 + 5&= 9&= 3^2\
9 + 7&= 25&= 5^2\
end{aligned}$$

In order to determine the next pattern he will perform the steps stated below.
$$25 + 11 = 36 = 6^2$$

Thus, from the above discussion the formula become as follows:
$$1 + 3 + 5 + 7 + ……+ (2n-1)= n^2$$
where , n is a real number.

$a$.
Given ,
$$begin{aligned}
x^{tfrac{-4}{3}} + 1 &= 17rightarrow(1)
end{aligned}$$

In order to evaluate x^{tfrac{-1}{3}}, we will first subtract one from both sides. Mathematically it is expressed as follows:
$$begin{aligned}
x^{tfrac{-4}{3}} + 1 -1&= 17-1\
x^{tfrac{-4}{3}}&= 16rightarrow(2)\
end{aligned}$$

Now, in equation (2), raise both sides to the exponents $dfrac{1}{4}$.
$$begin{aligned}
(x^{tfrac{-4}{3}})^{tfrac{1}{4}}&= (16)^{tfrac{1}{4}}\
x^{tfrac{-1}{3}}&= 2^{4(tfrac{1}{4})}\
&boxed{x^{tfrac{-1}{3}}= 2}
end{aligned}$$

$b$.
Given,
$$log A = a$$

In order to evaluate $dfrac{log A}{100}$, we will use $log$ property,
$$begin{aligned}
log (dfrac{m}{n})&= log m – log n\
end{aligned}$$
Thus,
$$begin{aligned}
log (dfrac{A}{100})&= log A – log 100\
&= log A – log 10^{2}\
&= log A – 2\
&boxed{log (dfrac{A}{100})= a – 2}\
end{aligned}$$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

a. 20 meters, because it travels 10 meters up and 10 meters down.

b.
$$
60% times 10 = 6
$$

Thus the ball will have traveled $20m+6m+6m=32m$

c.
$$
t(n)=20cdot 0.6^{n-1}
$$

d. The sum of the first 15 terms is then:

$$
20dfrac{1-0.6^{15}}{1-0.6}approx 50
$$

a. 20m

b. 32m

c. $t(n)=20cdot 0.6^{n-1}$

d. $50$

No, because the increments infinitely small and thus the increments will approximate a real number.

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

The sum of the first $n$ terms is then:

$$
20dfrac{1-0.6^{n}}{1-0.6}
$$

As $nrightarrow infty$, $0.6^nrightarrow 0$ and thus as $n$ becomes larger and larger, we then obtain that:

$$
20dfrac{1-0.6^{n}}{1-0.6}rightarrow 20dfrac{1}{1-0.6}=50
$$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

The sum of the first $n$ terms is then:

$$
20dfrac{1-0.6^{25}}{1-0.6}approx 50
$$

$$
20dfrac{1-0.6^{35}}{1-0.6}approx 50
$$

$$
20dfrac{1-0.6^{50}}{1-0.6}approx 50
$$

$$
20dfrac{1-0.6^{100}}{1-0.6}approx 50
$$

As $nrightarrow infty$, $0.6^nrightarrow 0$ and thus as $n$ becomes larger and larger, we then obtain that:

$$
20dfrac{1-0.6^{n}}{1-0.6}rightarrow 20dfrac{1}{1-0.6}=50
$$

$$
50
$$

To solve the following problems we must know the formula to calculating the sum of first $n$ numbers in the sequence:

$$
S_n = A_1cdotfrac{r^n-1}{r-1}
$$

$S_n$ – the sum of first $n$ numbers in the sequence

$A_1$ – first number in the sequence

$r$ – change rate

$n$ – index of numbers in sequence

a) Firstly we need to define our parameters of the formula from the task.\\
begin{tabular}{ >{centeringarraybackslash}m{2in} |>{centeringarraybackslash}m{2in} }
$A_1 = 200$ & $A_1 = 200$\
$r = frac{1}{2} = 0.5$ & $r = frac{1}{2} = 0.5$\
$n = 10$ & $n = 30$\
end{tabular}\\
Now we input the known variables into the formula.\\
begin{tabular}{ >{centeringarraybackslash}m{2in} |>{centeringarraybackslash}m{2in} }
$S_{10} = 200 cdot frac{0.5^{10}-1}{0.5-1}$ & $S_{30} = 200 cdot frac{0.5^{30}-1}{0.5-1}$\\
$S_{10} = 399frac{39}{64}$ & $S_{30} approx 400$\
end{tabular}

a) Firstly we need to define our parameters of the formula from the task.\\
begin{tabular}{ >{centeringarraybackslash}m{2in} |>{centeringarraybackslash}m{2in} }
$A_1 = 16$ & $A_1 = 16$\
$r = frac{3}{2} = 1.5$ & $r = frac{3}{2} = 1.5$\
$n = 10$ & $n = 30$\
end{tabular}\\
Now we input the known variables into the formula.\\
begin{tabular}{ >{centeringarraybackslash}m{2in} |>{centeringarraybackslash}m{2in} }
$S_{10} = 16 cdot frac{1.5^{10}-1}{1.5-1}$ & $S_{30} = 16 cdot frac{1.5^{30}-1}{1.5-1}$\\
$S_{10} = 1,813frac{9}{32}$ & $S_{30} approx 6,136,001$\
end{tabular}

The sequence in a) is decreasing so the difference between the two sums is small compared to the difference in b) where the sequence is increasing. To know what kind of sequence we are dealing with we just take a look at the change rate.\
center$1 1 – $ the sequence is increasing\
$-1 1 -$ the sequence is decreasing\

a) $S_{10} = 399frac{39}{64}$ , $S_{30} approx 400$ b) $S_{10} = 1,813frac{9}{32}$ , $S_{30} approx 6,136,001$

The sum of an arithmetic series is the sum of the first and last term multiplied by the number of terms divided by 2. The sum of a geometric series with $n$ terms is:

$$
adfrac{1-r^n}{1-r}
$$

with $a$ the first term and $r$ the ratio.

If $r>1$, then the sequence will be increasing and the sum will keep getting larger and larger. If $r<1$, then as $n$ becomes very large: $r^napprox 0$ and thus

$$
adfrac{1-r^n}{1-r}approx adfrac{1}{1-r}
$$

Thus the sum will be approximately $adfrac{1}{1-r}$.

As $nrightarrow infty$, $0.6^nrightarrow 0$ and thus as $n$ becomes larger and larger, we then obtain that:

$$
20dfrac{1-0.6^{n}}{1-0.6}rightarrow 20dfrac{1}{1-0.6}=50
$$

$adfrac{1}{1-r}$, 50

Let $r$ be the common ratio of the sequence. If $r>1$, then the sequence will be increasing and the sum will keep getting larger and larger. If $r<1$, then as $n$ becomes very large: $r^napprox 0$ and thus

$$
adfrac{1-r^n}{1-r}approx adfrac{1}{1-r}
$$

Thus the sum will be approximately $adfrac{1}{1-r}$.

$$
adfrac{1}{1-r}
$$

a.
$$
S(n)=dfrac{rar^{n-1}-ar^{1-1}}{r-1}=dfrac{ar^n-a}{r-1}
$$

b. Multiply numerator and denominator by $-1$:

$$
S(n)=dfrac{a-ar^n}{1-r}
$$

Factor out $a$ in the numerator:

$$
S(n)=dfrac{a(1-r^n)}{1-r}
$$

c. As $n$ becomes larger and larger: $r^n$ will become closer to zero since $0.6<1$.

d. Since $r^napprox 0$ for very large values of $n$ (if $r<1$), then:

$$
S(n)approxdfrac{a(1-0)}{1-r}=dfrac{a}{1-r}=S
$$

e.
$$
S=dfrac{20}{1-0.6}=dfrac{20}{0.4}=50
$$

a.
$$
Sigma_{n=1}^{infty} left(dfrac{1}{2}right)^n
$$

b. The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{0.5}{1-0.5}=1
$$

c. Zeno is correct, because you obtain 1 (which means that you crossed the whole way) only in the limit.

a. $Sigma_{n=1}^{infty} left(dfrac{1}{2}right)^n$

b. 1

c. Zeno is correct

a.
$$
Sigma_{n=1}^{infty} 0.63left(0.01right)^{n-1}
$$

b. The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{0.63}{1-0.01}=dfrac{0.63}{0.99}=dfrac{63}{99}=dfrac{7}{9}
$$

a. $Sigma_{n=1}^{infty} 0.63left(0.01right)^{n-1}$

b. $dfrac{7}{9}$

Rewrite the number as an infinite series:

$$
Sigma_{n=1}^{infty} 0.143left(0.001right)^{n-1}
$$

The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{0.143}{1-0.01}=dfrac{0.143}{0.999}=dfrac{143}{999}
$$

$$
dfrac{143}{999}
$$

Rewrite the number as an infinite series:

$$
Sigma_{n=1}^{infty} 0.9left(0.1right)^{n-1}
$$

The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{0.9}{1-0.1}=dfrac{0.9}{0.9}=1
$$

$$
1
$$

The sum of an infinite geometric series is (if $r<1$):

$$
S=dfrac{a}{1-r}
$$

with $a$ the first term of the sequence and $r$ the common ratio.

$$
S=dfrac{a}{1-r}
$$

It is given that the balls are numbered from $1$ to $49$. Thus, it explains.
$$begin{aligned}
text{ Favorable cases of choosing six balls} & = {}^{49}C_6\
text{ Total number of cases} & = {}^{49}C_{49} \
end{aligned}$$

Let $E_{1}$ be an event of choosing six balls. Therefore the probability of choosing six balls can be mathematically expressed as follows.
$$begin{aligned}
P(E_{1}) & = dfrac{text{ Favorable Cases}}{text{ Total number of cases}}\\
&= dfrac{{}^{49}C_6}{{}^{49}C_{49}}\\
&= dfrac{dfrac{49!}{(43!)(6!)}}{dfrac{49!}{(49!)(0!)(49!)}} \\
&= dfrac{49times 48 times times 47times 46times 45 times 44 times 43!}{(43!)(6times 5 times 4 times 3 times 2 times 1)}\\
&= 13, 983, 816\
end{aligned}$$

$13,983,816$

a. The graph is the graph of $y=sin(x)$ translated to the right by $pi$ units and stretched horizontally by factor 2.

b. The graph is the graph of $y=sin(x)$ stretched horizontally by factor $dfrac{1}{3}$ and stretched vertically by factor $10$ and translated down by 2 units.

c. The graph is the graph of $y=cos(x)$ translated to the left by $dfrac{pi}{4}$ units and stretched vertically by factor 5.

d. The graph is the graph of $y=cos(x)$ stretched horizontally by factor $dfrac{1}{2}$ and translated right by $dfrac{pi}{4}$ units.

Given patterns,
Pattern $1rightarrow$ $dfrac{x-1}{x-1}= 1$.

Pattern $2rightarrow$ $dfrac{x^2-1}{x-1}= x+1$.

Pattern $3rightarrow$ $dfrac{x^3-1}{x-1}= x^2 + x+ 1$.

Pattern $4rightarrow$ $dfrac{x^4-1}{x-1}= x^3 + x^2 + x+ 1$.

Checking the pattern 1.
$$begin{aligned}
x-1&= 1(x-1)\
x-1&= x-1rightarrow(text{Correct})
end{aligned}$$

Checking the pattern 2.
$$begin{aligned}
x^2-1&= (x-1)(x+1)\
x^2-1&= x(x+1)-1(x+1)\
x^2-1&=x^2+x-x-1\
x^2-1&= x^2-1 rightarrow(text{Correct})
end{aligned}$$

Checking the pattern 3.
$$begin{aligned}
x^3-1&= (x-1)(x^2+x+1)\
x^3-1&= x(x^2+x+1)-1(x^2+x+1)\
x^3-1&= x^3 + x^2+ x – x^2-x-1\
x^3-1&=x^3 – 1rightarrow{text{(Correct)}}\
end{aligned}$$

Checking the pattern 4.
$$begin{aligned}
x^4-1&= (x-1)(x^3+x^2+x+1)\
x^4-1&= x(x^3+x^2+x+1)- 1(x^3+x^2+x+1)\
x^4-1&= x^4 + x^3 +x^2 + x-x^3-x^2-x-1\
x^4-1&=x^4 – 1rightarrow{text{(Correct)}}\
end{aligned}$$

According to the justification, determining the value of $dfrac{x^5-1}{x-1}$ :
$$begin{aligned}
&boxed{dfrac{x^5-1}{x-1}= x-1(x^4 + x^3 +x^2+x+1)}\
end{aligned}$$

Determining the $n^{th}$ expression of the result for $dfrac{x^n-1}{x-1}$:
$$begin{aligned}
&boxed{dfrac{x^n-1}{x-1}=(x^{n-1}+x^{n-2}+………+x^3+x^2+x+1)}
end{aligned}$$
Thus, the above equation is the conjecture.

Given function,
$$y = x^3(x-2)(x+2)^2$$

Determining the values of $x$- intercept by taking $y = 0$:
$$begin{aligned}
x^3&=0\
&boxed{x=0}rightarrow(1)\
x-2&=0\
&boxed{x=2}rightarrow(2)\
x+2&=0\
&boxed{x=-2}rightarrow(3)\
end{aligned}$$

Determining the value of $y$- intercept by taking $x = 0$:
$$begin{aligned}
y&= (0)^3 (0-2)(0+2)^2\
&boxed{y= 0}\
end{aligned}$$

The function has degree 6. Since the degree is even, both ends go in the same direction.
Because the leading term is positive, both ends go upwards.

$f(x)$ $rightarrow$ $infty$ $text{as}$ $x$ $rightarrow -infty$.
$f(x)$ $rightarrow$ $infty$ $text{as}$ $x$ $rightarrow infty$.

$OB=OC=OA=26$

$MN=48$

$OM=ON=dfrac{22}{2}=11$

$$
YN=ZM=24
$$

$x^2+y^2=26^2$

$$
x^2+y^2=676
$$

We have to check if the $YNleq PN$.

The circle has the equation:

$y^2=676-11^2$

$y^2=676-121$

$y^2=555$

$y=sqrt{555}$

$$
y=PNapprox 23.56
$$

We compute $PN$:

Because $PN<YN$, the house will not be able to fit through the tunnel.

Solution to this example is given below

$$
begin{align*}
15% x+22%($2700-x)&=$513.01&&boxed{text{Given proportion}}\
end{align*}
$$

When $x=$ the amount earned during lunch
and $boldsymbol{(2700-x)}$ the amount earned during dinner.

$$
begin{align*}
0.15x+0.22($2700-x)&=$513.01&&boxed{text{Convert the percent to a decimal form}}\
0.15x+594-0.22x&=513.01&&boxed{text{Distributive property}}\
-0.07x+594&=513.01&&boxed{text{Combine like terms}}\
-7x+59400&=51301&&boxed{text{Multiply both sides by 100}}\
-7x+59400-59400&=51301-59400&&boxed{text{Subtract }59400 text{ from both sides}}\
-7x&=-8099&&boxed{text{Simplify}}\
frac{-7x}{-7}&=frac{-8099}{-7}&&boxed{text{Divide both sides by }-7}\
x&=$1157&&boxed{text{Simplify}}\
end{align*}
$$

Betty made $boldsymbol{$1157}$ during lunch.

$$
begin{align*}
&boxed{{color{#c34632}x=$1157} }&&boxed{text{Final solution}}\
end{align*}
$$

$$
color{#4257b2} text{ }x=$1157
$$

Given quadratic function,
$$begin{aligned}
y &= x^{2} – x + 12tag{1} \
end{aligned}$$
In order to determine the intersecting points of the functions given in the table , we will sketch the graph of the quadratic function and we will plot the values given in the table.

From the above-sketched graph we can say that the intersecting points are $(-5, 42)$ and $(1, 12)$.

$(-5, 42)$ ; $(1, 12)$

Rewrite the total distance as an infinite series:

$$
Sigma_{n=1}^{infty} 25left(1-0.05right)^{n-1}=Sigma_{n=1}^{infty} 25left(0.95right)^{n-1}
$$

The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{25}{1-0.95}=dfrac{25}{0.05}=500
$$

$500$ miles

Rewrite the number as an infinite series:

$$
Sigma_{n=1}^{infty} 0.27left(0.01right)^{n-1}
$$

The sum of an infinite geometric series is:

$$
S=dfrac{a}{1-r}=dfrac{0.27}{1-0.01}=dfrac{0.27}{0.99}=dfrac{27}{99}=dfrac{3}{11}
$$

$$
dfrac{3}{11}
$$

Let an infinite series of geometric progression with common ratio $r$ as given below,

$a, ar , ar^2, ar^3, ar^4 ………. ar^{n-1}$.
where,
$a rightarrow$ First term of the geometric progression.
$ar rightarrow$ Second term of the geometric progression, and so.

Using,
$$begin{aligned}
a_n&= acdot r^{n-1}
end{aligned}$$

where, $n= 1,2,3,4,……..$

The sum of the geometric progression is given as follows:
$$begin{aligned}
S_n&= dfrac{a(1- r^n)}{1 – r}\\
text{According to the given series,}\
S_n&= dfrac{a+ ar +ar^2+ ar^3+ar^4 ……….+ ar^{n-1}}{1-r}\\
&= dfrac{a(1 + r + r^2 + r^3 + r^4+ ………… r^{n-1})}{1-r}\\
&=a cdot dfrac{1- r^n}{1-r}rightarrow(2)\
&end{aligned}$$

From the expression of sum of geometric progression concluded in equation (2), we can say that the value of $S_n$ is convergent if and only if $q^n rightarrow 0$. It is why the sum of an infinite geometric series contain the
conditions ,
$$0 < |r| < 1$$

Rewrite the sum as a series

$$
Sigma_{n=1}^{5} 3left(dfrac{1}{3}right)^{n-1}
$$

The sum of an finite geometric series is:

$$
S(n)=dfrac{a(1-r^n)}{1-r}=dfrac{3left( 1-left( dfrac{1}{3}right)^5right)}{1-dfrac{1}{3}}=dfrac{121}{27}
$$

$$
dfrac{121}{27}
$$

From the given figure the value of $x$ can be evaluated by using Law of Cosines which is mathematically stated below.
$$begin{aligned}
a^{2} &= b^{2} + c^{2} – 2bccdot cos A\
end{aligned}

Now, according to the given figure apply the law stated above, the value of $x$ will be calculated as follows.
$$begin{aligned}
x^{2}&= (6)^{2} + (4)^{2} – 2times 6 times 4 cos 130^{o} \
x^{2}&= 36 + 16 – 48 times -0.64\
x&= sqrt{82.72}\
x&= 9.10 text{ units} \
end{aligned}$$

Also, the area of the triangle will be given by using Law of sines. Mathematically which is stated as follows.
$$begin{aligned}
text{ Area}&= dfrac{1}{2}times btimes c times sin A\
end{aligned}$$

Thus, applying the law stated above in the given figure . The area is calculated as follows.
$$begin{aligned}
text{ Area}&= dfrac{1}{2}times (6)times (4 ) sin (130^{o} \
&= 0.5 times 24 times 0.76\
&= 9.12 text{ square units}\
end{aligned}$$

$a.$ $9.10 text{ units}$
$b.$ $9.12 text{ square units}$

It is given that , \
begin{align*}
sinbeta&= -dfrac{4}{5}tag{1} \
intertext{ $pi$ $leq$ $beta$$leq$ $dfrac{3pi}{2}$} \\
end{align*}

Since, we know that.\
begin{align*}
sin^{2}(x) + cos^{2}(x) &= 1 \
intertext{ The above formula can be used to determine the value of the $cos$ as well.}\
intertext{ Mathematically it is stated as follows.} \
cos^{2}(x)&= 1 – sin^{2}(x)\
cos(x)&=pm sqrt{1- sin^{2}(x)}\
end{align*}

Thus,

$$
begin{align*}
cos(beta)&= pm sqrt{1- left(-dfrac{4}{5}right)^{2}} \\
&= pm sqrt{1- dfrac{16}{25}}\\
&= pm sqrt{dfrac{25 – 16}{25}}\\
&= pm sqrt{dfrac{9}{25}}\\
&= pm dfrac{3}{5}\\
end{align*}
$$

But , $cos(x)$ in third quadrant is only negative. Thus, the value of $cos(x)$ will be given as follows.

$$
begin{align*}
cos(beta)&= -dfrac{3}{5}tag{2}\
end{align*}
$$

Also,

$$
begin{align*}
tan(x)&=dfrac{sin(x)}{cos(x)}\\
end{align*}
$$

Thus,

$$
begin{align*}
tan(beta)&= dfrac{dfrac{-4}{5}}{dfrac{-3}{5}} \\
tan(beta)&= dfrac{4}{3}\\
end{align*}
$$

At last, we will use.

$$
begin{align*}
cot(x)&= dfrac{1}{tan(x)}\
end{align*}
$$

So,

$$
begin{align*}
cot(beta)&= dfrac{1}{dfrac{4}{3}}\\
cot(beta)&= dfrac{3}{4}\\
end{align*}
$$

$a$.
In order to determine which region appear the disease is spreading faster, we will find the range of the two regions.

Determining the range of region 1:
$$begin{aligned}
text{ Range}, r_1&= dfrac{text{Highest frequency – Lowest frequency}}{text{Highest frequency – Lowest frequency}}\\
&= dfrac{170-120}{21-7}\\
&=dfrac{50}{14}\\
&approx 3.57\

end{aligned}$$

Determining the range of region 2:
$$begin{aligned}
text{ Range}, r_2&= dfrac{text{Highest frequency – Lowest frequency}}{text{Highest frequency – Lowest frequency}}\\
&= dfrac{62-22}{19-7}\\
&=dfrac{40}{12}\\
&approx 3.33\
end{aligned}$$

Since, $r_1 > r_2$. The disease is spreading faster in region 1.

$b$.
A linear model can be a good choice to represent the outbreak data.

$c$.
Since the average rate of change is higher in the first region and after June 19th there are already more infected people in Region A, it is more useful to set up the hospital in Region A.

$d$.
Determining the regression model for each region:
$$begin{aligned}
text{region A}, y&= 3.62x + 92.98\
text{region B}, y&= 3.33x – 3.93\
end{aligned}$$
For, $x = 20$.
$$begin{aligned}
text{region A}, y&= 3.62(20) + 92.98\
&boxed{yapprox135}rightarrow(text{Region A})\
text{region B}, y&= 3.33x – 3.93\
&boxed{yapprox63}rightarrow(text{Region B})\

end{aligned}$$

$e$.
Very confident.

According to the diagram, there are $6$ ways that Carrie can earn an ice cream $(2H)$.

$4$ is the number of total attempts she has and $2$ represents the $2$ hits.

Different possible outcomes are not equally likely. For example, there is only one way she can get $4$ hits and there are $6$ ways of her getting $2$ hits.

The probability of her getting an ice cream is $dfrac{6}{16}=dfrac{3}{8}$ because there is $6$ ways out of $16$ of her getting $2$ hits.

Her trial is trying to hit and her success is hitting the ball. She had $4$ trials in the game because she tried to hit $4$ times.

Since $65%$ people is intolerant, we can let “success” (being intolerant) have a probability of $s=0.65$ and “failure” (not intolerant) havea probability of $f=1-0.65=0.35$.

For a binomial distribution, the probability of exactly $x$ successes out of $n$ trials is:
$$P(x)=_nC_xs^xf^{n-x},$$
where $_nC_x=dfrac{n!}{x!(n-x)!}$ and $n!=1cdot2cdotdots n$.

There are $n=5$ adults and at least $3$ need to be intolerant. Therefore, we need to find $P(3)+P(4)+P(5)$.

Let us find $P(3)$:
$$begin{align*}
P(3)&=_5C_30.65^3cdot0.35^{5-3}\
&=dfrac{5!}{3!(5-3)!}cdot0.275cdot0.1225\
end{align*}$$

Let us rewrite factorials and multiply:
$$begin{align*}
&=dfrac{1cdot2cdot3cdot4cdot5}{1cdot2cdot3cdot1cdot2}cdot0.034\
&=10cdot0.034\
&=0.34
end{align*}

Let us find $P(4)$:
$$begin{align*}
P(4)&=_5C_40.65^4cdot0.35^{5-4}\
&=dfrac{5!}{4!(5-4)!}cdot0.18cdot0.35\
end{align*}$$

Let us rewrite factorials and multiply:
$$begin{align*}
&=dfrac{1cdot2cdot3cdot4cdot5}{1cdot2cdot3cdot4cdot1}cdot0.06\
&=5cdot0.063\
&=0.315
end{align*}

Let us find $P(5)$:
$$begin{align*}
P(5)&=_5C_50.65^5cdot0.35^{5-5}\
&=dfrac{5!}{5!(5-5)!}cdot0.12cdot1\
end{align*}$$

Let us rewrite factorials and multiply:
$$begin{align*}
&=dfrac{1cdot2cdot3cdot4cdot5}{1cdot2cdot3cdot4cdot5cdot1}cdot0.12\
&=0.12\
end{align*}

Therefore, the final probability is $0.34+0.315+0.12=0.775$.

$0.775$ or $77.5%$

If half of the outcomes have an equal number of heads and tails ($2H$ and $2T$), then the game is fair. Since there are $16$ outcomes and only $6$ of the have $2$ heads and $2$ tails, the game is not fair.