#### (a)

You can align one side of tracking paper with a given line.

Because, tracing paper is square, and the square has parallel lines,

you can draw a line aligned with the opposite side.

#### (b)

You can find a midpoint of the given line and align that point with

a corner of tracing paper.

Because tracing paper is square and the square has right angles,

the conclusion is that the line through midpoint aligned with one side

of tracing paper is actually perpendicular bisector.

#### (c)

On the following picture, there is formed a triangle $ABC$.

We can conclude that this triangle with, equal two sides,

$overline{AC}$ and $overline{AB}$.

#### (d)

First, you need to use a compass and graph semicircle from point $A$ to lines $overline{AC}$ and $overline{AB}$.

Then, from points of intersecting with a semicircle and lines,
$overline{AC}$ and $overline{AB}$ graph circle bends

inside the angle.

Through their point of intersection and point $A$, graph

a line using a ruler and that line will represent the angle bisector.

#### (b)

You can measure the same distance with

a straightedge from points $A$ and $B$ on

the right side and on that way mask that points

as $C$ and $D$ and then draw a line through them.

#### (c)

You can use a compass for this. You can measure

how long is $overline{AB}$ and then put on paper

twice that measure and draw a line through

the beginning and final point and that line is

$overline{EF}$ which is twice longer than $overline{AB}$.

#### (d)

Draw a line and on that line put circle bend from the angle
$measuredangle X$ and then graph another line at the end
of that circle bend.

That is angle $Y$.

#### (a)

On the following picture, there is graphed a circle

with radius $r$ and center $H$.

#### (c)

On the following picture, there is graphed an equilateral

triangle that is inscribed in a circle $H$.

First, you need to give the length of one

side of the triangle $overline{AB}$.

Then you need to place a compass point

on $A$ and measure the distance to the point $B$.

Then, swing an arc of this size above the segment.

After that, without changing the span on the compass,

place the compass point on $B$ and swing the same arc,

intersecting with the first arc.

Finally, label the point of intersection as the third vertex

of the equilateral triangle.

Find an equation representing the situation. Use the exponential equation:

$$
y=ab^x
$$

where $a$ is the initial value and $b$ is the constnt multiplier.

From the given, $a=2.79$ (current cost of $$2.79$) and $b=100%+8%=108%=1.08$ (increased 8% per year). With $x$ as the number of years and $y$ as the cost of a loaf in dollars, the equation is:

$$
y=2.79(1.08)^x
$$

$textbf{a.}$

$x=5$ represents 5 years later, so the cost will be:

$$
begin{align*}
y&=2.79(1.08)^{ 5}\
&approx 4.10to color{#c34632}$4.10
end{align*}
$$

$textbf{b.}$

5 years ago is represented by $x=-5$. So, the cost then was:

$$
begin{align*}
y&=2.79(1.08)^{-5}\
&approx 1.90to color{#c34632}$1.90
end{align*}
$$

a. $$4.10$

b. $$1.90$

#### (a)

Percentage of students who do not carry a backpack at school is:

$$
p=dfrac{39}{100}=39%
$$

#### (b)

The probability that a junior is carring a backpack is following:

$$
p=dfrac{14}{100}=0.14
$$

#### (c)

The required probability is the following:

$$
p=dfrac{14+16}{100}=dfrac{30}{100}=0.3
$$

#### (d)

We can notice that:

$$
dfrac{8}{11}nedfrac{16}{22}nedfrac{18}{22}nedfrac{19}{35}
$$

The conclusion is that there is a relationship between graduating class and carrying a backpack.

a) $39%$; b) $0.14$; c) $0.3$; d) There is an association.

We need to find length of $DE$.

Since two triangles are similar, so :

$$
dfrac{AB}{AD}=dfrac{BC}{DE} text{Where},
$$

$$
because AB=5 AD=AB+BD=5+4=9 BC=7
$$

$$
therefore dfrac{AB}{AD}=dfrac{BC}{DE} therefore dfrac{5}{9}=dfrac{7}{DE}
$$

$$
therefore DE=dfrac{7cdot9}{5}=dfrac{63}{5}=12.6
$$

$$
text{color{#4257b2}$$DE=12.6$$}
$$

We need to solve the following equations as follows:

$$
color{#4257b2}text{(a)} 8a+a-3=6a-2a-3
$$

Isolate the variables on the left side as follows:

$$
8a+a-6a+2a=-3+3 5a=0
$$

Divid both sides of equation by $5$ as follows.

$$
dfrac{5a}{5}=dfrac{0}{5} a=0
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=0$ in the equation as follows:

$$
((8cdot0)+0-3=(6cdot0)-(2cdot0)-3 0+0-3=0-0-3 -3=-3
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(b)} 8(3m-2)-7m=0
$$

Multiply from the left to the right as follows:

$$
24m-16-7m=0
$$

Isolate the variables on the left side as follows:

$$
24m-7m=16 17m=16
$$

Divid both sides of equation by $17$ as follows.

$$
dfrac{17m}{17}=dfrac{16}{17} m=0.9411
$$

$text{color{#4257b2}Check:}$ substitute the value of $m=0.9411$ in the equation as follows:

$$
8((3cdot0.9411)-2)-(7cdot0.9411)=0 6.58-6.58=0
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(c)} dfrac{x}{2}+1=6
$$

Isolate the variables on the left side as follows:

$$
dfrac{x}{2}=6-1 dfrac{x}{2}=5
$$

Multiply both sides of equation by $2$ as follows.

$$
dfrac{x}{2} cdot2=5cdot2 x=10
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=10$ in the equation as follows:

$$
dfrac{10}{2}+1=6 5+1=6 6=6
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(d)} |x-3|+5=11
$$

Isolate the variables on the left side as follows:

$$
|x-3|=11-5 |x-3|=6
$$

$$
x-3=6 x=6+3 x=9
$$

$text{color{#4257b2}Check:}$ substitute the value of $m=9$ in the equation as follows:

$$
|9-3|+5=11 |6|+5=11 11=11
$$

Since the left side is equal the right side, so the answer is correct.

$$
text{color{#4257b2}(a) $0$ (b) $m=0.9411$
\ \
(c) $x=10$ (d) $x=9$}
$$

a) The temperature of a cup never gains that $y$-value or get lower;

b) It will be more like a straight line at the beginning.

$textbf{a.}$

First, we find the boundary point by replacing the inequality symbol with $=$ then solve for the corresponding equation:

$$
begin{align*}
2&=2m-8 \
10&=2m \
5&=m
end{align*}
$$

Next, pick a test point except for the boundary point, say $m=0$ to substitute to the inequality:

$$
begin{align*}
2&<2(0)-8 \
2&<0-8 \
2&not{5
$$

On a number line, use an open circle for the boundary point 5 then draw an arrow pointing to the right:

$textbf{b.}$

First, we find the boundary point by replacing the inequality symbol with $=$ then solve for the corresponding equation:

$$
begin{align*}
dfrac{1}{3}x-1&=-3 \
dfrac{1}{3}x&=-2 \
x&=-6
end{align*}
$$

Next, pick a test point except for the boundary point, say $x=0$ to substitute to the inequality:

$$
begin{align*}
dfrac{1}{3}(0)-1&leq -3 \
0-1&leq -3\
-1&leq -3
end{align*}
$$

The test point is not a solution which means all points on this side of the boundary point are not part of the solution, which is to the right of it. So, the solution is to the left of the boundary point:

$$
color{#c34632}xleq -6
$$

On a number line, use a filled circle for the boundary point $-6$ then draw an arrow pointing to the left:

$textbf{c.}$

First, we find the boundary point by replacing the inequality symbol with $=$ then solve for the corresponding equation:

$$
begin{align*}
5(2x-8)+24&=3(4+2x) \
10x-40+24&=12+6x \
10x-16&=12+6x\
10x&=28+6x\
4x&=28\
x&=7
end{align*}
$$

Next, pick a test point except for the boundary point, say $x=0$ to substitute to the inequality:

$$
begin{align*}
5(2(0)-8)+24&>3(4+2(0)) \
5(-8)+24&>3(4) \
-40+24&>12 \
-16&not{>}12
end{align*}
$$

The test point is not a solution which means all points on this side of the boundary point are not part of the solution, which is to the left of it. So, the solution is to the right of the boundary point:

$$
color{#c34632}x>7
$$

On a number line, use an open circle for the boundary point 7 then draw an arrow pointing to the right:

$textbf{d.}$

First, we find the boundary point by replacing the inequality symbol with $=$ then solve for the corresponding equation:

$$
begin{align*}
5+2k&=k-2+k \
5+2k&=2k-2 \
5&=-2
end{align*}
$$

The variable was eliminated and the resulting equation is false for all $k$-values. The corresponding inequality $5<-2$ is also false so the solution is the empty set (no solution):

$$
color{#c34632}{text{O}}
$$

Represent this as simply the number line:

a. $m>5$

b. $xleq -6$

c. $x>7$

d. No solution

#### (a)

We can use a graph method. On the following picture,

there are graphed both equations and solution is their

point of intersecting, we can notice that the solution

is point $(1,-1)$, or $x=1$ and $y=-1$.

#### (b)

We can use the graph method. On the following picture,

there are graphed both equations and solution is their

point of intersection, we can notice that the solution is point

$left(-2,dfrac{1}{2} right)$, or $m=-2$ and $n=dfrac{1}{2}$.

a) $x=1$, $y=-1$; b) $m=-2$, $n=dfrac{1}{2}$

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

$$
y=mx+b
$$

or in this case,

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

The two terms represent the two points$(8,1056)$ and $(13,116)$ so the slope (sequence generator) is:

$$
m=dfrac{Delta y}{Delta x }=dfrac{116-1056}{13-8}=dfrac{-940}{5}=-188
$$

Using any point, say $(13,116)$, solve for $t(0)$:

$$
begin{align*}
116&=-188(13)+t(0)\
116&=-2444+t(0)\
2560&=t(0)
end{align*}
$$

So, the equation for the arithmetic sequence is:

$$
color{#c34632}t(n)=-188n+2560
$$

Find $t(5)$ using the equation we found:

$$
begin{align*}
t(5)&=-188(5)+2560\
&=-940+2560\
&=color{#c34632}1620
end{align*}
$$

$t(n)=-188n+2560$

$$
t(5)= 1620
$$

#### (a)

$bigodot A$ and $bigodot B$ have the same radius because of each of them passes through the center of the other circle.

#### (b)

On the following picture, there are graphed described circles.

#### (c)

On the following picture, there is graphed required quadrilateral $ABCD$.

We can conclude that Patricia is correct because

those are circles with the same radius and because

points $C, D$ are on circles, their distances from points

$A$ and $B$ are equal because of $A$ and $B$ are centers

of circles.

#### (d)

From the following picture, we can conclude that a triangle $triangle ACD$

is a triangle with equal sides, while $triangle ACD$ has two equal sides,

$overline{AC}$ and $overline{AD}$.

#### (e)

We can conclude that the line segments $overline{CD}$ and $overline{AB}$

are perpendicular line and their point of intersect is

a midpoint of both segments.

a) If you use GeoGebra;

b) You can use GeoGebra;

c) $ABCD$ is a rhombus;

d) $triangle ACD$ is equilateral and $triangle ACD$ has two equal sides;

e) they are perpendicular lines and their point of intersect is a midpoint of both of them.

#### (a)

If you are already given a length of the segment $overline{AB}$,

next thing is to fix the compass opening to a distance somewhat

bigger than $dfrac{1}{2}$ the length of the segment $AB$, but not

more than the length of $AB$. After that put the needle of compass

at $A$ and draw an arc, then put the needle at $B$ and also draw an arc.

Finally, draw the line joining the two points where the two arcs intersect.

The point of intersection of the line and the segment is required midpoint.

Now, through the midpoint, using a straightedge, it is possible to construct

the perpendicular line on that way you align one side of straightedge with the segment.

#### (b)

On the following picture, there is constructed a perpendicular bisector of $overline{KM}$ using previous steps.

#### (a)

According to $SSS$, those triangles are congruent

because it is already given that they have two equivalent

sides and $BD$ is a side of both triangles.

#### (b)

The conclusion is that $overline{BD}$ splits $measuredangle ABC$
on two equal angles.

#### (c)

First place the compasses point on the angle’s vertex $R$, then adjust the compasses to a median wide setting. The exact width is not important. After that, without changing the compasses width, draw an arc across each leg of the angle, place the compasses on the point where one are crosses a leg and draw an are in the interior of an angle.

Repeat this for the other leg so that the two are cross.

Finally, using a ruler draw a line from the vertex to the point where the areas cross, that line is the angle bisector.

a) According to $SSS$, triangles are congruent;

b) It splits $measuredangle ABC$ into two equal angles;

c) Use a compass and a ruler

On the following picture, there are given segments $overline{PQ}$ and $overline{ST}$ and $measuredangle V$, and also a triangle $triangle ACE$ whose sides $overline{AC}$ and $overline{AE}$ are congruent to $overline{ST}$ and $overline{PQ}$ and angle $measuredangle A$ is congruent to angle $measuredangle V$.

You can use a ruler and compass to construct a triangle $triangle ACE$.

She needs to set the compasses width equal to the segment $k$ and set its point at $P$ and draw arcs on lines $m$ and $n$.

That two points of intersecting arcs and lines $m$ and $n$ are two vertexes $M$ and $N$ of the square, the first vertex is in $P$.

Then, construct a perpendicular line through point $M$ on the line $m$ and through the point $N$ on line $n$.

Point of intersect of those perpendicular lines is the fourth vertex of a square and label it as $K$.

On that way, we get square $PMKN$.

#### (a)

The multiplier for the geometric sequence in this situation is $3$.

#### (b)

If $y$ represents the number of all students in this school who

heard the rumor at $3$ p.m, then, $dfrac{y}{3}$ will represent

the number of students who heard the rumor at $2$ p.m.

#### (c)

An equation for a geometric sequence that represents the fraction

of students that have heard the rumor after $n$ hours is:

$$
a_{n}=a(3)^n
$$

#### (d)

Number of students who heard about the rumor at $3$ p.m. is:

$$
a_{8}=a(3)^7
$$

So, required fraction will be:

$$
dfrac{a_{8}}{3^7}=dfrac{acdot3^7}{3^7}=a
$$

#### (e)

We will substitute $3$ for $a$ in the previous equation and get:

$$
a_{8}=3(3^7)=6561
$$

a) $3$; b) $dfrac{y}{3}$; c) $a_{n}=a(3)^n$; d) $a_{8}=a(3)^7$; e) $6561$

#### (a)

Conclusion is that Kenny should use columns.

#### (b)

For, example, let’s examine flavor Fudgy Fantasy.

$$
dfrac{5883}{2240}nedfrac{2771}{5446}nedfrac{672}{985}nedfrac{519}{1302}
$$

We can conclude that there is an association between

celebration and cupcake flavor to Kenny.

We need to write the below items for the following function.

$$
f(x)=8(x-3)+4
$$

$$
{color{#4257b2}f(-2)}=8(-2-3)+4=8(-5)+4=-40+4={color{#4257b2}-36}
$$

$$
{color{#4257b2}f(-0.5)}=8(-0.5-3)+4=8(-3.5)+4=-28+4={color{#4257b2}-24}
$$

$$
{color{#4257b2}f(8)}=8(8-3)+4=8(5)+4=40+4={color{#4257b2}44}
$$

$$
{color{#4257b2}fleft(dfrac{3}{4}right)}=8left(dfrac{3}{4}-3right)+4=8left(dfrac{3-12}{4}right)+4=8left(dfrac{-9}{4}right)+4
$$

$$
=left(dfrac{8cdot-9}{4}right)+4=(2cdot-9)+4=-18+4={color{#4257b2}-14}
$$

(a) $f(-2)=-36$ (b)$f(-0.5)=-24$

(c) $f(8)=44$ (d)$fleft(dfrac{3}{4}right)=-14$

We need to solve the following inequalities.

$$
color{#4257b2}text{(a)} 4x-1geq7
$$

Isolate the variables on the left side as follows:

$$
4xgeq7+1 4xgeq8
$$

divide both sides of inequality by $4$ as follows.

$$
xgeq dfrac{8}{4} xgeq2
$$

$$
color{#4257b2}text{(b)} 3-2xleq x+6
$$

Isolate the variables on the left side as follows:

$$
-2x-xleq6-3 -3xleq3
$$

Divide both sides of inequality by $-3$ as follows.

$$
xgeqdfrac{3}{-3} xgeq-1
$$

$$
color{#4257b2}text{(c)} 2(x-5)leq8
$$

Multiply from the left to the right as follows:

$$
2x-10leq8
$$

Isolate the variables on the left side as follows:

$$
2xleq8+10 2xleq18
$$

divide both sides of inequality by $2$ as follows.

$$
xleq dfrac{18}{2} xleq9
$$

$$
color{#4257b2}text{(d)} dfrac{1}{2} xgeq5
$$

Multiply both sides of inequality by $2$ as follows.

$$
dfrac{2}{2} xgeq5cdot2 xgeq10
$$

$$
text{color{#4257b2}(a) $xgeq2$ (b) $xgeq-1$
\ \
(c) $xleq9$ (d) $xgeq10$}
$$

#### (a)

In this case, graph of $f(x)$ will be translated $3$ units down.

#### (b)

In this case, graph of $f(x)$ will be translated $2$ units up.

a) Translation $3$ units down; b) Translation $2$ units up

We need to solve the following equations as follows:

$$
color{#4257b2}text{(a)} 3x+5=16
$$

Isolate the variables on the left side as follows:

$$
3x=16-5 3x=11
$$

Divid both sides of equation by $3$ as follows.

$$
dfrac{3x}{3}=dfrac{11}{3} x=3.6667
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=3.6667$ in the equation as follows:

$$
(3cdot3.6667)+5=16 11+5=16 16=16
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(b)} 3x+5=-16
$$

Isolate the variables on the left side as follows:

$$
3x=-16-5 3x=-21
$$

Divid both sides of equation by $3$ as follows.

$$
dfrac{3x}{3}=dfrac{-21}{3} x=-7
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=-7$ in the equation as follows:

$$
(3cdot-7)+5=-16 -21+5=-16 -16=-16
$$

Since the left side is equal the right side, so the answer is correct.

(a) $x=3.6667$ (b) $x=-7$

$$
text{color{#4257b2}How many kind of fruit did he but?}
$$

Assume the number of apples is $color{#4257b2}x$ and number of pairs is $color{#4257b2}11-x$

Total cost of fruit can be represented as follows:

$$
0.6x+0.35(11-x)=5.6
$$

Multiply from left to right as follows:

$$
0.6x+3.85-0.35x=5.6
$$

Isolate the variables on th left side as follows:

$$
0.6x-0.35x=5.6-3.85 0.25x=1.75
$$

Divide both side by $0.25$ as follows:

$$
dfrac{0.25x}{0.25}=dfrac{1.75}{0.25} x=7
$$

Number of apples is $color{#4257b2}7$ pieces.

Number of pairs is $11-7=color{#4257b2}4$ pieces

$$
Largecolor{#4257b2}7text{ and} 4
$$

One possible option for starting expression could be the following:

$$
6(m-2)
$$

Because, after applying the Distributive property, we get:

$$
6(m-2)=6m-12
$$

$$
6(m-2)
$$

#### (a)

If $l$ and $m$ are parallel, that means that $measuredangle a$ and $measuredangle c$ are angles

with equal measure.

Also, $measuredangle b$ and $measuredangle c$ are cross angles and they are equal,

so and $measuredangle a$ and $measuredangle b$ are equal angles.

#### (b)

On the following picture, there is constructed a line parallel to $l$ through

the point $P$ using transversal $f$.

#### (c)

In this case, you need to construct the angle with vertex $P$

which is equal to the angle made of lines $f$ and $l$, where

one leg is $f$ and the other is actually lining parallel to $l$.

a) $measuredangle a$ and $measuredangle c$ and $measuredangle b$ are equal angles;

b) Use GeoGebra;

c) The second leg of the angle $measuredangle P$ will be required parallel line.

#### (a)

Square is a geometric figure with four equal sides and all straight angles.

#### (b)

First, start with a given line segment $AB$, that will be one side of the square.

Then, extend the line $AB$ to the right and set the compasses on $B$ and any

convenient width scribe an arc on each side of $B$, creating the two points $F$ and $G$.

With the compasses on $G$ and any convenient width, draw an arc above the point $B$.

Then, without changing the compasses width, place the compasses on $F$ and draw

an arc above $B$, crossing the previous arc and creating point $H$.

After that, draw a line from $B$ through $H$, $BH$ will become the second side.

Set the compasses on $A$ and set its width to $AB$.

This width will be held unchanged as we create the square’s other three sides.

Then draw an arc above point $A$. Without changing the width, move the compasses

to the point $B$.

Draw an arc across $BA$ creating point $C$- a vertex of the square.

Without changing the width, move the compasses to $C$. Draw an arc to the left of $C$

across the existing arc, creating point $D$- a vertex of the square.

Draw the lines $CD$ and $AD$ and we got square $ABCD$.

We have that $measuredangle ABC=90^{circ}$ by construction.

Then, segments $BC$, $CD$ and $DA$ are the same length as $AB$

because they were drawn with the same compass width-$AB$.

So, $ABCD$ is a square.

a) It has four equal sides and four right angles;

b) Use the compass and a ruler;

c) According to construction, $ABCD$ is a square.

First, we will construct an isosceles triangle with side length $overline{AB}$.

Draw a line and on that line pick two points, $K$ and $M$.

Then, set the compasses width to be equal to the length of $overline{AB}$.

After that, from both points, we picked on the line draw two arcs without changing the width.

Their point of intersecting will be the third vertex of the required triangle,

while two others will be the first two points packed on the line.

Now, we will construct triangles with side lengths $3$, $4$ and $5$ units.

First, draw a segment whose length is $3$ units, $overline{AB}$.

From the point, $B$ draw an arc with compasses whose width is $4$ units.

Point of intersect of those two arcs will be $C$, the third vertex.

On the same way construct the triangle with side lengths $2$, $3$ and $6$ units.

Draw an isosceles triangle with side length $overline{AB}$, with side lengths $3,4$ and $5$;

$2,3$ and $6$ units

#### (a)

We can notice that the center of the smaller circle

is $B$ and its radius is $overline{BC}$.

We can notice that the center of a larger circle is point

$A$ and its radius is $overline{AC}$.

#### (b)

Because of triangles $triangle ABC$ and $triangle GHJ$

have all equal sides, we conclude

that they are congruent, according to the $SSS$.

#### (c)

On the following picture, there are graphed $triangle ABC$ and $triangle GHJ$.

A transformation that maps $GH$ and $overline{AB}$ is a rotation
about the point with a given angle.

#### (d)

The transformation which maps point $J$ to the point $C$ is

a translation by the vector.

#### (e)

According to the previous transformation, we can confirm that those

triangles are congruent according to the $SSS$.

a) $B$, $overline{BC}$; $A$, $overline{AC}$;

b) They are congruent;

c) rotation about point;

d) Translation by vector;

e) It is demonstrated.

Here, the right strategy is $b)$ in order to find the line parallel to $m$ through the point $Q$.

#### (a)

We can conclude that the annual multiplier is $0.46$ and annual percent decrease is $46%$.

#### (b)

The equation of a function that describes this situation is the following, where $x$
represent the time in years.

$$
y=20(0.46)^x
$$

subsection*{(c)}
There is the following table of the first $5$ values.\
begin{center}
begin{tabular}{ |c|c| c| c| c| c |c| }
hline
$x$ & $y$ \
hline
$1$ & $9.2$ \
hline
$2$ & $4.232$ \
hline
$3$ & $1.95$ \
hline
$4$ & $0.897$ \
hline
$5$ & $0.41$ \
hline
end{tabular}
end{center}
On the following picture,there is a sketch of a previous function.\

a) $0.46$, $46%$; b) $y=20(0.46)^x$; c) Use graphing calculator.

#### (a)

On the following picture, there are graphed points $A$ and $B$ and $M$, the midpoint.
Its coordinates are $M(1.5,5)$.

#### (b)

The equation of the line that passes through points $A$ and $B$ is the following:

$$
y+1=dfrac{11+1}{6+3}(x+3)
$$

$$
y=dfrac{12}{9}(x+3)-1
$$

$$
y=dfrac{4}{3}x+3
$$

#### (c)

The distance between points $M$ and $B$ is the following:

$$
MB=sqrt{(11-5)^2+(sqrt{a}6-1.5)^2}=7.5
$$

a) $M(1.5, 5)$; b) $y=dfrac{4}{3}x+3$; c) $7.5$

Here, the first term is $a_{1}=15$, the constant difference between terms

is $-8$, explicit formula for calculating $n$th term of this sequence is following:

$$
a_{n}=15+(n-1)(-8)
$$

Now, we will substitute $-225$ for $a_{n}$ and find $n$ on the following way:

$$
-225=15-8n+8
$$

$$
8n=248
$$

$$
n=31
$$

So, the conclusion is that there are shown $31$ terms

of the arithmetic sequence.

$31$ terms

We need to solve the following inequalities.

$$
color{#4257b2}text{(a)} 3x-5le7+2x
$$

Isolate the variables on the left side as follows:

$$
3x-2xle7+5 xle12
$$

$$
color{#4257b2}text{(b)} |x|-3le7
$$

Isolate the variables on the left side as follows:

$$
|x|le7+3 |x|le10
$$

$$
xle10
$$

$$
color{#4257b2}text{(c)} 5(2-x)+6ge16
$$

Multiply from left to right as follows:

$$
10-5x+6ge16
$$

Isolate the variables on the left side as follows:

$$
-5xge16-6-10 -5xge16-16 -5xge0
$$

Divide both sides by$-5$as follows:

$$
dfrac{-5x}{-5}ledfrac{0}{-5} xle0
$$

$$
color{#4257b2}text{(d)} |x+2|ge3
$$

Isolate the variables on the left side as follows:

$$
x+2ge3 xge3-2 xge1
$$

(a) $xle12$ (b) $xle10$

(c) $xle0$ (d) $xge1$

We need to write the below items for the following function.

$$
f(x)=2^x -3
$$

$$
{color{#4257b2}f(-2)}=2^{(-2)}-3=dfrac{1}{2^2}-3=dfrac{1}{4}-3=dfrac{1-12}{4}={color{#4257b2}dfrac{-11}{4}}
$$

$$
{color{#4257b2}f(-0.5)}=2^{(-0.5)}-3=dfrac{1}{2^{0.5}}-3=dfrac{1}{sqrt{2}}-3={color{#4257b2}dfrac{1-3sqrt{2}}{sqrt{2}}}
$$

$$
{color{#4257b2}f(8)}=2^{(8)}-3=256-3={color{#4257b2}253}
$$

$$
{color{#4257b2}fleft(dfrac{3}{4}right)}=2^{(0.75)}-3=1.682-3={color{#4257b2}-1.318}
$$

(a) $f(-2)=dfrac{-11}{4}$ (b)$f(-0.5)=dfrac{1-3sqrt{2}}{sqrt{2}}$

(c) $f(8)=253$ (d)$fleft(dfrac{3}{4}right)=-1.318$

We need to solve the following equations as follows:

$$
color{#4257b2}text{(a)} |x-3|+3=4
$$

Isolate the variables on the left side as follows:

$$
|x-3|=4-3 |x-3|=1 x-3=1
$$

Add both sides of equation by $3$ as follows.

$$
x-3+3=1+3 x=4
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=4$ in the equation as follows:

$$
|4-3|+3=4 |1|+3=4 4=4
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(b)} (x-5)^3=8
$$

Evaluate the exponent as follows:

$$
(x-5)^3=2^3
$$

Since the power of the left side is equal the power in the right side, so:

$$
x-5=2
$$

Isolate the variables on the left side as follows:

$$
x=2+5 x=7
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=7$ in the equation as follows:

$$
(7-5)^3=8 2^3=8 8=8
$$

Since the left side is equal the right side, so the answer is correct.

$$
color{#4257b2}text{(c)} dfrac{5}{9}-dfrac{x}{3}=dfrac{4}{9}
$$

Subtract both sides of equation by $dfrac{5}{9}$ as follows.

$$
dfrac{5}{9}-dfrac{5}{9}-dfrac{x}{3}=dfrac{4}{9}-dfrac{5}{9} -dfrac{x}{3}=dfrac{-1}{9}
$$

Multiply both sides of equation by $-3$ as follows.

$$
-3 cdot-dfrac{x}{3}=-3 cdot -dfrac{1}{9} x=dfrac{1}{3}
$$

$text{color{#4257b2}Check:}$ substitute the value of $x=dfrac{1}{3}$ in the equation as follows:

$$
dfrac{5}{9}-dfrac{dfrac{1}{3}}{3}=dfrac{4}{9} dfrac{5}{9}-dfrac{1}{9}=dfrac{4}{9}
$$

$$
dfrac{4}{9}=dfrac{4}{9}
$$

Since the left side is equal the right side, so the answer is correct.

(a) $x=4$ (b) $x=7$ (c) $x=dfrac{1}{3}$

#### (b)

Solution for $y$ is the following:

$$
3y=2x-10
$$

$$
y=dfrac{2x-10}{3}
$$

#### (c)

Yes, the slope of each line is $dfrac{2}{3}$ and it confirms our statement

that those lines are parallel.

a) Lines are parallel; b) $y=dfrac{2}{3}x-dfrac{10}{3}$; c) slope$=dfrac{2}{3}$

Both situations represent an exponential function since there is a constant multipier so we use:

$$
y=acdot b^x
$$

where $a$ is the initial value (at $x=0$) and $b$ is the multiplier.

$textbf{a.}$

Let $x$ be the number of years and $y$ be the value of the car, in dollars. From the given, $a=23500$ and $b=100%-15%=85%=0.85$, so the equation is:

$$
color{#4257b2}y=23500cdot (0.85)^x
$$

Substitute $x=15$ for 15 years old and solve for $y$

$$
begin{align*}
y&=23500cdot (0.85)^{15} \
&approx 2052.82tocolor{#c34632}$2052.82
end{align*}
$$

$textbf{b.}$

Let $x$ be the number of years and $y$ be the population. From the given, $a=14365112$ and $b=100%+12%=112%=1.12$, so the equation is:

$$
color{#4257b2}y=14365112cdot (1.12)^x
$$

Substitute $x=20$ for 20 years from now and solve for $y$

$$
begin{align*}
y&=14365112cdot (1.12)^{20} \
&approx color{#c34632}138570081
end{align*}
$$

a. about $$2052.82$

b. about 138570081

Let $x$ represent the time it took to both teams to mow the entire lawn.

The rate for the first team is $dfrac{1}{10}$ and the rate for the second team is $dfrac{1}{15}$.

We need to solve the following equation for $x$ in order to find $x$:

$$
dfrac{1}{10}+dfrac{1}{15}=dfrac{1}{x}
$$

$$
dfrac{3+2}{30}=dfrac{1}{x}
$$

$$
5x=30
$$

$$
x=6
$$

So, it will take $6$ hours to both teams to mow the entire lawn.

$6$ hours

a) If the first team can mow the lawn in 10 hours, then in one hour it will mow $dfrac{1}{10}$.

b) If the second team can mow the lawn in 15 hours, then in one hour it will mow $dfrac{1}{15}$.

$dfrac{1}{10}+dfrac{1}{15}=dfrac{3}{30}+dfrac{2}{30}=dfrac{5}{30}=dfrac{1}{6}$

d) Since $dfrac{1}{6}$ from the lawn is done in one hour, they will need 6 hours to perform the job.

$$
dfrac{1}{x}=dfrac{1}{10}+dfrac{1}{15}
$$

e)Let’s note:

$x$=the amount of time (in hours) they need to mow the lawn if they work together.

We write the equation:

$dfrac{1}{x}=dfrac{3}{30}+dfrac{2}{30}$

$dfrac{1}{x}=dfrac{5}{30}$

$dfrac{1}{x}=dfrac{1}{6}$

$$
x=6
$$

Let’s note:

$x$=the amount of time (in minutes) they need to staple the programs if they work together

$$
dfrac{1}{x}=dfrac{1}{30}+dfrac{1}{10}
$$

$dfrac{1}{x}=dfrac{1}{30}+dfrac{3}{30}$

$dfrac{1}{x}=dfrac{4}{30}$

$4x=30$

$x=dfrac{30}{4}=dfrac{15}{2}=7.5$ minutes

We solve the equation to determine $x$:

$7.5$ minutes

Let’s note:

$x$=the amount of time (in minutes) his sister needs to wash the truck alone

$$
dfrac{1}{6}=dfrac{1}{8}+dfrac{1}{x}
$$

$dfrac{1}{x}=dfrac{1}{6}-dfrac{1}{8}$

$dfrac{1}{x}=dfrac{4}{24}-dfrac{3}{24}$

$dfrac{1}{x}=dfrac{1}{24}$

$x=24$ minutes

We solve the equation to determine $x$:

$24$ minutes

Let $x$ represent the number of students in $7$th grade and $y$ represents the number

of students in $8$th grade.

We get the following system of equations which we need to solve for $x$ and $y$ and get:

$$
dfrac{40}{100}x+dfrac{90}{100}y=dfrac{72}{100}cdot1000
$$

$$
x+y=1000
$$

We got the following:

$$
40x+90y=72000
$$

$$
x+y=1000Rightarrow x=1000-y
$$

Now, substitute $x$ from the second equation in the first and get:

$$
40(1000-y)+90y=72000
$$

$$
50y=32000
$$

$$
y=640Rightarrow x=360
$$

The conclusion is that there are $360$ students in $7$th grade and $640$ in $8$th grade.

$360$ in $7$th, $640$ in $8$th

We have that $dfrac{50}{3cdot2}=dfrac{50}{6}$ streamers are hanged for $1$ hour by $1$ student.

We will have the following:

$$
15cdotdfrac{50}{6}=125
$$

So, the conclusion is that $5$ students will hang $125$ streamers for $3$ hours.

$125$ streamers

#### (a)

Required system of equations in this situation is following:

$$
3.15g+1.85r=4.50
$$

$$
g+r=12
$$

#### (d)

A new system of equations for described situation in this part is following:

$$
3.15g+1.85r=3
$$

$$
g+r=15
$$

(a) $3.15g+1.85r=4.50, g+r=12$; (b) $3.15g+1.85r=3, g+r=15$

Mixture problems involve creating a mixture from two or more things, and then determining some quantity, for example percentage of the resulting mixture. For instance, following example can represent mixture problem with percentage:

You need a $15%$ acid solution for a certain test, but your supplier only ships a $10%$ solution and a $30%$ solution. Rather than pay the hefty surcharge to have the supplier make a $15%$ solution, you decide to mix $10%$ solution with $30%$ solution, to make your own $15%$ solution. You need $10$ liters of the $15%$ acid solution. How many liters of $10%$ solution and $30%$ solution should you use?

Let $x$ represent the sum of money invested in stocks and

$y$ represent the sum of money invested in bonds.

Using the information in the task, we get the following system

of equations and we will solve it for $x$ and $y$:

$$
x+y=10000Rightarrow x=x=1000-y
$$

$$
0.03x+0.05y=430
$$

We will substitute $x$ from the first equation in the second

and solve it for $y$:

$$
0.03(10000-y)+0.05y=430
$$

$$
0.02y=130
$$

$$
y=6500
$$

$$
x=10000-6500=3500
$$

So,
$$
3,500$ is invested is stocks and
$$
6,500$ is invested

in bonds.

$$
3,500$;
$$
6,500$

Let $x$ be the required pounds of coffee.

We get the following equation, which we will solve for $x$:

$$
6x+3cdot4=4.75(3+x)
$$

$$
6x+12=14.25+4.75x
$$

$$
1.25x=2.25
$$

$$
x=1.8
$$

So, the conclusion is that there should be mixed $1.8$ pound of

coffee costing $$6$ per pound.

$1.8$ pound

We can notice that the required equation is:

$$
y=5(1.5)^x
$$

#### (b)

On the following picture, there are graphed given points and the best

fit line:

We can notice that the required equation is:

$$
y=0.5(0.4)^x
$$

$y=5(1.5)^x$; $y=0.5(0.4)^x$

On the following picture, there are graphed $triangle ABC$ and rotated $triangle ABC$.

We can notice that vertices of rotated $triangle ABC$ have the following coordinates:

$$
A'(-2,3), B'(-9,4), C'(-7,6)
$$

Which means that the correct answer is $textbf{a)}$.

First, we find the boundary line equation using the $y=mx+b$ form where $m$ is the slope and $b$ is the $y$-intercept. From the graph, the intercepts are $(0.5,0)$ and $(0,-1)$ so the slope is:

$$
m=dfrac{Delta y}{Delta x }=dfrac{-1-0}{0-0.5}=dfrac{-1}{-0.5}=2
$$

Since $b=-1$, then the equation is:

$$
y=2x-1
$$

Since the line is dashed, the points on the line itself are not solutions in the inequality. So, either use $$.

We can then assign a point that is on the shaded region, say $(0,0)$ and determine what inequality to use:

$$
begin{align*}
y&=2x-1\
0&stackrel{?}{=}2(0)-1\
0&stackrel{?}{=}0-1\
0&stackrel{?}{=}-1
end{align*}
$$

The appropriate inequality to use is $>$ since $0>-1$. Hence, the inequality for the graph is:

$$
color{#c34632} y>2x-1
$$

$$
y>2x-1
$$

$textbf{a.}$

Graphically, the $x$-intercepts are where the graph intersects the $x$-axis:

$$
color{#c34632}(-2,0)text{ and }(0,0)
$$

and the $y$-intercept is where the graph intersects the $y$-axis:

$$
color{#c34632} (0,0)
$$

$textbf{b.}$

Graphically, the $x$-intercepts are where the graph intersects the $x$-axis:

$$
color{#c34632}(-3,0)text{ and }(5,0)
$$

and the $y$-intercept is where the graph intersects the $y$-axis:

$$
color{#c34632} (0,3)
$$

$textbf{c.}$

Using a table, the $x$-intercepts are where $y=0$:

$$
color{#c34632}(-1,0)text{ and }(1,0)
$$

and the $y$-intercept is where $x=0$:

$$
color{#c34632} (0,-1)
$$

a. $x$-intercepts: $(-2,0)$ and $(0,0)$ ; $y$-intercept: $(0,0)$

b. $x$-intercepts: $(-3,0)$ and $(5,0)$ ; $y$-intercept: $(0,3)$

c. $x$-intercepts: $(-1,0)$ and $(1,0)$ ; $y$-intercept: $(0,-1)$

$textbf{a.}$

Find the boundary point(s). Replace the inequality symbol with $=$ so we have the equation:

$$
|x|+3=5
$$

Subtract 3 from both sides:

$$
|x|=2
$$

Since the right side is positive, then there are two solutions: $x=-2$ and $x=2$. These are the boundary points. Hence, the number line will be divided into 3 regions:

$$
x< -2,-2<x2
$$

For each region, assign a test point. If the test point satisfies the inequality, the region it belongs to is a solution.

For $x< -2$, I used $x=-3$:

$$
begin{align*}
|-3|+3&stackrel{?}{<}5\
6&not{<}5
end{align*}
$$

For $-2<x<2$, I used $x=0$:

$$
begin{align*}
|0|+3&stackrel{?}{<}5\
3&2$, I used $x=3$:

$$
begin{align*}
|3|+3&stackrel{?}{<}5\
6&not{<}5
end{align*}
$$

So, the solution of the inequality is:

$$
color{#c34632}-2<x<2
$$

On a number line, use open circles for both $-2$ and $2$ since they are not part of the solution:

$textbf{b.}$

First, we find the boundary point by replacing the inequality symbol with $=$ then solve for the corresponding equation:

$$
begin{align*}
5(2x+1)&=30\
2x+1 &=6\
2x&=5\
x&=2.5
end{align*}
$$

Next, pick a test point except for the boundary point, say $x=0$ to substitute to the inequality:

$$
begin{align*}
5(2(0)+1) &geq 30\
5(1) &geq 30\
5&geq 30
end{align*}
$$

The test point is not a solution which means all points on this side of the boundary point are not part of the solution, which is to the left of it. So, the solution is to the right of the boundary point:

$$
color{#c34632}xgeq 2.5
$$

On a number line, use a filled circle for 2.5 since it is part of the solution:

$textbf{c.}$

Multiply both sides by the LCD which is $2x$:

$$
2-5x=3xcolor{white}tag{1}
$$

Add $5x$ to both sides:

$$
2=8x
$$

Divide both sides by 8:

$$
dfrac{1}{4}=x
$$

or

$$
color{#c34632}x=dfrac{1}{4}
$$

On a number line, represent the solution as a point:

$textbf{d.}$

Adding $x$ to both sides, we obtain:

$$
-5>3
$$

which is false for all values of $x$. Hence, we can conclude that the inequality has $text{textcolor{#c34632}{no solution}}$. The graph is simply the number line itself:

$textbf{e.}$

Subtract 1 from both sides:

$$
3sqrt{4-x}=12
$$

Divide both sides by 3:

$$
sqrt{4-x}=4
$$

Square both sides:

$$
4-x=16
$$

Subtract 4 from both sides:

$$
-x=12
$$

Multiply both sides by $-1$:

$$
color{#c34632}x=-12
$$

On a number line, represent the solution as a point:

$textbf{f.}$

Find the boundary point(s). Replace the inequality symbol with $=$ so we have the equation:

$$
|x+1|=4
$$

Since the right side is positive, then there are two solutions. We write and solve the two equations:

$$
begin{align*}
x+1&=-4 & x+1&=4\
x&=-5 & x&=3
end{align*}
$$

So, the boundary points are $x=-5$ and $x=3$. Hence, the number line will be divided into 3 regions:

$$
xleq -5,-5 leq xleq 3,text{ and }xgeq 3
$$

For each region, assign a test point. If the test point satisfies the inequality, the region it belongs to is a solution.

For $xleq -5$, I used $x=-6$:

$$
begin{align*}
| -6+1 |&stackrel{?}{leq}4\
5&not{leq} 4
end{align*}
$$

For $-5 leq xleq 3$, I used $x=0$:

$$
begin{align*}
|0+1 |&stackrel{?}{leq}4\
1&leq 4hspace{5mm}checkmark
end{align*}
$$

For $xgeq 3$, I used $x=4$:

$$
begin{align*}
| 4+1 |&stackrel{?}{leq}4\
5&not{leq} 4
end{align*}
$$

So, the solution of the inequality is:

$$
color{#c34632}-5 leq xleq 3
$$

On a number line, use filled circles for both $-5$ and $3$ since they are part of the solution:

a. $-2<x<2$

b. $xgeq 2.5$

c. $x=-2$, $x=dfrac{1}{3}$

d. no solution

e. $x=-12$

f. $-5 leq xleq 3$

$$
text{color{#4257b2}How many toppings did he get?}
$$

Assume the number of toppings is $x$, so:

$$
(3 cdot 1.19)+(0.49 cdot x)=4.55 3.57+0.49 x=4.55
$$

Isolate the variables on the left side as follows:

$$
0.49 x=4.55-3.57 0.49 x=0.98
$$

Divide both of side by $0.49$ as follows:

$$
x=dfrac{0.98}{0.49} x=2
$$

The number of toppings did he get is $2$ toppings.

$$
text{color{#4257b2}Large$Large2$ toppings.}
$$

We need to multiply the following polynomials.

$$
color{#4257b2}text{(a)} (2a+b)(a-3b)
$$

Use distributive property to simplify as follows:

$$
{color{#4257b2}2a}(a)+{color{#4257b2}2a}(-3b)+{color{#4257b2}b}(a)+{color{#4257b2}b}(-3b)
$$

$$
2a^2-6ab+ab-3b^2
$$

Rearrange the tiles to group like terms as follows:

$$
2a^2-3b^2+(ab-6ab)
$$

$$
2a^2-3b^2-5ab
$$

$$
color{#4257b2}text{(b)} (x+2)(x^2-2x+5)
$$

Use distributive property to simplify as follows:

$$
{color{#4257b2}x}(x^2)+{color{#4257b2}x}(-2x)+{color{#4257b2}x}(5)+{color{#4257b2}2}(x^2)+{color{#4257b2}2}(-2x)+{color{#4257b2}2}(5)
$$

$$
x^3-2x^2+5x+2x^2-4x+10
$$

Rearrange the tiles to group like terms as follows:

$$
x^3+(2x^2-2x^2)+(5x-4x)+10
$$

$$
x^3+0+x+10
$$

$$
color{#4257b2}x^3+x+10
$$

(a) $2a^2-3b^2-5ab$ (b) $x^3+x+10$

#### (a)

The conclusion is that the equation has only one solution because,

it represent $x$-intercept on the graph.

#### (b)

On the following picture, there are graphed functions $y=1$ and $y=0.5^x-3$.

The solution of the given equation is the point of intersecting those lines,

that point has coordinates $(-2,1)$, which means that solution is $x=-2$.

#### (c)

You can substitute $-2$ for $x$ and $1$ for $y$ in order to verify answer:

$$
1=(0.5)^{-2}-3
$$

$$
1=1
$$

So, the conclusion is that the solution is correct.

a) One solution; b) $x=-2$; c) Substitute $-2$ for $x$ and $1$ for $y$

#### (a)

On the following picture, there are graphed functions

$y=20x+1$ and $y=3^x$.

We can notice that there is a point of intersecting, $(0,1)$,

which means that there is a solution for $x=0$ and $y=1$.

If we substitute $0$ for $x$ and $1$ for $y$ in the previous

equations, we get:

$$
1=10cdot0+1=1
$$

$$
1=3^{circ}=1
$$

So, we proved that they are solutions.

#### (b)

It must be coordinates of a point because it can be

solved using graph, solution will be the point of intersection.

#### (c)

On the following picture, there are graphed functions:

$y=20x$ and $y=3^x-1$.

We can notice that in this case, the solution is not the same

as before, here solution is a point $(0,0)$, or $x=0$ and $y=0$.

#### (d)

Karen could not solve the system algebraically because of one equation

is linear and the second one is exponential.

a) $(0,1)$; b) The coordinates; c) $(0,0)$; d) One equation is linear, the second is exponential.

#### (a)

The conclusion is that $y=3.5-dfrac{1}{2}x$ also has infinitely

many solutions.

#### (b)

If Mason graph both functions, point of their intersect will be

the solution, on that way a graph will help Mason to solve

the equation.

#### (c)

On the following picture, there are graphed functions $y=3.5-dfrac{1}{2}x$ and $y=0.5^x-3$.

We can notice that there are two solutions, there are two points

of intersect $(-3,5)$ and $(13,-3)$.

a) infinitely many solutions; b) solution is point of intersect; c) $(-3,5), (13,-3)$

On the following picture, there are graphed functions

$y=2cdotleft(dfrac{3}{4} right)^x+7$ and $y=0.7(4x+38)-6$.

We can notice that the point of intersect is $(-3.11,11.893)$

or $x=-3.11$ and $y=11.893$ are solutions.

$(-3.11, 11.893)$

We can use a graph in order to solve this system.

On the following picture, there are graphed functions

$y=2^x-3$ and $y=2x+1$.

We can notice that there are two solutions, there are two points

of intersecting $(3.445, 7.89)$ and $(-1.863,-2.725)$.

$(3.445, 7.89)$, $(-1.863,-2.725)$.

#### (a)

On the following picture, there are graphed functions

$f(x)$ and $g(x)$:

#### (b)

From the previous graph, we can notice that the estimated

point of intersection is $(-2,1)$, so, we will make a table of

values to prove this.

The following table is a table of values for $f(x)$.

begin{center}
begin{tabular}{ |c|c| c| c| c| c |c| }
hline
$x$ & $f(x)$ \
hline
$-3$ & $1.04$ \
hline
$-2$ & $1.11$ \
hline
$-1$ & $1.33$ \
hline
$0$ & $1$ \
hline
$1$ & $4$ \
hline
end{tabular}
end{center}
The following table is a table of values for $g(x)$.\
begin{center}
begin{tabular}{ |c|c| c| c| c| c |c| }
hline
$x$ & $g(x)$ \
hline
$-3$ & $2$ \
hline
$-2$ & $0$ \
hline
$-1$ & $-2$ \
hline
$0$ & $-4$ \
hline
$1$ & $-6$ \
hline
end{tabular}
end{center}

#### (c)

Based on the previous part, the approximate solution to the original equation is $(-2,1)$.

a) Use graphing calculator; b) $(-2,1)$; c) $(-2,1)$

$textbf{a.}$

Let $b$ be the multiplier. There are 5 years from 2005 to 2010 so we can write the equation:

$$
60b^5=45
$$

Divide both sides by 60:

$$
b^5=0.75
$$

Take the fifth root of both sides (or raise them by $1/5$):

$$
b=sqrt[5]{0.75}approx color{#c34632}0.944
$$

To find the percent decrease, subtract the multiplier from 1, then convert to percent:

$$
1-0.944=0.056to color{#c34632}5.6%
$$

$textbf{b.}$

Let $x$ be the number of years since 2005 and $y$ be the value of the stock.

Use the exponential equation:

$$
y=ab^x
$$

where $a=60$ (the initial value in 2005, or $x=0$) and $b=0.944$ (constant multiplier). So, the equation is:

$$
y=60(0.944)^x
$$

or

$$
color{#c34632}f(x)=60(0.944)^x
$$

$textbf{c.}$

Year 2020 is represented by $x=15$. Using the function from part (b),

$$
begin{align*}
f(15)&=60(0.944)^{15}\
&approx 25.28to color{#c34632}$25.28
end{align*}
$$

a. multiplier: $approx 0.944to 5.6%$ decrease

b. $f(x)=60(0.944)^x$

c. $$25.28$

Here, the conclusion is that the right matches are the following:

$a-iv$; $b-ii$; $c-v$; $d-i$; $e-iii$

Those answers are because $b$ is the only histogram with narrow

range, $c$ has a uniform distribution, $d$ has a lot of data at the edges,

those are reasons for the previous answers.

$a-iv$; $b-ii$; $c-v$; $d-i$; $e-iii$

Let $x$ represent rate, according to the given information, it is:

$$
40=2cdot3cdot x
$$

$$
x=dfrac{40}{6}=dfrac{20}{3}
$$

So, now we get the following equation, where $n$ is a number

of employees:

$$
dfrac{20}{3}cdot ncdot1=60
$$

$$
n=dfrac{180}{20}
$$

$$
n=9
$$

So, $9$ employees would be needed to build $60$ planters

in one day.

$9$ employees

We need to solve the following inequalities.

$$
color{#4257b2}text{(a)} 6x-1<11
$$

Isolate the variables on the left side as follows:

$$
6x<11+1 6x<12
$$

Divide both sides of inequality by $6$ as follows.

$$
x< dfrac{12}{6} x<2
$$

$$
color{#4257b2}text{(b)} dfrac{1}{3} xgeq2
$$

Multiply both sides of inequality by $3$ as follows.

$$
xgeq 2cdot3 xgeq6
$$

$$
color{#4257b2}text{(c)} 9(x-2)>18
$$

divide both sides of inequality by $9$ as follows.

$$
x-2>dfrac{18}{9} x-2>2
$$

Divide both sides of inequality by $2$ as follows.

$$
x>2+2 x>4
$$

$$
color{#4257b2}text{(d)} 5-dfrac{x}{4}leqdfrac{1}{2}
$$

Subtract both sides of inequality by $5$ as follows.

$$
-dfrac{x}{4}leqdfrac{1}{2}-5 -dfrac{x}{4}leqdfrac{1-10}{2} -dfrac{x}{4}leqdfrac{-9}{2}
$$

Multiply both sides of inequality by $-4$ as follows.

$$
xgeqdfrac{-9}{2} cdot -4 xgeq9cdot2 xgeq18
$$

$$
text{color{#4257b2}(a) $x 4$ (d) $xge18$}
$$

The piecewise function tells use that for $xgeq 0$, the graph is the exponential graph $y=2^x$ and for $x<0$, the graph is $y=|x|$.

When $x=0$, $y=2^0=1$ for the first piece so $(0,1)$ should be a filled circle since $geq$ was used. When $x=0$, $y=|0|=0$ for the second piece so $(0,0)$ should be an open circle since $<$ was used. So, the piecewise function should look like:

Hint: The piecewise function tells use that for $xgeq 0$, the graph is the exponential graph $y=2^x$ and for $x<0$, the graph is $y=|x|$.

We need to solve the following equation.

$$
color{#4257b2}text{(a)} dfrac{1}{8}=dfrac{x}{12}
$$

Multiply both of sides by $12$ as follows:

$$
dfrac{1}{8} cdot 12=x x=dfrac{12}{8}
$$

$$
x=dfrac{3}{2} x=1.5
$$

$$
color{#4257b2}text{(b)} dfrac{x}{10}=dfrac{12}{15}
$$

Multiply both of sides by $10$ as follows:

$$
x=dfrac{12}{15} cdot 10 x=dfrac{120}{15}
$$

$$
x=dfrac{24}{3} x=8
$$

$$
text{color{#4257b2}Large(a) $x=1.5$ (b) $x=8$}
$$

$$
text{color{#4257b2}How far away from her house does he live?}
$$

Since she traveled $(40)$ miles after $(10)$ minutes from grandmothers house, and traveled $(34)$ miles after $(22)$ minutes from her house at constant rate, so she traveled $(6)$ miles with $(12)$ minute at constant rate.

$$
dfrac{text{Distance}}{text{Time}}=dfrac{6}{12}=dfrac{x}{90}
$$

$$
x=dfrac{90cdot6}{12}=dfrac{540}{12}=45text{ miles}
$$

$$
text{color{#4257b2}$x=45$ miles}
$$

$$
text{color{#4257b2}We need to get the width and area of rectangle.}
$$

Assume that the length of rectangle is $x$ which equal$=11$ mm, and the width is $y$, and diagonal is $k$ which equal $=61$ mm

$$
k^2=x^2+y^2 y^2=k^2-x^2
$$

$$
y^2=(61)^2-(11)^2 y^2=3721-121 y^2=3600
$$

Use square root for both sides to get the value of $y$ as follows:

$$
sqrt{y^2}=sqrt{3600} y=60text{ mm}
$$

The width of rectangle is $60$ mm.

$text{color{#4257b2}Area}$ $=$ length $cdot$ width $=11cdot60=660$ sq.mm.

$$
text{color{#4257b2}Width $=60$ mm Area $=660$ sq. mm}
$$

In the following table, there is a number of hours which \
show how long one special candle last.\
begin{center}
begin{tabular}{ |c|c| c| c| c| c |c| }
hline
$x$ & $y$ \
hline
$1$ & $3$ \
hline
$2$ & $3.25$ \
hline
$3$ & $4$ \
hline
$4$ & $3.70$ \
hline
end{tabular}
end{center}
On the following picture, there is graphed data from\
the table and a best fit line.\

The equation of the best fit line is $y=0.285x+2.775$.

Here, we can conclude that special candle will last

between $3$ and $4$ hours.

Between $3$ and $4$ hours special candle will last.

#### (a)

Here, we can notice that data is increasing and there are several

points which are outliers.

#### (b)

Here, we can notice that the required slope is $14.65$.

#### (c)

We got that $R^2=0.863041$ and it shows how strong is linear
relationship between
$x$ and $y$ values.

#### (d)

$y$-intercept represents actually coefficient $b$ in linear

equation $y=ax+b$.

#### (e)

We will substitute $80, 150$ and $200$ for $x$ in equation of

best fit line and get that required costs are:

$2567$, $3592.5$ and $4325$, respectively.

#### (f)

We will substitute $80$ for $x$ in equation of a best fit line

and get that predicted value is $2567$, which means that

the actual cost of the well was $2567+363=2930$, because

the residual is difference between observed and predicted value.

#### (g)

We can see from the graph, because of graphed data, that a linear

model was the best choice for predicting well cost using well depth.

a) There are several outliers; b) $14.65$; c) $R^2=0.863041$

d) it represents coefficient $b$ in $y=ax+b$; e) $2567$, $3592.5$, $4325$;

f) $2930$; g) It was the best choice

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

$$
begin{align*}
& text{ For x=2 we solve this}\
&frac{2-4}{3}=frac{2}{15}\\
&frac{-2}{3}=frac{2}{15}\\
&boxed{color{#c34632}text{The sides are not equal}}\
end{align*}
$$

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

$$
begin{align*}
& text{ For x=2 we solve this}\
&left(2-2right)^2<0\\
&0<0\\
&boxed{color{#c34632}text{False}}\
end{align*}
$$

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

$$
begin{align*}
& text{ For x=2 we solve this}\
&left|3cdot :2-8right|ge :-1\\
&|6-8|ge-1\
&|-2|ge-1 \
&2ge-1 \
&boxed{color{#c34632}text{True}}\
end{align*}
$$

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

$$
begin{align*}
& text{ For x=2 we solve this}\
&sqrt{2+2}=16\\
&sqrt{4}=16\
&2=16 \\
&boxed{color{#c34632}text{The sides are not equal}}\
end{align*}
$$

$$
color{#4257b2} text{a)The sides are not equal}
$$

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

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

$$
color{#4257b2} text{d)The sides are not equal}
$$

Here, we can conclude that $DB=AC$, both triangles

have the right angle and the same side, $AB$, so,

according to the $SSA$ condition, those

triangles are congruent.

Let $x$ represent the weight of the coin.

According to the given information, we get the following

inequality with absolute value:

$$
|x-2.5|leq0.03cdot2.5
$$

$$
-0.03cdot2.5leq x-2.5leq0.03cdot2.5
$$

$$
2.5-0.03cdot2.5leq xleq2.5+0.03cdot2.5
$$

$$
2.425leq xleq2.575
$$

We can notice that $2.35inleft[2.425,2.575 right]$,

which means that the coin is legitimate.

On the following picture, there are graphed functions

$y=-left(dfrac{1}{3} right)^x$ and $y=-2.5x+3$.

From the graph, we can notice that the solution is point of

intersection, it is $(1.296,-0.241)$, so, this is exact values which

are solutions for $x$ and $y$.

$(1.296,-0.241)$

$$
text{color{#4257b2}We need to get the time needed to paint the room together.}
$$

Assume Susan paint $dfrac{1}{2}$ room per hour, and Jaime paint $dfrac{1}{3}$ room per hour.

Let represent rate together as follows:

$$
dfrac{1}{2}+dfrac{1}{3}=dfrac{1}{x}
$$

$$
dfrac{1}{x}=dfrac{2+3}{6}=dfrac{5}{6}
$$

$$
dfrac{1}{x}=dfrac{5}{6} x=dfrac{6}{5}=1.2
$$

The time needed to paint the room together is $1.2$ hour

Since the hour $=60$ minutes, so the time needed is equal:

$$
1.2 cdot (60)=72text{ Minutes}
$$

$$
72text{ Minutes}
$$

The error is that when Mr. Greer used the distributive property, he forgot to multiply 8 by $-3$.

$$
text{color{#4257b2}We need to resolve the following equation}
$$

$$
4x=8(2x-3)
$$

Use distributive property to simplify the equation as follows:

$$
4x=(8cdot2x)-(8cdot3) 4x=16x-24
$$

Isolate the variables on the left side as follows:

$$
4x-16x=-24 -12x=-24 12x=24
$$

Divide both sides by $12$ as follows:

$$
x=dfrac{24}{12} x=2
$$

The error is that when Mr. Greer used the distributive property, he forgot to multiply 8 by $-3$. The correct solution is $x=2$.

#### (b)

The conclusion is that the required equation is:

$$
y=10(2.3)^x
$$

We can see it from the following picture.

#### (c)

First, the exponential function which describes this situation is the following:

$$
y=42000(1.25)^x
$$

Now, we will substitute $5$ for $x$ in the previous equation and get:

$$
y=42000(1.25)^5
$$

$$
y=128173.83
$$

So, in five years, population will be $128173.83$.

#### (d)

On the following picture, there are graphed points $(2000,25)$ and $(2010,60)$ and the best fit line.

We can notice that annual multiplier is $1.09$ and percent increase is $9%$ every year.

a) Use a graphing calculator; b) $y=10(2.3)^x$;

c) $128173.83$; d) $1.09, 9%$

$$
text{color{#4257b2}How many student attended the play?}
$$

Assume number of adults is $x$ and number of students is $150+x$

and total cost is $=4730$ dollars.

$$
4730=5 x+3(150+x) 4730=5x+450+3x
$$

Isolate the variables on the left side as follows:

$$
8x=4730-450 8x=4280
$$

Divide both of sides by $8$ as follows:

$$
x=dfrac{4280}{8} x=535
$$

The number of students attended the play is

$$
150+x=150+535={color{#4257b2}685text{ students}}
$$

$$
Large{color{#4257b2}685text{ students}}
$$

$text{color{#4257b2}We need to get the length of $FG$ and $BC$}$.

Since the $FG$ is mid segment in triangle $ADE$, So $FG=dfrac{1}{2} DE$

$$
FG=dfrac{1}{2} cdot7=dfrac{7}{2} FG=3.5text{ cm}
$$

Since the $DE$ is mid segment in triangle $ABC$, So $DE=dfrac{1}{2} BC$

$$
BC=2 DE=2cdot7=14 BC=14text{ cm}
$$

$$
text{color{#4257b2}$$FG=3.5text{ cm} BC=14text{ cm} $$}
$$

$$
text{color{#4257b2}We need to get the time needed to paint the wall together.}
$$

Assume RJ rate in paint is $dfrac{1}{45}$ wall per hour, and Delaney rate in paint is $dfrac{1}{30}$ wall per hour.

Let represent rate together as follows:

$$
dfrac{1}{45}+dfrac{1}{30}=dfrac{1}{x}
$$

$$
dfrac{1}{x}=dfrac{30+45}{30cdot45}=dfrac{75}{1350}
$$

$$
dfrac{1}{x}=dfrac{75}{1350} x=dfrac{1350}{75}=18
$$

The time needed to paint the wall together is $18$ minutes

$$
text{color{#4257b2}Large$18$ minutes}
$$

#### (a)

Here we have two pairs of sides which are equal in the length and the

included angles are equal in measurement, which means that those

triangles are congruent according to the $SAS$ condition.

#### (b)

Here, we can conclude that angles in those triangles are $62^{circ}$, $45^{circ}$ and $73^{circ}$.

They have and one pair of sides equal length and the included angles

are equal in measurement, which means that those triangles are

congruent according to the $ASA$ condition.

#### (a)

We can notice that this is an arithmetic sequence where the first term is

$a_{1}=7$ and the constant difference is $d=-3$, so, its equation is following:

$$
a_{n}=7+(n-1)(-3)
$$

#### (b)

We can notice that this is a geometric sequence where first term

$a_{1}=2000$ and common ration is $q=3$, so, its equation is following:

$$
a_{n}=2000(3)^{n-1}
$$

a) $a_{n}=7+(n-1)(-3)$; b) $a_{n}=2000(3)^{n-1}$

We can write two expressions, one for Shu Min and one for Wei taking $x$ as number of months.

Shu Min : $1980 + 30x$

Wei : $2590 + 20x$

Add the expressions and set them equal to 5000.

$$
1980+30x + 2590 + 20x = 5000
$$

$$
50x = 5000-1980-2590
$$

$$
50x = 430
$$

$$
x = 8.6text{ months}
$$

$text{color{#4257b2}We need to select the expression which equivalent to the following equation}$.

$$
dfrac{1}{2}(6x-14)+5x=2-3x+8
$$

Use distributive property to simplify as follows:

$$
left(dfrac{1}{2} cdot 6xright)-left(dfrac{1}{2} cdot 14right)+5x=2-3x+8
$$

$$
3x-7+5x=10-3x
$$

The equivalent expression is first one (a).

$$
text{color{#4257b2}(a) $3x-7+5x=10-3x$}
$$

#### (a)

On the following picture, there are graphed given points and the best fit line.

We can notice that the required equation is:

$$
y=-6(3)^x
$$

#### (b)

On the following picture, there are graphed given points and the best fit line.

We can notice that the required equation is:

$$
y=7(0.5)^x
$$

a) $y=-6(3)^x$; b) $y=7(0.5)^x$

#### (a)

On the following picture, there is rotated $ABCDE$.

We can notice that coordinates of the point $A’$ are $A'(-1,-2)$.

#### (b)

On the following picture, there is reflected $ABCDE$.

We can notice that coordinates of the point $C’$ are $C’ (4,-4)$.

#### (c)

Point $B$ has coordinates $B(4,1)$, which means $x=4$, $y=1$.

Now, we have the following:

$$
x-1=4-1=3
$$

$$
y+3=1+3=4
$$

So, the coordinates which correspond to the point $B’$ are $B'(3,4)$.

a) $A'(-1,-2)$; b) $C'(4,-4)$; c) $B'(3,4)$

#### (a)

Here, for this relationship, the domain is set $D=[-2,2]$ while

the range is $R=[-4,4]$.

We can conclude that this relationship is not a function because

for the same value of $x$ from the domain there are two corresponding

values of $y$ from the range.

#### (b)

Here, for this realtionship the domain is set of all real numbers

while range is set $R=left{4 right}$.

We can conclude that this relationship is a function because for

every value of $x$ from the domain there is exactly one correspond point on graph, or $y$-value, and this is constant function.

#### (c)

Here, for this realtionship, the domain is set $D=[0,4]$ while

the range is set $R=[1,5]$.

We can conclude that this relationship is a function because for

every value of $x$ from the domain there is exactly one correspond point on graph, or $y$-value, and this is constant function.

#### (d)

Here, for this realtionship, the domain is set $D=(0,+infty)$, while
the range is set $R=Bbb/{R}left{ 0right}$.

We can conclude that this relationship is a function because for

every value of $x$ from the domain there is exactly one correspond point on graph, or $y$-value, and this is constant function.

a) $D=[-2,2]$, $R=[-4,4]$, is not a function; b) $D=Bbb{R}, R=left{ 4right}$, it is a function.

c) $D=[0,4]$, $R=[1,5]$, it is not a function;

d) $D=(0,+infty)$, $R=Bbb{R}/left{ 40right}$, it is not a function.

We can notice that this graph passes through the points $(0,1)$

and $(-1,15)$, so, we will sibstitute those values for $x$ and $y$

in the equation $y=ab^x$ and find $a$ and $b$ on that way:

$$
1=ab^{circ}Rightarrow a=1
$$

$$
15=ab^{-1}Rightarrowdfrac{1}{15}
$$

So, the required equation of this function is $y=left( dfrac{1}{15}right)^x$.

$y=left( dfrac{1}{15}right)^x$

Here, we have a finite number of data, so, as well we can

calculate the median and $IQR$ as the mean and standard

deviation.

So, we got that mean is $50.6913$ and the standard deviation

is $2.63$.

mean$=50.69$; standard deviation$=2.63$

We need to solve for $x$.

$$
2(12x+7)=30x-4
$$

Use distributive property to simplify as follows:

$$
(2cdot12x)+(2cdot7)=30x-4 24x+14=30x-4
$$

Isolate the variables on the left side as follows:

$$
24x-30x=-4-14 -6x=-18
$$

Divide both sides by $-6$ as follows:

$$
x=dfrac{-18}{-6} x=3
$$

$$
text{color{#4257b2}Large$$x=3$$}
$$

Let $x$ represent a number of gallons of powder blue and

$y$ represent a number of gallons of spring blue.

We get the following system of equations which we will

solve for $x$ and $y$:

$$
0.02x+0.1y=0.04
$$

$$
0.98x+0.9y=0.96
$$

We will solve it using a graph.

On the following picture, there are graphed both equations

and the solution is a point of intersection.

We can notice that the point of intersection is $(0.75,0.25)$,

which means that Antoine should use $0.75$ gallons of powder

blue and $0.25$ gallons of spring blue.

$0.75$ gallons of powder blue and $0.25$ of spring blue.

We need to solve the following inequalities.

$$
color{#4257b2}text{(a)} 5x-7ge2x+5
$$

Isolate the variables on the left side as follows:

$$
5x-2xge5+7 3xge12
$$

Divide both sides of inequality by $3$ as follows.

$$
xgedfrac{12}{3} xge4
$$

$$
color{#4257b2}text{(b)} 6x-29>4x+12
$$

Isolate the variables on the left side as follows:

$$
6x-4x>12+29 2x>41
$$

Divide both sides of inequality by $2$ as follows.

$$
x>dfrac{41}{2} x>20.5
$$

$$
color{#4257b2}text{(c)} x+6le6
$$

Isolate the variables on the left side as follows:

$$
xle6-6 xle0
$$

$$
color{#4257b2}text{(d)} |2x-7|>31
$$

Equal $2x-7>31$ as follows:

$$
2x-7>31 2x>31+7 2xdfrac{38}{2} x>19
$$

Equal $-2x+7>31$ as follows:

$$
-2x+7>31 -2x>31-7 -2x>24 x19$ or $x<-12$

(a) $x geq4$ (b) $x >20.5$

(c) $x leq0$ (d) $x>19$ or $x<-12$

On the following picture, there are graphed funcitions

$y=2^x$ and $y=5-x$.

The solution to the given equation is the point of intersecting

of the previous function.

So, using the previous picture, we can notice that the point

of an intersection, the solution is a point $(1.7,3.3)$.

$(1.7,3.3)$

When they have the same amount of money in their bank account?

Let represent total amount of Zachary money by $718-14x$

Let represent total amount of Christian money by $212+32x$

Where their have the same amount of money, so the two equations are equals as follows:

$$
718-14x=212+32x
$$

Isolate the variables on the left side as follows:

$$
-14x-32=212-718 -64x=-506
$$

Divide both sides of equation by $-64$ as follows.

$$
x=dfrac{-506}{-64} x=11text{ months}
$$

They have the same amount of money in their accounts after $11$ months.

$$
text{color{#4257b2}After $11$ months}
$$

Two lines are parallel if they have the same slope but different $y$-intercepts. So, we first rewrite the given line in slope-intercept form: $y=mx+b$

$$
begin{align*}
3x+2y&=10\
2y&=-3x+10\
y&=-dfrac{3}{2}x+5
end{align*}
$$

So, the slope of the parallel line is $m=-dfrac{3}{2}$. Since it goes through $(4,-7)$, we solve for the $y$-intercept $b$:

$$
begin{align*}
y&=mx+b\
-7&=-dfrac{3}{2}(4)+b\
-7&=-6+b\
-1&=b
end{align*}
$$

So, the equation of the line is:

$$
color{#c34632}y=-dfrac{3}{2}x-1
$$

$$
y=-dfrac{3}{2}x-1
$$

Let $y$ represents the number of tiles.

Let $x$ represents the figure number.

Equation: $y – 11 = 2 left( x – 3right)$

$$
begin{align*}
y – 11 & = 2x – 6 && {text {equation for the situation}} \
y – 11 + 11 & = 2x – 6 + 11 && {text {add 11 to both sides of the equation}} \
y & = 2x + 5 && {text {equation for the value of $y$}} \\
y & = 2 left(50right) + 5 && {text {substitute for the value of $x$}} \
& = 100 + 5 && {text {evaluate}} \
y & = 105 text { tiles} && {text {number of tiles for the figure}}
end{align*}
$$

$$
text{color{#4257b2}(a) What the relation between $DE$ and $AB$?}
$$

$DE$ is mid segment of $AB$ $DE=dfrac{1}{2}AB$

$$
text{color{#4257b2}(b) What the relation between $angle CDE$ and $angle DAB$?}
$$

$angle CDE$ is equal $angle DAB$ $CD=AD CE=BE DE||AB$

$$
text{color{#4257b2}(c) What the value of $x$ if $DE=4x+7$ and $AB=34$?}
$$

$$
because DE=dfrac{1}{2}AB therefore 2(4x+7)=34
$$

Use distributive property to simplify as follows:

$$
(2cdot4x)+(2cdot7)=34 8x+14=34
$$

Isolate the variables on the left side as follows:

$$
8x=34-14 8x=20
$$

Divide both of sides by $8$ as follows:

$$
x=dfrac{20}{8} x=2.5
$$

(a) $DE=dfrac{1}{2}AB$ (b) $angle CDE=angle DAB$ (c) $x=2.5$

#### (a)

On the following picture, there is graphed quadrilateral $ABCD$.

#### (b)

On the following picture, there are labeled points $E, F, G$ and $H$ on $ABCD$.

Coordinates of the midpoints are the following: $E(-4,1)$; $F(0,5)$; $G(3,1)$; $H(-1,-3)$.

Now, we will find the slopes:

$$
text{slope }EF=dfrac{3-1}{0+4}=1
$$

$$
text{slope }FG=dfrac{1-5}{3-0}=-dfrac{4}{3}
$$

$$
text{slope }GH=dfrac{-3-1}{-1-3}=1
$$

$$
text{slope }EH=dfrac{-3-1}{-1+4}=-dfrac{4}{3}
$$

We can notice that slope $EF=$slope $GH$ and slope $FG=$ slope $EH$, which means

that those sides are parallel, so, $EFGH$ is a parallelogram.

#### (d)

We will calculate $EF$ and $FG$:

$$
EF=sqrt{(0+4)^2+(5-1)^2}=sqrt{32}
$$

$$
FG=sqrt{(3-0)^2+(1-5)^2}=sqrt{25}=5
$$

We got that $EFne FG$, which means that $EFGH$ is not a rhombus.

a) Use GeoGebra; b) Use GeoGebra; c) $EFGH$ is a parallelogram;

d) $EFGH$ is not a rhombus

We have that $HE$ is the midsegment of triangle $ABD$, so,

$HE$ is parallel to $DB$.

$GF$ is also a midsegment of triangle $BCD$, so, $GF$ is parallel

to $BD$.

$HE$ is parallel to $BD$ and $BD$ is parallel to $GF$,

by transitivity, $HE$ is parallel to $GF$.

On the same way, $HG$ is parallel to $EF$, which means that

$EFGH$ is parallelogram.

The following table is a table of a relative frequency.\
begin{center}
begin{tabular}{ |c|c| c| c| c| c |c| }
hline
$$ & $text{Taco shack}$ & $text{competitor}$ \
hline
$25text{ grams cheese}$ & $30%$ & $29%$ \
hline
end{tabular}
end{center}
We can notice that there is almost no difference between\
the amount of cheese at Taco Shack and at the competitor.\
So, there is no association.\
The conclusion is that the Taco Shack owner does not need\
to adjust the amount of cheese.\