a.) In the revenue function,

$R(x)=(30-2x)(100+20x)$

price per T-shirt = $30 – 2x$

number of T-shirt sold = $100+20x$

We shall describe what each factor in revenue expression represents.

Consider the revenue function,

$R(x)=(30-2x)(100+20x)$

where $R(x)$ is the revenue and $x$ is the number of times at which the price is reduced. Remember that revenue is the product of the T-shirt price and the total number of T-shirt sold. Practically, if the price is reduced, the quantity of T-shirt sold will increase while the price of each shirt will decrease.

b.) $R(x)=(30-2x)(100+20x)=0$

This implies that,

$30-2x=0$ or $100+2x=0$

$x=15$ or $x=-5$

Since the number of times can only take positive values, we shall discard the negative value

$x=15$ times

Find the value of $x$ where $R(x)=0$

begin{table}[]
begin{tabular}{|c|c|}
hline
begin{tabular}[c]{@{}c@{}}Number of \ Price Reductionend{tabular} & begin{tabular}[c]{@{}c@{}}Revenue (in dollars)\ $R(x)=(30-2x)(100+2x)$end{tabular} \ hline
0 & 3000 \ hline
1 & 3360 \ hline
2 & 3540 \ hline
3 & 3840 \ hline
4 & 3960 \ hline
5 & 4000 \ hline
6 & 3960 \ hline
7 & 3840 \ hline
end{tabular}
end{table}

e.) price per T-shirt = $30-2x$

$;;;;;;;=30-2(5)$

price per T-shirt = 20 dollars

From the table, for maximum revenue, $x=5$. Substitute it to the expression for the price per shirt.

number of T-shirt sold = $100+20x$

$=100+20(5)$]

number of T-shirt sold = 200 T-shirts

To determine the number of T-shirt sold for maximum revenue, substitute $x=5$ to the expression for number of T-shirt sold.

b.) The maximum distance from its starting point is at 180 cm, and this occurred at $t=0.25 s$

c.) distance = 180 cm

time = 0.25 s

speed = $dfrac{180 cm}{0.25 s}=720 cm/s$

We shall calculate the speed of the puck in the first 0.25 s. Remember the formula

$speed = dfrac{distance}{time}$

d.) The problem, the puck travels up to 2.5 s. Therefore, the domain is all real numbers within the close interval [0,2.5]. Alternatively, this can be written as

${t in bold{R} ;|; 0 leq t leq 2.5 }$

The distance from the edge is restricted by the size of the table. Thus, the range is

$$
{d in bold{R} ;|; 0 leq d leq 180 }
$$

${t in bold{R} ; |; 0 leq t leq 2.5 s }$, ${d in bold{R} ;| ;0 leq d leq 180 cm }$

$cos 40^circ=dfrac{y}{15}$

$$
y= 15 cos 40^circ
$$

We shall calculate the value of $theta$.

First, we need to calculate $y$. Remember that

$cos alpha = dfrac{adjacent;side}{hypotenuse}$

$sin theta = dfrac{15 cos 40^circ}{22}$

$theta = sin^{-1}left(dfrac{15 cos 40^circ}{22}right)$

$$
theta = 31.5^circ
$$

Now we can calculate $theta$ using the formula

$$
sin theta = dfrac{opposite;side}{hypotenuse}
$$

$tan 52^circ = dfrac{9}{x}$

$$
x=dfrac{9}{tan 52^circ}
$$

Use the formula

$$
tan alpha = dfrac{opposite;side}{adjacent;side}
$$

$tan (180^circ-theta) = dfrac{9}{11+9/tan 52^circ}$

$180^circ-theta= tan^{-1}left( dfrac{9}{11+9/tan 52^circ}right)$

$$
theta = 153.5^circ
$$

Now, that we have both legs of the bigger right triangle, we can calculate $theta$ using the formula

$tan alpha=dfrac{opposite;side}{adjacent;side}$

Also remember that straight angle has a measure of $180^circ$.

a.) $31.5^circ$ b.) $153.5^circ$

$sin35=dfrac{7}{9+x}$

$9+x=dfrac{7}{sin35}$

$x=dfrac{7}{sin35}-9$

$$
x=3.2
$$

We can use the sin function to determine the length of $9+x$. From which we can then solve for $x$. Sine is the opposite side length of the angle divided by the hypotenuse.

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$y=f(x)+c$ & begin{tabular}[c]{@{}l@{}}vertical translation of\ $c$ units upwardend{tabular} \ hline
$y=f(x+d)$ & begin{tabular}[c]{@{}l@{}}horizontal translation of $d$ units\ to the leftend{tabular} \ hline
$y=acdot f(x)$ & begin{tabular}[c]{@{}l@{}}vertical stretching by a factor of $a$end{tabular} \ hline
$y=f(kx)$ & begin{tabular}[c]{@{}l@{}}horizontal compression by $dfrac{1}{|k|}$end{tabular} \ hline
$y=-f(x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in\ the $x$-axisend{tabular} \ hline
$y=f(-x)$ & begin{tabular}[c]{@{}l@{}}reflecting the function in \ the $y$-axisend{tabular} \ hline
end{tabular}
end{table}

a) $y=-f(x)implies$ reflection in the $x$-axis

b) $y=3f(x)implies$ vertical stretching by factor 3

c) $f(x)+4implies$ translation 4 units upward

d) $-2f(x-3)implies$ reflection in $x$-axis, vertical stretching by factor 2 and translation 3 units to the right

a) reflection in $x$-axis

b) vertical stretch by factor 3

c) translation 4 units up

d) reflection in $x$-axis, vertical stretch by factor 2, translation 3 units to the right

The graphs have been plotted in the answers.

Sketch the triangle.

The problems mentioned that the angle of elevation is $32^circ$, thus the other angle must be $90^circ-32^circ=58^circ$

$dfrac{40}{sin 58^circ}=dfrac{x}{sin 32^circ}$

$x=dfrac{40}{sin 58^circ}cdot sin 32^circ$

$x=25$ m

Use sine law.

$dfrac{A}{sin alpha}=dfrac{B}{sin beta}$

Here, A must be the opposite side of $alpha$ and B is the opposite side of $beta$.

$x=25$ m

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$f(x)+k$ & shift $f(x)$ up $k$ units \ hline
$f(x)-k$ & shift $f(x)$ down $k$ units \ hline
$f(x+k)$ & shift $f(x)$ left $k$ units \ hline
$f(x-k)$ & shift $f(x)$ right $k$ units \ hline
$kcdot f(x)$ & multiply $y$-values by $k$ \ hline
$f(kx)$ & divide $x$-values by $k$ \ hline
$-f(x)$ & reflect $f(x)$ over $x$-axis \ hline
$f(-x)$ & reflect $f(x)$ over $y$-axis \ hline
end{tabular}
end{table}

The list of known transformations of functions with their description is shown.

We shall apply this transformation to

$$
y=f(x)=x^2
$$

$color{#c34632} y=x^2$

$color{#4257b2}y=x^2+2$

$color{#c34632} y=x^2$

$color{#4257b2}y=x^2-2$

$color{#c34632} y=x^2$

$color{#4257b2}y=(x+2)^2$

$color{#c34632} y=x^2$

$color{#4257b2}y=(x-2)^2$

$color{#c34632} y=x^2$

$color{#4257b2}y=2x^2$

Multiple $y$-values by 2

$color{#c34632} y=x^2$

$color{#4257b2}y=(2x)^2$

Divide $x$-values by 2

$color{#c34632} y=x^2$

$color{#4257b2}y=-x^2$

Reflect over $x$-axis.

$color{#c34632} y=x^2$

$color{#4257b2}y=(-x)^2$

Reflect over $y$-axis. Since $y=x^2$ is symmetric with respect to $y$-axis, the graph is the same.

begin{table}[]
defarraystretch{1.5}%
begin{tabular}{|l|l|}
hline
Transformation & Description \ hline
$f(x)+k$ & shift $f(x)$ up $k$ units \ hline
$f(x)-k$ & shift $f(x)$ down $k$ units \ hline
$f(x+k)$ & shift $f(x)$ left $k$ units \ hline
$f(x-k)$ & shift $f(x)$ right $k$ units \ hline
$kcdot f(x)$ & multiply $y$-values by $k$ \ hline
$f(kx)$ & divide $x$-values by $k$ \ hline
$-f(x)$ & reflect $f(x)$ over $x$-axis \ hline
$f(-x)$ & reflect $f(x)$ over $y$-axis \ hline
end{tabular}
end{table}