$x^4+3x^3=4x^2+12x$

$x^4+3x^3-4x^2-12x=0$

$(x^4+3x^3)+(-4x^2-12x)=0$

$x^3(x+3)-4x(x+3)=0$

$(x+2)(x^3-4x)=0$

$x(x+3)(x^2-4)=0$

$x(x+3)(x+2)(x-2)=0$

$x=0$ or $x=-3$ or $x=-2$ or $x=2$, so, we can conclude that $textbf{the correct answer is (d)}$.

$f(x)=a(x+1)(x-1)(x-4)$

$36=a(2+1)(2-1)(2-4)$

$36=a(3)(1)(-2)$

$a=-6$

So, $textbf{the equation is}$

$f(x)=-6(x+1)(x-1)(x-4)=-6(x^2-1)(x-4)=-6(x^3-4x^2-x+4)=-6x^3+24x^2+6x-24$

So, $textbf{the correct answer is}$ (b).

$2-3x < x-5$

$2-4x < -5$

$-4x dfrac{7}{4}$

$-2$ is not greater than $dfrac{7}{4}$, so, $textbf{the correct answer is (a)}$.

$-10 leq 3x +5 leq 8$

$-15leq 3xleq3$

$-5leq xleq1$

$textbf{The correct answer is c)}$.

Using the graph $f(x) 2$.

$textbf{The correct answer is a)}$.

$h(t)=-5t^2+3.5t+10$

$-5t^2+3.5t+10 > 10$

$-5t^2+3.5t > 0$

$t(-5t+3.5) > 0$

$textbf{The critical points are}$ $t=0$ and $t=0.7$.

Test:$-1: (-1)(-5(-1)+3.5) > 0 Rightarrow (-1)(5+3.5) > 0$ FALSE

Test $0.5: (0.5)(-5(0.5)+3.5) > 0 Rightarrow (0.5)(-2.5+3.5) > 0$ TRUE

Test $1: (1)(-5(1)+3.5) > 0 Rightarrow (1)(-5+3.5) > 0$ FALSE

So, $textbf{the solution is}$ $tin (0,0.7)$

$textbf{The correct answer is (b)}$.

From the given information, the function must have a maximum at $x=0$ and a minimum at $x=2$.
Choice a) is not a posiible set of zeros only for the function.If the function has zeros only at $x=0$ and $x=1$, then the function has to increase to $(0, 0)$ and turn there and begin to decrease. Sometime before $x=1$, the function must turn again and increase to get back to the $x$-axis before $x=1$.
But this contradicts the given information that the function is decreasing for $0<x<2$.

$textbf{The correct answer is a)}$.

$f(x)=2x^3-4x^2+6x$

$f(-0.05)=2(-0.05)^3-4(-0.05)^2+6(-0.05)=-0.31025$

Find the rate of change using the points $(0,0)$ and $(-0.05,-0.31025)$.

$m=dfrac{-0.31025-0}{-0.05-0}=6.2$

$textbf{So, the correct answer is (c)}$.

The graph has vertical asymptotes at $x^2-3x=0$
$x(x-3)=0$

$x=0$ and $x=3$

Test $x=-1$: The function will be $(+)div(-)(-)=(+)$

So, the function is poitive to the left of zero.

Test $x=1$: The function will be $(+)div(+)(-)=(-)$

So,the function is negative between $0$ and $3$.

Test $x=4$: The function will be $(+)div(+)(+)=(+)$

So, the function is positive to the right of $3$.

$textbf{The function that matches this is c)}$.

The function will have $textbf{a horizontal asymptote}$ of $0$ and $textbf{vertical asymptote}$ of $x=-5$ and $x=2$.

$textbf{So, the correct answer is (c)}$.

To have an $textbf{oblique asymptote}$, the degree of the numerator must be greater than the degree of the denominator by exactly $1$. Neither choice (a) nor choice (c) meets this condition.While it seems that choice (b) meets the condition, it does not because the numerator can be factored and the function simplified,

$g(x)=dfrac{(x+3)(x-3)}{x-3}=x+3$.

$textbf{The correct answer is (d)}$.

Only functions $a$ and $b$ are undefined at $x=3$.
Examine the behaviour of these functions for $-2<x<3$.
For choice a: Test $x=0:(+)div(+)=(+)$

For choice b: Test $x=0: (+)div(-)=(-)$

So, choice a is positive for $-2<x<3$

$textbf{The correct answer is a)}$.

Choose a value that is very close to and to the left of $dfrac{3}{5}$.

Try $0.599$:

$(2-3(0.599))div(5(0.599)-3)=-40.6$

So, the function approaches $-infty$,

$textbf{The correct answer is d)}$.

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

$3-2x=3x(x+2)$

$3-2x=3x^2+6x$

$3x^2+8x-3=0$

$(3x-1)(x+3)=0$

$x=dfrac{1}{3}$ or $x=-3$.

$textbf{The correct answer is (c)}$.

Any of the steps listed can be used to begin solving the rational equation.

$textbf{The correct answer is (d)}$.

$2x-3 leq dfrac{2}{x}$

$2x-3-dfrac{2}{x} leq 0$

$dfrac{2x^2}{x}-dfrac{3x}{x}-dfrac{2}{x} leq 0$

$dfrac{2x^2-3x-2}{x} leq 0$

$dfrac{(2x+1)(x-2)}{x} leq 0$

$textbf{So, the correct answer is (a)}$.

$x-3>dfrac{6}{x-2}$

$x-3-dfrac{6}{x-2}>0$

$dfrac{x(x-2)}{x-2}-dfrac{3(x-2)}{x-2}-dfrac{6}{x-2}>0$

$dfrac{x^2-2x-3x+6-6}{x-2}>0$

$dfrac{x^2-5x}{x-2}>0$

$dfrac{x(x-5)}{x-2}>0$

The critical points are $0,5$ and $2$.

Test $-1: (-)(-)div(-)=(-)$ FALSE

Test $1: (+)(-)div(-)=(+)$ TRUE

Test $3 : (+)(-)div(+)=(-)$ FALSE

Test $6: (+)(+)div(+)=(+)$ TRUE

So, the solution is $(0, 2)$ and $(5, infty)$.

$textbf{The correct answer is d)}$.

Let $y=f(x)$.

$dfrac{G(a+h)-G(a)}{h}=dfrac{G(1.001)-G(1)}{0.001}=dfrac{0.9985-1}{0.001}=-1.5$

$textbf{So, the correct answer is (b)}$.

$dfrac{s(a+h)-s(a)}{h}=dfrac{s(3.001)-s(3)}{0.001}=dfrac{-7.009009-(-7)}{0.001}=-9$

$textbf{So, the correct answer is (b)}$.

$P=3(dfrac{5pi}{12})+2cdot3=dfrac{5pi}{4}+6$ m.

$textbf{So, the correct answer is (b)}$.

$20(dfrac{pi}{180})=dfrac{pi}{9}$

$135(dfrac{pi}{180})=dfrac{3pi}{4}$

$-270(dfrac{pi}{180})=-dfrac{3pi}{2}$

Each of the pairs of angles are equivalent, $textbf{so, the correct answer is (d)}$.

$tan^{-1}(dfrac{7}{4})=1.0517$

$pi-1.0517=2.09$

$textbf{The correct answer is (c)}$.

Since $sin theta=-dfrac{sqrt{3}}{2}$, the value for $r$ is $2$ and the value for $a=sqrt{3}$. So, let’s find $b$:

$(sqrt{3})^2+b^2=2^2$

$3+b^2=4$

$b^2=1$

$b=1$

The angle is in either in the third or fourth quadrant since the $sin$ is negative.

So, $cos theta=dfrac{1}{2}$ or $-dfrac{1}{2}$.

$tan theta=sqrt{3}$ when $cos theta=-dfrac{1}{2}$ or $-sqrt{3}$ when $cos theta=dfrac{1}{2}$.

$textbf{The correct answer is (a)}$.

$x=sin^{-1}0.5$

$x=dfrac{pi}{6}$ and $x=dfrac{5pi}{6}$.

$textbf{So, the correct answer is (d)}$.

The graph shown has been $textbf{stretched vertically}$ by a factor of $3$, since the minimum is $-4$ and the maximum is $2$, a difference of $6$.The graph has also been $textbf{compressed horizontally}$ by a factor of $dfrac{1}{2}$, as the period shown is only $pi$.The graph has been $textbf{translated}$ down $1$ unit.The equation for the graph shown is:

$y=3sin(2x)-1$.

$textbf{So, the correct answer is (b)}$.

The equation shows a horizontal stretch by a factor of $3$ and a horizontal translation of $2pi$ units to the left.

$textbf{The correct answer is d)}$.

The maximum heigth represented by a choice b) is $45-5=36$, not $41$ as required.
The maxmum height represented by choice c) is $18-23=-5$, not $41$ as required.
The maximum height represented by choice d) is $41-36=5$, not $41$ as required.
The maximum heigth represented by choice a) is $18+23=41$, as required.The minimum height represented by choice a) is $-18+23=5$, as required.

$textbf{The correct answer is a)}$.

Using a graphing calculator, the function is decreasing in both intervals which means the instaneous rate of change is negative in both intervals.

$textbf{So, the correct answer is (c)}$.

$P(0)=23.7cos(dfrac{pi}{6}(0-7)+24.1)=3.575$

$P(4)=23.7cos(dfrac{pi}{6}(4-7)+24.1)=24.1$

So, $textbf{the rate of change is }dfrac{24.1-3.575}{4-0}=5.13125$

$P(1)=23.7cos(dfrac{pi}{6}(1-7)+24.1)=0.4$

$P(7)=23.7cos(dfrac{pi}{6}(7-7)+24.1)=47.8$

So, $textbf{the rate of change is }dfrac{47.8-0.4}{7-1}=7.9$

$P(16)=23.7cos(dfrac{pi}{6}(16-7)+24.1)=24.1$

So, $textbf{the rate of change is }dfrac{24.1-47.8}{16-7}=-2.63$

$P(10)=23.7cos(dfrac{pi}{6}(10-7)+24.1)=24.1$

$P(18)=23.7cos(dfrac{pi}{6}(18-7)+24.1)=44.62$

So, $textbf{the rate of change is }dfrac{44.62-24.1}{18-10}=2.565$

$textbf{The correct answer is (b)}$.

#### (a)

length$=50-2x$

width$=40-2x$

$V=lwh$

$=(50-2x)(40-2x)x$

#### (b)

$6000=(50-2x)(40-2x)x$

$6000=(2000-80x-100x+4x^2)x$

$6000=2000x-180x^2+4x^3$

$0=4x^3-180x^2+2000x-6000$

$0=x^3-45x^2+500x-1500$

The possible reasonable solutions are $pm1, pm 2, pm 3, pm4,pm5,pm6,pm10,pm12, pm15, pm20$.

Use synthetic division to determine the solution…

So, $x=5$.

Now solve $x^2-40x+300=0$

$(x-10)(x-30)=0$

So, $x=10$ or $30$, but $30$ does not make sense in the context of the problem. So , $x=5$ or $x=10$.

#### (c)

Using a graphing calculator, find the relative maximum value.It occurs when $x=7.4$cm.

#### (d)

Using a graphing calculator, the range is $3<x<12.8$

#### (a)

$f(x): 0=x^2-5x+6$

$0=(x-3)(x-2)$

So, $x=3$ and $x=2$

$g(x): 0=x-3 Rightarrow x=3$

$dfrac{f(x)}{g(x)}: 0=dfrac{x^2-5x+6}{x-3}$

The function is undefined at $x=3$, so, the zero is $x=2$.

$dfrac{g(x)}{f(x)}: 0=dfrac{x-3}{x^2-5x+6}$

The function is undefined at $2$ and $3$, so, there are no zeros.

#### (b)

$dfrac{f(x)}{g(x)}:$ hole at $x=3$, no asymptotes

$dfrac{g(x)}{f(x)}:$ hole at $x=3$, asymptotes at $x=2$ and $y=0$
#### (c)

$dfrac{f(x)}{g(x)}:y=x-2$

$dfrac{g(x)}{f(x)}:y=-x$.

#### (a)

Vertical compressions and streches do not affect location of zeros; maximum and minimumm values are multiplied by the scale factor, but locations are unchanged; instantaneous rates of change are multiplied by the scale factor;
Horizontal compressions and strecthes move locations of zeros, maximus, and minimums toward or away from the $y$-axis by the reciprocal of the scale factor; instantaneous rates of change are multiplied by the reciprocal of scale factor; Vertical translations change location of zeros or remove them; maximum and minimum values are increased or decreased by the amount of the translation, but locations are unchanged; instantaneous rates of change are unchanged; Horizontal translations move location of zeros by the same amount as the translation;
maximum and minimum values are unchanged, but locations are moved by the same amount as the translation; instantaneous rates of change are unchanged, but locations are moved by the same amount as the translation.

#### (b)

$y=cos x$: Vertical compressions and stretches do not affect location of zeros; maximum and minimum values are multiplied by the scale factor,but locations are unchanged; instantaneous rates of change are multiplied by the scale factor; Horizontal compressions and stretches move locations of zeros,maximus, and minimums toward or away from the $y$-axis by the reciprocal of the scale factor; instantaneous rates of change are multiplied by the reciprocal of scale factor; Vertical translations change location of zeros or remove them; maximum and minimum values are increased or decreased by the amount of the translation, but locations are unchanged; instantaneous rates of change are unchanged; Horizontal translations move location of zeros by the same amount as the translation; maximum and minimum values are unchanged, but locations are moved by the same amount as the translation; instantaneous rates of change are unchanged, but locations are moved by the same amount as the translation.

$y=tan x$: Vertical compressions and stretches do not affect location of zeros; instantaneous rates of change are multiplied by the scale factor;
Horizontal compressions and stretches move locations of zeros toward or away from the $y$-axis by the reciprocal of the scale factor; instantaneous rates of change are multiplied by the reciprcal of scale factor; Vertical translations change location of zeros or remove them; instantaneous rates of change are unchanged; Horizontal translations move location of zeros by the same amount as translation; instantaneous rates of change are unchanged, but locations are moved by the same amount as the translation.