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Questions 1 - 10

Find the derivative of the function $f(x)=x^{3}+5$ .

$f'(x)=3x^{2}$

$f'(x)=3x^{2}+5$

$f'(x)=3x^{3}$

$f'(x)=\frac{1}4{}x^{4}$

None of the above

Explanation

For any function $f(x)=Ax^{n}$ , the first derivative $f'(x)=nAx^{n-1}$ .

Therefore, taking each term of :

$f(x)=x^{3}+5$

$f'(x)=3x^{3-1}+0(5)$

$f'(x)=3x^{2}$

Geo_8_sec_6_graphik_3

If each mark on the graph represents units, what is the equation of the circle?

Explanation

Since the circle is centered at we use the most basic form for the equation for a circle:

Given the circle has a radius of marks, which represent units each, the circle has a radius of units.

We then plug in for :

and simplify: .

Solve for .

$2e^5e^{-3}\ln(e^x)=2x^2$

$x=\ln(2e)$

$x=2\ln(e^2)$

Explanation

First, let's begin by simplifying the left hand side.

$e^5e^{-3}$ becomes and $\ln(e^x)$ becomes . Remember that $\ln(e)=1$ , and the in that expression can come out to the front, as in $x\cdot \ln(e)$ .

Now, our expression is

From this, we can cancel out the 2's and an x from both sides.

Thus our answer becomes:

Convert the polar coordinates to rectangular form:

$\left(10, \frac{4\pi}{3}\right)$

$(-5, -5\sqrt3)$

$(-5\sqrt3, -5)$

$(5\sqrt3, -5)$

$(5, 5\sqrt3)$

Explanation

To convert polar coordinates $(r, \theta)$ to rectangular coordinates ,

$x=r\cos\theta$

$y=r\sin\theta$

Using the information given in the question,

$x=10\cos \frac{4\pi}{3}=10\left(-\frac{1}{2}\right)=-5$

$y=10 \sin \frac{4\pi}{3}=10\left(-\frac{\sqrt3}{2}\right)=-5\sqrt3$

The rectangular coordinates are $(-5, -5\sqrt3)$

Let

$A=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}$ $B=\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}$

Determine the sum .

$\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

DNE

$\begin{pmatrix} 1&-1 \ -6&7 \end{pmatrix}$

$\begin{pmatrix} 4&2 \ 9&-7 \end{pmatrix}$

Explanation

Since the dimensions of the two matrices are equal the sum of the two matrices exists.

To find the sum, add each component entry from the first matrix to the same component entry of the second matrix.

$A+B=\begin{pmatrix} 5&1 \ 3&-7 \end{pmatrix}+\begin{pmatrix} 4&2 \ 9& 0 \end{pmatrix}}=\begin{pmatrix} 9&3 \ 12&-7 \end{pmatrix}$

Write the equation in polar form

$r = 2+3 \sin \theta$

$r = 3 \sin \theta$

$r = 4 + 3 \sin \theta$

$r = 4 - 3 \sin \theta$

$r = 2 - 3 \sin \theta$

Explanation

First re-arrange the original equation so that the 4 is factored out on the right side, and put and next to each other:

Make the substitutions and $y = r \sin \theta$ :

$(r^2 - 3 r \sin \theta )^2 = 4r^2$ take the square root of both sides

$r^2 - 3 r \sin \theta = 2 r$ divide both sides by r

$r - 3 \cos \theta = 2$ add $3 \cos \theta$ to both sides

$r = 2 + 3 \cos \theta$

Write the parametric equation for the line y = 5x - 3.

x = 5t - 3

y = t

x = t

y = 5t - 3

x = 5t - 3

y = 5t - 3

x = t

y = t

Explanation

In the equation y = 5x - 3, x is the independent variable and y is the dependent variable. In a parametric equation, t is the independent variable, and x and y are both dependent variables.

Start by setting the independent variables x and t equal to one another, and then you can write two parametric equations in terms of t:

x = t

y = 5t - 3

Find $\overrightarrow{PQ}$ if and .

Explanation

To find the direction vector going from to , subtract the x and y-coordinates of from .

$\overrightarrow{PQ}=<-5-4, 2-1>= <-9, 1>$

Find where .

Explanation

In order to find the derivative we will need to use the power rule on each term. The power rule states,

$f(x)=x^n \rightarrow f'(x)=nx^{n-1}$ .

Applying this rule we get the following.

$y'=(4*5)x^{5-1}+(3*3)x^{3-1}+(6*2)x^{2-1}+(1)'$

Which of the following functions is represented by this graph?

Cosine

y = sin(x)

y = cos(x)

y = tan(x)

y = sec(x)

y = csc(x)

Explanation

This graph is the graph of y = cos x. The domain of this function is all real numbers. The range of this function is $-1\leq y\leq 1$ . The period of this function is $2\pi$ .

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