Q: Given the point $$(-4,\sqrt{12})$$ on the terminal side of angle $$\theta$$, what is $$\tan\theta$$?

Use the definition.$$\tan\theta=\frac{y}{x}$$Substitute.$$\tan\theta=\frac{\sqrt{12}}{-4}$$Simplify.$$\sqrt{12}=2\sqrt{3}$$$$\tan\theta=-\frac{2\sqrt{3}}{4}$$$$\tan\theta=-\frac{\sqrt{3}}{2}$$Final answer: $$-\frac{\sqrt{3}}{2}$$

Q: Given the rectangular coordinate $$(-2,-2\sqrt{3})$$, what is an equivalent polar coordinate with $$0\le\theta$$<$$2\pi$$ and $$r$$<$$0$$?

Compute the radius.$$r=\sqrt{(-2)^2+(-2\sqrt{3})^2}$$$$r=\sqrt{4+12}$$ $$r=4$$Find the angle.$$\tan\theta=\frac{-2\sqrt{3}}{-2}=\sqrt{3}$$Reference angle:$$\frac{\pi}{3}$$Point is in Quadrant III.$$\theta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}$$For negative radius, add $$\pi$$.$$\theta=\frac{4\pi}{3}+\pi=\frac{7\pi}{3}$$Bring into interval $$0\le\theta$$<$$2\pi$$.$$\frac{7\pi}{3}-2\pi=\frac{\pi}{3}$$Final answer: $$(-4,\frac{\pi}{3})$$

Q: The range of the function $$f(x)=3\sec(2x-\pi)-2$$ is:

The range of $$\sec x$$ is $$(-\infty,-1]\cup[1,\infty)$$Multiply by $$3$$.$$(-\infty,-3]\cup[3,\infty)$$Shift down by $$2$$.$$(-\infty,-5]\cup[1,\infty)$$Final answer: $$(-\infty,-5]\cup[1,\infty)$$

Question 1

A geometric sequence has $$a_3=18$$ and $$a_5=162$$. Which explicit formula represents the sequence?

Accepted Answer

For a geometric sequence, $$a_n=a_1r^{n-1}$$Use the given terms.$$a_3=a_1r^2=18$$$$a_5=a_1r^4=162$$Divide.$$\frac{a_5}{a_3}=r^2=\frac{162}{18}=9$$$$r=3$$Substitute back.$$a_1\cdot3^2=18$$$$a_1=2$$So,$$a_n=2\cdot3^{n-1}$$Final answer: $$2\cdot3^{n-1}$$

Question 2

A geometric sequence has $$a_3=18$$ and $$a_5=162$$. Which explicit formula represents the sequence?

Accepted Answer

For a geometric sequence,$$a_n=a_1r^{n-1}$$Use the given terms.$$a_3=a_1r^2=18$$$$a_5=a_1r^4=162$$Divide.$$\frac{a_5}{a_3}=r^2=\frac{162}{18}=9$$$$r=3$$Substitute back.$$a_1\cdot3^2=18$$$$a_1=2$$So,$$a_n=2\cdot3^{n-1}$$Final answer: $$2\cdot3^{n-1}$$

Question 3

A bacteria culture triples every $$6$$ hours. Write an equation to model the population $$y$$ after $$t$$ days, given an initial amount $$y_0$$.

Accepted Answer

There are $$4$$ periods of $$6$$ hours in one day.So, the number of growth cycles is $$4t$$The model is $$y=y_0\cdot3^{4t}$$Final answer: $$y=y_0\cdot3^{4t}$$

Question 4

A bacteria culture triples every $$6$$ hours. Write an equation to model the population $$y$$ after $$t$$ days, given an initial amount $$y_0$$.

Accepted Answer

There are $$4$$ periods of $$6$$ hours in one day.So, the number of growth cycles is $$4t$$The model is $$y=y_0\cdot3^{4t}$$Final answer: $$y=y_0\cdot3^{4t}$$

Question 5

Is the function $$f(x)=2\sin(5x)+3\cos(5x)$$ sinusoidal on its domain?

Accepted Answer

Both terms have the same angular frequency. $$5$$A sum of sine and cosine with the same frequency can be written as a single sinusoidal function.Final answer: $$	ext{Yes, same frequency}$$

Question 6

Given the point $$(-4,\sqrt{12})$$ on the terminal side of angle $$	heta$$, what is $$	an	heta$$?

Accepted Answer

Use the definition.$$	an	heta=\frac{y}{x}$$Substitute.$$	an	heta=\frac{\sqrt{12}}{-4}$$Simplify.$$\sqrt{12}=2\sqrt{3}$$$$	an	heta=-\frac{2\sqrt{3}}{4}$$$$	an	heta=-\frac{\sqrt{3}}{2}$$Final answer: $$-\frac{\sqrt{3}}{2}$$

Question 7

Given the rectangular coordinate $$(-2,-2\sqrt{3})$$, what is an equivalent polar coordinate with $$0\le	heta$$<$$2\pi$$ and $$r$$<$$0$$?

Accepted Answer

Compute the radius.$$r=\sqrt{(-2)^2+(-2\sqrt{3})^2}$$$$r=\sqrt{4+12}$$ $$r=4$$Find the angle.$$	an	heta=\frac{-2\sqrt{3}}{-2}=\sqrt{3}$$Reference angle:$$\frac{\pi}{3}$$Point is in Quadrant III.$$	heta=\pi+\frac{\pi}{3}=\frac{4\pi}{3}$$For negative radius, add $$\pi$$.$$	heta=\frac{4\pi}{3}+\pi=\frac{7\pi}{3}$$Bring into interval $$0\le	heta$$<$$2\pi$$.$$\frac{7\pi}{3}-2\pi=\frac{\pi}{3}$$Final answer: $$(-4,\frac{\pi}{3})$$

Question 8

The range of the function $$f(x)=3\sec(2x-\pi)-2$$ is:

Accepted Answer

The range of $$\sec x$$ is $$(-\infty,-1]\cup[1,\infty)$$Multiply by $$3$$.$$(-\infty,-3]\cup[3,\infty)$$Shift down by $$2$$.$$(-\infty,-5]\cup[1,\infty)$$Final answer: $$(-\infty,-5]\cup[1,\infty)$$

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