Question

1.The table provides various values, including all minimums and maximums, of a cosine function f (x) on the interval [0, 2π].

Answer 1

Problem 1

Answer: D) 2pi

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Explanation:

The given table we have is

$\begin{array}{|c|c|c|c|c|c|c|} \cline{1-7}\text{Angle} & 0 & \pi/6 & \pi/2 & 5\pi/6 & 3\pi/2 & 2\pi\\ \cline{1-7}f(x) & 1 & 0 & -1 & 0 & 3 & 1\\ \cline{1-7}\end{array}$

If you were to graph all those points and draw a cosine curve through them, then the function you should get is

$y = 2\cos\left(x+\frac{\pi}{2}\right)+1$

The B coefficient is 1, so that leads to a period of

T = 2pi/B

T = 2pi/1

T = 2pi

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Problem 2

Answer: C) 343.134 yards

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Explanation:

Draw out a triangle with these properties

• side a = 270
• side b = 170
• angle C = 100

We'll use the law of cosines to find side c

c^2 = a^2 + b^2 - 2*a*b*cos(C)

c^2 = 270^2 + 170^2 - 2*270*170*cos(100)

c^2 = 117740.902710

c = sqrt(117740.902710)

c = 343.133943

c = 343.134

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Problem 3

Answer: A) 2

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Explanation:

If you started with y = sin(x) and applied the transformations stated, then you should end up with $y = 2\sin\left(2(x+\frac{\pi}{4})\right)$

From here, plug in x = 0 to find the y intercept.

$y = 2\sin\left(2(x+\frac{\pi}{4})\right)\\\\y = 2\sin\left(2(0+\frac{\pi}{4})\right)\\\\y = 2\sin\left(\frac{\pi}{2}\right)\\\\y = 2*1\\\\y = 2$