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2 years ago

Choose the graph of y = 2 cos x

Mathematics

2 answers:

drek231 [11]2 years ago

5 0

Y = cos x begins at (0,1), descends to (pi/2,0), descends to (pi,-1), ascends to (3pi/2,0), and then ascends to (2pi,1). This covers only one full period.

The graph of y = 2cost x is the same, except that the amplitudes (y-values) are 2x as great as before:

(0,2), (pi/2, 0), and so on.

Butoxors [25]2 years ago

5 0

Answer:

The graph is in the attachment.

Step-by-step explanation:

The cosine function is periodic. This means that every 2pi, the image is the same.

For y = 2 cos x, the amplitude of the function is given by 2.

The domain of the function is (-∞,∞).

The image of the function is [-2, 2].

For example:

y(0) = 2 cos 0 = 2 . 1 = 2.

y(pi/2) = 2 cos(pi/2) = 2 * 0 = 0.

y(pi) = 2 cos(pi) = 2 * -1 = -2.

y(3pi/2) = 2 cos(3pi/2) = 2 * 0 = 0.

y(2pi) = 2 cos(2pi) = 2 * 1 = 2.

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bija089 [108]

Part A:

The average rate of change refers to a function's slope. Thus, we are going to need to use the slope formula, which is:

$m = \dfrac{y_2 - y_1}{x_2 - x_1}$

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You can see that we are given the x-values for our interval, but we are not given the y-values, which means that we will need to find them ourselves. Remember that the y-values of functions refers to the outputs of the function, so to find the y-values simply use your given x-value in the function and observe the result:

$h(0) = 3(5)^0 = 3 \cdot 1 = 3$

$h(1) = 3(5)^1 = 3 \cdot 5 = 15$

$h(2) = 3(5)^2 = 3 \cdot 25 = 75$

$h(3) = 3(5)^3 = 3 \cdot 125 = 375$

Now, let's find the slopes for each of the sections of the function:

Section A

$m = \dfrac{15 - 3}{1 - 0} = \boxed{12}$

Section B

$m = \dfrac{375 - 75}{3 - 2} = \boxed{300}$

Part B:

In this case, we can find how many times greater the rate of change in Section B is by dividing the slopes together.

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It is 25 times greater. This is because $3(5)^x$ is an exponential growth function, which grows faster and faster as the x-values get higher and higher. This is unlike a linear function which grows or declines at a constant rate.

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arlik [135]

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Suppose a spherical balloon grows in such a way that after t seconds, its volume is V = 4 sqrt(t) cm3. What is the volume of the

Arisa [49]

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