To find the value of q, we need to find d(-8). Put another way, we need to find the value of d(x) when x = -8
So this means q = 0. Note that -0 is just 0.
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The value of r will be a similar, but now we use f(x) this time.
Plug in x = 0
Therefore, r = 2.
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For s, we plug x = 10 into f(x)
So s = 3.
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Finally, plug x = 10 into d(x) to find the value of t
A shortcut you could have taken is to note how d(x) = -f(x), so this means
d(10) = -f(10) = -9 since f(10) = 9 was found in the previous section above.
Whichever method you use, you should find that t = -3.
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<h3>In summary:</h3><h3>q = 0</h3><h3>r = 2</h3><h3>s = 3</h3><h3>t = -3</h3>
Answer:
A.oposite
Step-by-step explanation:
I'm not sure but it's help
Given:
The function is:
To find:
The range of the given function.
Solution:
We have,
The range of secant inverse function is:
The range of the given function in interval notation is:
![Range=\left[0,\dfrac{\pi}{2}\right)\text{ and }\left( \dfrac{\pi}{2}, \pi\right ]](https://tex.z-dn.net/?f=Range%3D%5Cleft%5B0%2C%5Cdfrac%7B%5Cpi%7D%7B2%7D%5Cright%29%5Ctext%7B%20and%20%7D%5Cleft%28%20%5Cdfrac%7B%5Cpi%7D%7B2%7D%2C%20%5Cpi%5Cright%20%5D)
Therefore, the correct option is C.
Rather than solve the entire problem for you, I'll give you some hints to help you get started:
1. The amplitude of your sinusoidal graph is |3|, or just 3.
2. Because of that, your graph begins at the point (0,3).
3. Because this is the cosine function, your graph descends from (0,3) to y=3 and then begins to ascend (back to y=3).
4. The coefficient of x is "one half pi," or pi/2. Call this "b".
5. The period of your function is 2pi/b. Here, b=pi/2.
Dividing, [2pi]/[pi/2] = 4.
6. Mark your horizontal axis as follows: x=0, 4, 8, 12, 16, ...
7 Draw one cycle of the cosine function with amplitude 3. It must begin at (0,3) and end at (4,3) (which covers one period).
8. Draw another cycle or two, beginning at (4,3) and ending at (8,3), and so on.
