Answer:
1 cause its going to be one sometimes
<h3>
Answer: 17</h3>
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Explanation:
We'll start things off by computing the inner function u(2)
Plug x = 2 into the u(x) function
u(x) = -x-1
u(2) = -2-1
u(2) = -3
This tells us that w(u(2)) is the same as w(-3). I replaced u(2) with -3.
We'll plug x = -3 into the w(x) function
w(x) = 2x^2-1
w(-3) = 2(-3)^2 - 1
w(-3) = 2(9) - 1
w(-3) = 18-1
w(-3) = 17
Therefore, w(u(2)) = 17
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Here's a slightly different approach:
Let's find what w(u(x)) is in general
w(x) = 2x^2 - 1
w(u(x)) = 2(u(x))^2 - 1
w(u(x)) = 2(-x-1)^2 - 1
Then we can plug in x = 2
w(u(x)) = 2(-x-1)^2 - 1
w(u(2)) = 2(-2-1)^2 - 1
w(u(2)) = 2(-3)^2 - 1
w(u(2)) = 2(9) - 1
w(u(2)) = 18 - 1
w(u(2)) = 17
Answer:
first 60 digits of pi: 3.14159265358979323846264338327950288419716939937510582097494
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Step-by-step explanation:
Let us revise some logarithmic rules
∵ 10 log(x) + 5 log(64)
- At first re-write 10 log(x)
∴ 10 log(x) = 
- Then re-write 5 log(64)
∴ 5 log(64) = 
∴ 10 log(x) + 5 log(64) =
+ 
- Use the 3rd rule above to make it single logarithm
∵
+
= ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(64)^{5}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%2864%29%5E%7B5%7D%5D)
∵ 64 = 2 × 2 × 2 × 2 × 2 × 2
∴ We can write 64 as 
∴ 
- Multiply the two powers of 2
∴ 
∴ 10 log(x) + 5 log(64) = ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
The expression into a single logarithm is ![log[(x)^{10}][(2)^{30}]](https://tex.z-dn.net/?f=log%5B%28x%29%5E%7B10%7D%5D%5B%282%29%5E%7B30%7D%5D)
Learn more:
You can learn more about the logarithmic functions in brainly.com/question/11921476
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