<h3>
Answer: Choice B</h3>
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Explanation:
The rule we use is
![\Large a^{m/n} = \sqrt[n]{a^m} = \left(\sqrt[n]{a}\right)^m](https://tex.z-dn.net/?f=%5CLarge%20a%5E%7Bm%2Fn%7D%20%3D%20%5Csqrt%5Bn%5D%7Ba%5Em%7D%20%3D%20%5Cleft%28%5Csqrt%5Bn%5D%7Ba%7D%5Cright%29%5Em)
where 'a' is the base, m stays in the role of the exponent, and n plays the role of the root index (eg: n = 3 is a cube root, n = 4 is a fourth root, and so on).
So for instance,
![\Large 2^{3/4} = \sqrt[4]{2^3} = \left(\sqrt[4]{2}\right)^3](https://tex.z-dn.net/?f=%5CLarge%202%5E%7B3%2F4%7D%20%3D%20%5Csqrt%5B4%5D%7B2%5E3%7D%20%3D%20%5Cleft%28%5Csqrt%5B4%5D%7B2%7D%5Cright%29%5E3)
or in this case,
![\Large t^{5/8} = \sqrt[8]{t^5} = \left(\sqrt[8]{t}\right)^5](https://tex.z-dn.net/?f=%5CLarge%20t%5E%7B5%2F8%7D%20%3D%20%5Csqrt%5B8%5D%7Bt%5E5%7D%20%3D%20%5Cleft%28%5Csqrt%5B8%5D%7Bt%7D%5Cright%29%5E5)
What’s the question asking??
I'm taking the liberty of editing your function as follows: <span>w=4r^2+6s^2, where I use " ^ " to indicate exponentiation.
The partial of w with respect to r is 8r. That with respect to s is 12s.</span>
(-4, 6), (0, 2), (4, -2) are all ordered pairs that have the sum as 2.
-4+6 is the same thing as 6-4 if there's more positive than negative
0+2=2 because 0 plus a number equals to that number
4-2=2 because you're adding a negative number with the positive number together to get the answer
Just be aware that I might not make any sense at all
Answer:
a.
Step-by-step explanation:
