<span>Center(0,0)
a^2=16
a=4
b^2=9
b=3
c^2=a^2+b^2=16+9=25
c=√25=5</span>
Answer:
The answer is 5
Step-by-step explanation:
I got it from goggle....your welcome
So g(x) doesn't come in
ok
remember pemdas
inside first
evaluate f(4)
f(4)=4^2+1=16+1=17
now we have
[17]^2=289
answer s 289
Answer:
![\sqrt[5]{2^4}](https://tex.z-dn.net/?f=%5Csqrt%5B5%5D%7B2%5E4%7D)
Step-by-step explanation:
Maybe you want 2^(4/5) in radical form.
The denominator of the fractional power is the index of the root. Either the inside or the outside can be raised to the power of the numerator.
![2^{\frac{4}{5}}=\boxed{\sqrt[5]{2^4}=(\sqrt[5]{2})^4}](https://tex.z-dn.net/?f=2%5E%7B%5Cfrac%7B4%7D%7B5%7D%7D%3D%5Cboxed%7B%5Csqrt%5B5%5D%7B2%5E4%7D%3D%28%5Csqrt%5B5%5D%7B2%7D%29%5E4%7D)
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In many cases, it is preferred to keep the power inside the radical symbol.
Answer:
i dont know lol but anyways
Step-by-step explanation: