Answer:
29/40
Step-by-step explanation:
Finding the inverse function of

Remember that when you compose

with its inverse

you'll get the identity function:
![\mathsf{(f\circ f^{-1})(x)=x}\\\\ \mathsf{f\big[f^{-1}(x)\big]=x}\\\\](https://tex.z-dn.net/?f=%5Cmathsf%7B%28f%5Ccirc%20f%5E%7B-1%7D%29%28x%29%3Dx%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3Dx%7D%5C%5C%5C%5C)
So if

then
![\mathsf{f\big[f^{-1}(x)\big]=10-[f^{-1}(x)]^2}\\\\ \mathsf{x=10-[f^{-1}(x)]^2}\\\\ \mathsf{[f^{-1}(x)]^2=10-x}\\\\ \mathsf{f^{-1}(x)=\pm \sqrt{10-x}}\\\\ \mathsf{f^{-1}(x)=-\sqrt{10-x}~~~or~~~f^{-1}(x)=\sqrt{10-x}}](https://tex.z-dn.net/?f=%5Cmathsf%7Bf%5Cbig%5Bf%5E%7B-1%7D%28x%29%5Cbig%5D%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7Bx%3D10-%5Bf%5E%7B-1%7D%28x%29%5D%5E2%7D%5C%5C%5C%5C%20%5Cmathsf%7B%5Bf%5E%7B-1%7D%28x%29%5D%5E2%3D10-x%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D%5Cpm%20%5Csqrt%7B10-x%7D%7D%5C%5C%5C%5C%20%5Cmathsf%7Bf%5E%7B-1%7D%28x%29%3D-%5Csqrt%7B10-x%7D~~~or~~~f%5E%7B-1%7D%28x%29%3D%5Csqrt%7B10-x%7D%7D)
The sign of the inverse depends on the domain of

itself, and where it's invertible.
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Tags: <em>inverse function definition identity domain algebra</em>
The answer is 9 I’m pretty positive
I hope this helps you
3 1/4= 3.4+1/4=13/4
1 1/3=1.3+1/3=4/3
1 3/4=1.4+3/4=7/4
Volume =13/4×4/3×7/4
Volume =91/12