Can you guys evaluate this expression

[bucknaged]

This is grade 12 college level math, through correspondence. I'm really only doing this as personal choice. I guess, I'm just wishing to develop some math skills as i have been out school for awhile.

The book doesn't give you many examples or explanations.
Any help is appreciated

 

[bucknaged]

Problem written out

(2^-3+5^0)^1/2/2^-5

(1/2^3+1) ^1/2/1/2^5
=1/8+1/1/32 =(4+1)^1/2
=2+1=3

Sorry, i didn't realize it was upside down. I can always submit a new attachment and rotate it. This problem will be hard to read

 

[bucknaged]

This is better

 

[soroban]

Hello, bucknaged!

\(\displaystyle \dfrac{(2^{-3} + 5^0)^{\frac{1}{2}}}{2^{-5}} \)

We have: .\(\displaystyle \dfrac{\left(\frac{1}{2^3} + 1\right)^{\frac{1}{2}}}{2^{-5}} \;=\;\dfrac{\left(\frac{1}{8} + 1\right)^{\frac{1}{2}}}{2^{-5}} \;=\;\dfrac{\left(\frac{9}{8}\right)^{\frac{1}{2}}}{2^{-5}} \;=\;\left(\dfrac{9}{8}\right)^{\frac{1}{2}}\cdot 2^5 \;=\;\dfrac{9^{\frac{1}{2}}}{8^{\frac{1}{2}}}\cdot 2^5 \)

. . . . . . . . \(\displaystyle =\;\dfrac{(3^2)^{\frac{1}{2}}}{(2^3)^{\frac{1}{2}}} \cdot 2^5 \;=\; \dfrac{3}{2^{\frac{3}{2}}}\cdot 2^5 \;=\;3\cdot 2^{\frac{7}{2}} \;=\; 3\cdot 2^3\cdot 2^{\frac{1}{2}} \;=\; 24\sqrt{2}\)