$\bf 343^{\frac{2}{3}}+36^{\frac{1}{2}}-256^{\frac{3}{4}}\qquad \begin{cases}
343=7\cdot 7\cdot 7\\
\qquad 7^3\\
36=6\cdot 6\\
\qquad 6^2\\
256=4\cdot 4\cdot 4\cdot 4\\
\qquad 4^4
\end{cases}\\\\\\ (7^3)^{\frac{2}{3}}+(6^2)^{\frac{1}{2}}-(4^4)^{\frac{3}{4}}
\\\\\\
\sqrt[3]{(7^3)^2}+\sqrt[2]{(6^2)^1}-\sqrt[4]{(4^4)^3}\implies \sqrt[3]{(7^2)^3}+\sqrt[2]{(6^1)^2}-\sqrt[4]{(4^3)^4}
\\\\\\
7^2+6-4^3\implies 49+6-64\implies -9$

to see what you can take out of the radical, you can always do a quick "prime factoring" of the values, that way you can break it in factors to see who is what.

8 0

3 years ago

I really need help, answer as many as you can please

amid [387]

5^2 -7•4+(36-2^(5) so do (36-2^5 first put that in the calculator which gets u 32.Then bring down 5^2-7•4+32 then do 5^2 and 7×4 which 25 -28+32 add them together u get 60 so 25 -60 is 35

6 0

3 years ago

HELP will give brainliest

mr Goodwill [35]

Answer:it is a

Step-by-step explanation:

is a

8 0

2 years ago

Read 2 more answers

Put the following algebraic expression in standard simplified form. 7-12m-19

sammy [17]