Question

Evaluate cube root of 7 multiplied by the square root of 7 over sixth root of 7 to the power of 5.

a. 7 to the −1st power
b. 1
c. 7
d. 7 to the power of 5 over 3

Answer 1

I hope u can understanx this. Write it all as exponential and than use the theorems to evaluate, the answer is 1

Answer 2

Answer:

The answer is the option b

$1$

Step-by-step explanation:

we have

$(\sqrt[3]{7} *\sqrt{7})/\sqrt[6]{7^{5}}$

we know that

$(\sqrt[3]{7} *\sqrt{7})/\sqrt[6]{7^{5}}=(7^{\frac{1}{3}} *7^{\frac{1}{2}})/7^{\frac{5}{6}}\\ =7^{\frac{5}{6}}/7^{\frac{5}{6}}\\=1$