<img src="https://tex.z-dn.net/?f=%28%5Csqrt%5B3%5D%7B7x%5E6%7D%29%5E5" id="TexFormula1" title="(\sqrt[3]{7x^6})^5" alt="(\sqrt[

All categories

3 years ago

3]{7x^6})^5" align="absmiddle" class="latex-formula">
How would I rewrite this using rational exponents? I know the answer is $7^\frac{5}{3}x^{10}$ , but I don't know how to get the $x^{10}$ .

Mathematics

1 answer:

Readme [11.4K]3 years ago

5 0

Answer:

When you cube root x to the power of 6, x becomes squared, and the exponent, 5, makes it x to the power of 10.

Step-by-step explanation:

You might be interested in

Can someone tell me if this is right

Ivan

No you have to right

2:1

4:2

6:3

8:4

These are correct

3 0

3 years ago

Find the limit if it exists lim x→0 sqrtx+7-sqrt7 over x

Triss [41]

Answer:

$\frac{1}{ 2\sqrt{7} }$

Step-by-step explanation:

$\lim_{x\to 0} \frac{ \sqrt{x + 7} - \sqrt{7} }{x} \\ \\ = \lim_{x\to 0} \frac{( \sqrt{x + 7} - \sqrt{7}) }{x} \times \frac{( \sqrt{x + 7} + \sqrt{7}) }{( \sqrt{x + 7} + \sqrt{7}) } \\ \\ = \lim_{x\to 0} \frac{( \sqrt{x + 7} )^{2} - (\sqrt{7})^{2} }{x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{( {x + 7} - {7}) }{x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{ {\cancel x}}{\cancel x( \sqrt{x + 7} + \sqrt{7})} \\ \\ = \lim_{x\to 0} \frac{ {1}}{\sqrt{x + 7} + \sqrt{7}} \\ \\ = \frac{1}{ \sqrt{0 + 7} + \sqrt{7} } \\ \\ = \frac{1}{ \sqrt{7} + \sqrt{7} } \\ \\ = \frac{1}{ 2\sqrt{7} }$