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worty [1.4K]
3 years ago
14

Help.........................​

Mathematics
1 answer:
garri49 [273]3 years ago
7 0
<h3>Answer: Choice A</h3>

x^2\left(\sqrt[4]{x^2}\right)

=====================================================

Explanation:

The fourth root of x is the same as x^(1/4)

I.e,

\sqrt[4]{x} = x^{1/4}

The same applies to x^10 as well

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4}

Multiply the exponents 10 and 1/4 to get 10/4

\sqrt[4]{x^{10}} = \left(x^{10}\right)^{1/4} = x^{10*1/4} = x^{10/4}

\sqrt[4]{x^{10}} = x^{10/4}

-----------------------

If we have an expression in the form x^(m/n), with m > n, then we can simplify it into an equivalent form as shown below

x^{m/n} = x^a\sqrt[n]{x^b}

The 'a' and 'b' are found through dividing m/n

m/n = a remainder b

'a' is the quotient, b is the remainder

-----------------------

The general formula can easily be confusing, so let's replace m and n with the proper numbers. In this case, m = 10 and n = 4

m/n = 10/4 = 2 remainder 2

We have a = 2 and b = 2

So

x^{m/n} = x^a\sqrt[n]{x^b}

turns into

x^{10/4} = x^2\sqrt[4]{x^2}

which means

\sqrt[4]{x^{10}} = {x^2} \sqrt[4]{x^2}

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