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Margarita [4]
3 years ago
5

Simplifying square roots problem below.

Mathematics
1 answer:
Mars2501 [29]3 years ago
6 0
Break x⁴ into x³ and x¹, watch the step below
\sqrt[3]{ x^{4} }
= \sqrt[3]{ x^{3+1} }

Remember that
\sqrt{ x^{m+n} } = \sqrt{ x^{m} }  \sqrt{ x^{n} }

Use the property above to simplify the root
= \sqrt[3]{ x^{3+1} }
=(\sqrt[3]{ x^{3} })( \sqrt[3]{x^{1}} )
= x ∛x

The answer is first option
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Step-by-step explanation:

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the nth term of sequence B is written as:

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Then we will heave:

3*n^2 - 1 = 30 - x^2

To find the pair, we could isolate one of the variables, then input different integers in the other variable and see if the outcome is also an integer.

Let's isolate n.

3*n^2 = 30 - x^2 + 1

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n^2 = (31 - x^2)/3

n = √(  (31 - x^2)/3)

Now let's input different values for x, and see if the outcome is also an integer, notice that x is in a negative term inside a square root, then we have only a few values of x such that the equation can be true.

Then let's start with x = 1.

n(1) = √(  (31 - 1^2)/3) = √(30/3) = √10

We know that √10 is not an integer.

now with x = 2,

n(2) = √(  (31 - 2^2)/3)  = √( (31 - 4)/3) = √(27/3) = √9 = 3

then if x = 2, we have n = 3.

Both of them are integers, then we get:

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Step-by-step explanation:

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