Mathematics: Which expression is equivalent to StartFraction StartRoot 2 EndRoot Over RootIndex 3 StartRoot 2 EndRoot EndFraction? One-fourth RootIndex 6 StartRoot 2 EndRoot StartRoot 2 EndRoot StartFraction StartRoot 2 EndRoot Over 2 EndFraction Answer is RootIndex 6 StartRoot 2 EndRoot B on edg

Which expression is equivalent to StartFraction

Ответ:

ryan1kk

29.01.2022

Mathematics

The expression equivalent to given expression is $\sqrt[6]{2}$ .

Given expression is, $\frac{\sqrt{2} }{\sqrt[3]{2} }$

Expression can be written as,

$\frac{\sqrt{2} }{\sqrt[3]{2} } =\frac{2^{1/2} }{2^{1/3} } =2^{\frac{1}{2}-\frac{1}{3} } \\
\\
2^{\frac{1}{2}-\frac{1}{3} } =2^{\frac{1}{6} }=\sqrt[6]{2}$

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Ответ:

willasiacochran

29.01.2022

Mathematics

Question:

Which expression is equivalent to

$\frac{\sqrt{2}}{\sqrt[3]{2}}$

$\frac{\sqrt{2}}{\sqrt[3]{2}} =2^{\frac{1}{6}} = \sqrt[6]{2}$

Step-by-step explanation:

Given

$\frac{\sqrt{2}}{\sqrt[3]{2}}$

Required

Simplify

From laws of indices;

$\sqrt[n]{x} =x^{\frac{1}{n}}$

So, the expression can be rewritten as

$\frac{\sqrt{2}}{\sqrt[3]{2}} = \frac{{2^{\frac{1}{2}}}}{2^{\frac{1}{3}}}$

Also, from laws of indices

$\frac{x^a}{x^b} = x^{a-b}$

So, the expression is further solved to:

$\frac{\sqrt{2}}{\sqrt[3]{2}} =2^{\frac{1}{2} - \frac{1}{3}$

Solve exponents as fraction

$\frac{\sqrt{2}}{\sqrt[3]{2}} =2^{\frac{3-2}{6}}$

$\frac{\sqrt{2}}{\sqrt[3]{2}} =2^{\frac{1}{6}}$

This can be rewritten as

$\frac{\sqrt{2}}{\sqrt[3]{2}} =2^{\frac{1}{6}} = \sqrt[6]{2}$

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Ответ:

wdgyvwyv8840

21.04.2021

Mathematics

Complete question is;

Suppose we want to choose 5 colors, without replacement, from 16 distinct colors.

A) How many ways can this be done, if the order of the choices is taken into consideration?

B) How many ways can this be done, if the order of the choices is not taken into consideration?

A) 524160 ways

B) 4368 ways

Step-by-step explanation:

A) If the order of choices is taken into consideration, then it is a Permutation problem.

Formula for Permutation is;

P(n, r) = n!/(n - r)!

In the question, n = 16 and r = 5

Thus;

P(16, 5) = 16!/(16 - 5)!

P(16, 5) = 524160

B) If the order of choices is not taken into consideration, then it becomes a Permutation problem.

Formula for Combination is;

C(n, r) = n!/((r! × (n - r)!)

C(16, 5) = 16!/((5! × (16 - 5)!)

C(16, 5) = 4368

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Which expression is equivalent to StartFraction StartRoot 2 EndRoot Over RootIndex 3 StartRoot 2 EndRoot EndFraction? One-fourth RootIndex 6 StartRoot 2 EndRoot StartRoot 2 EndRoot StartFraction StartRoot 2 EndRoot Over 2 EndFraction Answer is RootIndex 6 StartRoot 2 EndRoot B on edg

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