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Zinaida [17]
3 years ago
7

Simplify:

3%7D%20%29%20%5E%7B%20%5Cfrac%7B3%7D%7B2%7D%20%7D%20" id="TexFormula1" title="(2x) ^{ \frac{1}{2} } \times (2x ^{3} ) ^{ \frac{3}{2} } " alt="(2x) ^{ \frac{1}{2} } \times (2x ^{3} ) ^{ \frac{3}{2} } " align="absmiddle" class="latex-formula">
​
Mathematics
2 answers:
wlad13 [49]3 years ago
5 0

Answer:

\huge\boxed{(2x)^\frac{1}{2}\times(2x^3)^\frac{3}{2}=4x^5}

Step-by-step explanation:

(2x)^\frac{1}{2}\times(2x^3)^\frac{3}{2}\qquad\text{use}\ (ab)^n=a^nb^m\\\\=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{3}{2}(x^3)^\frac{3}{2}\qquad\text{use}\ (a^n)^m=a^{nm}\\\\=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{2}{3}x^{(3)(\frac{3}{2})}=2^\frac{1}{2}x^\frac{1}{2}\times2^\frac{2}{3}x^\frac{9}{2}\\\\\text{use the commutative and associative property}\\\\=\left(2^\frac{1}{2}\times2^\frac{3}{2}\right)\left(x^\frac{1}{2}\times x^\frac{9}{2}\right)\qquad\text{use}\ a^n\times a^m=a^{n+m}

=2^{\frac{1}{2}+\frac{3}{2}}x^{\frac{1}{2}+\frac{9}{2}}=2^\frac{1+3}{2}x^{\frac{1+9}{2}}=2^\frac{4}{2}x^\frac{10}{2}=2^2x^5=4x^5

brilliants [131]3 years ago
5 0

Answer:

4x^5

Step-by-step explanation:

(2x)^\frac{1}{2} \times (2x^3)^\frac{3}{2} =

= (2x)^\frac{1}{2} \times (2x \times x^2)^\frac{3}{2}

= [(2x)^\frac{1}{2} \times (2x)^\frac{3}{2}] \times (x^2)^\frac{3}{2}

= (2x)^{\frac{1}{2} + \frac{3}{2}} \times x^{{2} \times \frac{3}{2}}

= (2x)^{\frac{4}{2}} \times x^{\frac{6}{2}}

= 2^2x^2 \times x^3

= 4x^5

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Step-by-step explanation:

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7 0
3 years ago
A scuba diver descends in the water at a rate of 23 1/2
Makovka662 [10]

The scuba diver's position relative to sea level after the 4.6 minutes is 5.1 ft.

Given:

A scuba diver descends in the water at a rate of 23 1/2  feet per minute for 2.6 minutes.

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The scuba diver's position relative to sea level after the 4.6 minutes is 5.1ft.

Learn more about ascending rate and descending rate here:

brainly.com/question/1477877?referrer=searchResults

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3 years ago
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Step-by-step explanation:

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The solution is where the lines intersect which is at the point (3,2).

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4 years ago
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