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3 years ago

Solve the equation.

Mathematics

1 answer:

Svetlanka [38]3 years ago

6 0

Answe

Step-by-step explanation:

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Its close to 1/2 I think this is an interesting question

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Step-by-step explanation:

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0.125in per hour

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Which of the following could be<br> the sum of exactly three<br> consecutive even integers?

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Step-by-step explanation:

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2 years ago

What is the simplified form of the following expression?

USPshnik [31]

Option C:

$$\frac{\sqrt[3]{100 x}}{5}=\sqrt[3]{\frac{4 x}{5}}$

Solution:

Given expression is

$$\sqrt[3]{\frac{4 x}{5}}$

Note: $\sqrt[3]{125}=\sqrt[3]{{5^3}} = 5$

To find the correct expression for the above simplified expression.

Option A: $\frac{\sqrt[3]{4 x}}{5}$

5 can be written as $\sqrt[3]{125}$ .

$$\frac{\sqrt[3]{4 x}}{5}=\frac{\sqrt[3]{4 x}}{\sqrt[3]{125} }$

$$=\sqrt[3]{\frac{4x}{125} }$

It is not the given simplified expression.

Option B: $\frac{\sqrt[3]{20 x}}{5}$

$$\frac{\sqrt[3]{20 x}}{5}=\frac{\sqrt[3]{20 x}}{\sqrt[3]{125} }$

$$=\sqrt[3]{\frac{20x}{125} }$

Cancel the common factor in both numerator and denominator.

$$=\sqrt[3]{\frac{4x}{25} }$

It is not the given simplified expression.

Option C: $\frac{\sqrt[3]{100 x}}{5}$

$$\frac{\sqrt[3]{100 x}}{5}=\frac{\sqrt[3]{100 x}}{\sqrt[3]{125} }$

$$=\sqrt[3]{\frac{100x}{125} }$

Cancel the common factor in both numerator and denominator.

$$=\sqrt[3]{\frac{4 x}{5}}$

It is the given simplified expression.

Option D: $\frac{\sqrt[3]{100 x}}{125}$

$$\frac{\sqrt[3]{100 x}}{125}=\frac{\sqrt[3]{100 x}}{5^3}$

It is not the given simplified expression.

Hence Option C is the correct answer.

$$\frac{\sqrt[3]{100 x}}{5}=\sqrt[3]{\frac{4 x}{5}}$

3 0

3 years ago

Please help!! I will give correct answer brainelest! 15 points!

4vir4ik [10]

Minimum is 40

First quartile is 43

Median is 61

Third quartile is 65

Maximum is 97

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3 years ago