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1 year ago

50 points for this one

Mathematics

2 answers:

ss7ja [257]1 year ago

7 0

Answer:

-1, 0

Step-by-step explanation:

Hello!

We can test each solution for by substituting the value for x, and see if the inequality is true.

$\frac{3(-1)^2}{(-1)^2+3} < 1$
$\frac34 < 1$

This inequality is true.

$\frac{3(-2)^2}{(-2)^2+3} < 1$
$\frac{12}{7} < 1$

This inequality is not true because 12/7 is greater than 1.

$\frac{3(0)^2}{(0)^2+3} < 1$
$0 < 1$

This is inequality is true.

$\frac{3(3)^2}{(3)^2+3} < 1$
$\frac{27}{12} < 1$

This inequality is not true because 27/12 is greater than 1.

The solutions that work are -1 and 0.

Georgia [21]1 year ago

6 0

Answer: A

Step-by-step explanation:

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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}}$

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