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3 years ago

How can you use proportional reasoning to solve problems involving percent?

Mathematics

1 answer:

sukhopar [10]3 years ago

7 0

Well percent is actually a decimal and can be transformed into a decimal because they're equal to a decimal

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Step-by-step explanation:

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Would a teaspoon be a good way to measure the capacity of a milk carton

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No, a teaspoon is too small to measure the capacity of a milk carton. You would use a quart to measure the capacity of a gallon.

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3 years ago

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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}}$

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3 years ago

If the mass of a material is 49 grams and the volume of the material is 7 cm^3, what would the density of the material be?

inysia [295]

The equation for density is:
$density = \frac{mass}{volume}$

To help you remember - my teacher told us to remember "We <span>♡ density," and the heart stands for </span> $\frac{m}{v}$ , or $\frac{mass}{volume}$ . Cheesy, but it works!<span>
</span>
Back to the problem:
We know the mass if 49g and the volume is 7 $cm^{3}$ , so plug these values into the equation for density:
$density = \frac{mass}{volume}\\
density = \frac{49\:g}{7\:cm^{3}} \\
density = 7\:g/cm^{3}$

Answer: The density is 7 g/ $cm^{3}$

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3 years ago

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Answer:

Step-by-step explanation:

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