Common Core: High School - Geometry : High School: Geometry

Study concepts, example questions & explanations for Common Core: High School - Geometry

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All Common Core: High School - Geometry Resources

6 Diagnostic Tests 114 Practice Tests Question of the Day Flashcards Learn by Concept

Example Questions

Example Question #12 : Modeling With Geometry

Objects in the real world can be described in geometric shapes. What geometric shape describes the rim of a basketball hoop?

Possible Answers:

None of the other answers

Cylinder

Circle

Sphere

Rectangle

Correct answer:

Circle

Explanation:

A basketball hoop is constructed of a pole, a backboard, a rim, and a net that is attached to the rim. Since the question is asking for the geometric shape that describes the rim, recall that the rim of the basketball hoop has the geometric shape of a circle.

Example Question #1 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2


If a balloon is filled with 58398 cubic meters of xenon with a density of 0.1629 kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to 2 decimal places

 

Possible Answers:

\(\displaystyle 358489.87 \text{ kilograms}\)

\(\displaystyle 58398.16 \text{ kilograms}\)

 

\(\displaystyle 58397.84 \text{ kilograms}\)

\(\displaystyle 4756.52 \text{ kilograms}\)

\(\displaystyle 9513.03 \text{ kilograms}\)

Correct answer:

\(\displaystyle 9513.03 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle \uptext{m}\)).

\(\displaystyle 0.1629 = \frac{m}{58398}\)

\(\displaystyle m = 9513.03\)

Thus the mass of the balloon is 9513.03 kilograms.

 

 

Example Question #2 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 53678\) cubic meters of xenon with a density of \(\displaystyle 0.9606\) kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

 

Possible Answers:

\(\displaystyle 55879.66 \text{ kilograms}\)

\(\displaystyle 51563.09 \text{ kilograms}\)

\(\displaystyle 53677.04 \text{ kilograms}\)

\(\displaystyle 25781.54 \text{ kilograms}\)

\(\displaystyle 53678.96 \text{ kilograms}\)

Correct answer:

\(\displaystyle 51563.09 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.9606 = \frac{m}{53678} \\\\m = 51563.09\)

Thus the mass of the balloon is \(\displaystyle 51563.09 \text{ kilograms}\).

 

 

Example Question #13 : Modeling With Geometry


If a balloon is filled with \(\displaystyle 58398\) cubic meters of xenon with a density of \(\displaystyle 0.1629\) kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places

 

Possible Answers:

\(\displaystyle 358489.87 \text{ kilograms}\)

\(\displaystyle 9513.03 \text{ kilograms}\)

\(\displaystyle 58398.16 \text{ kilograms}\)

\(\displaystyle 58397.84 \text{ kilograms}\)

\(\displaystyle 4756.52 \text{ kilograms}\)

Correct answer:

\(\displaystyle 9513.03 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle \uptext{m}\)).

\(\displaystyle 0.1629 = \frac{m}{58398}\)

\(\displaystyle m = 9513.03\)

Thus the mass of the balloon is \(\displaystyle 9513.03 \text{ kilograms}\).

 

 

Example Question #14 : Modeling With Geometry

If a balloon is filled with \(\displaystyle 58744\) cubic meters of xenon with a density of \(\displaystyle 0.1417\) kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places. 

Possible Answers:

\(\displaystyle 414565.98 \text{ kilograms}\)

\(\displaystyle 58743.86 \text{ kilograms}\)

\(\displaystyle 8324.02 \text{ kilograms}\)

\(\displaystyle 58744.14 \text{ kilograms}\)

\(\displaystyle 4162.01 \text{ kilograms}\)

Correct answer:

\(\displaystyle 8324.02 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle \uptext{m}\)).

\(\displaystyle \\0.1417 = \frac{m}{58744} \\\\m = 8324.02\)

Thus the mass of the balloon is \(\displaystyle 8324.02 \text{ kilograms}\).

 

 

Example Question #5 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 86938\) cubic meters of xenon with a density of \(\displaystyle 0.5102\) kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

Possible Answers:

\(\displaystyle 44355.77 \text{ kilograms}\)

\(\displaystyle 170399.84 \text{ kilograms}\)

\(\displaystyle 86938.51 \text{ kilograms}\)

\(\displaystyle 22177.88 \text{ kilograms}\)

\(\displaystyle 86937.49 \text{ kilograms}\)

Correct answer:

\(\displaystyle 44355.77 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.5102 = \frac{m}{86938} \\\\m = 44355.77\)

Thus the mass of the balloon is \(\displaystyle 44355.77 \text{ kilograms}\).

Example Question #6 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 72481\) cubic meters of neon with a density of \(\displaystyle 0.0364\) kilograms per cubic meter. How many kilograms of neon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

Possible Answers:

\(\displaystyle 72480.96 \text{ kilograms}\)

\(\displaystyle 2638.31 \text{ kilograms}\)

\(\displaystyle 72481.04 \text{ kilograms}\)

\(\displaystyle 1991236.26 \text{ kilograms }\)

\(\displaystyle 1319.15 \text{ kilograms}\)

Correct answer:

\(\displaystyle 2638.31 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.0364 = \frac{m}{72481} \\\\m = 2638.31\)

Thus the mass of the balloon is \(\displaystyle 2638.31 \text{ kilograms}\).

Example Question #1 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 78453\) cubic meters of water with a density of \(\displaystyle 0.011\) kilograms per cubic meter. How many kilograms of water does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

 

Possible Answers:

\(\displaystyle 862.98 \text{ kilograms }\)

\(\displaystyle 78452.99 \text{ kilograms}\)

\(\displaystyle 431.49 \text{ kilograms}\)

\(\displaystyle 78453.01 \text{ kilograms}\)

\(\displaystyle 7132090.91 \text{ kilograms}\)

Correct answer:

\(\displaystyle 862.98 \text{ kilograms }\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.011 = \frac{m}{78453} \\\\m = 862.98\)

Thus the mass of the balloon is \(\displaystyle 862.98 \text{ kilograms }\)

Example Question #8 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 77003\) cubic meters of water with a density of \(\displaystyle 0.4689\) kilograms per cubic meter. How many kilograms of water does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

 

Possible Answers:

\(\displaystyle 77002.53 \text{ kilograms}\)

\(\displaystyle 18053.35 \text{ kilograms}\)

\(\displaystyle 77003.47 \text{ kilograms}\)

\(\displaystyle 164220.52 \text{ kilograms}\)

\(\displaystyle 36106.71 \text{ kilograms}\)

Correct answer:

\(\displaystyle 36106.71 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume.

Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.4689 = \frac{m}{77003} \\\\m = 36106.71\)Thus the mass of the balloon is \(\displaystyle 36106.71 \text{ kilograms}\).

Example Question #9 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

If a balloon is filled with \(\displaystyle 97999\) cubic meters of xenon with a density of \(\displaystyle 0.183\) kilograms per cubic meter. How many kilograms of xenon does the balloon contain?

Round your answer to \(\displaystyle 2\) decimal places.

Possible Answers:

\(\displaystyle 8966.91 \text{ kilograms}\)

\(\displaystyle 97999.18 \text{ kilograms}\)

\(\displaystyle 17933.82 \text{ kilograms}\)

\(\displaystyle 535513.66 \text{ kilograms}\)

\(\displaystyle 97998.82 \text{ kilograms}\)

Correct answer:

\(\displaystyle 17933.82 \text{ kilograms}\)

Explanation:

In order to solve this problem, we need to use an equation that involves, density, mass, and volume. Here is the equation that we need to use.

\(\displaystyle d = \frac{m}{v}\)

Since we are given the density, and volume, we can plug those values in, and then solve for the mass (\(\displaystyle m\)).

\(\displaystyle \\0.183 = \frac{m}{97999} \\\\m = 17933.82\)

Thus the mass of the balloon is \(\displaystyle 17933.82 \text{ kilograms}\).

All Common Core: High School - Geometry Resources

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