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Common Core: High School - Geometry : Modeling with 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 #431 : High School: Geometry

Objects in the real world can be described in geometric shapes. What geometric shape describes a coffee mug without a handle?

Possible Answers:

Sphere

None of the other answers.

Prism

Rectangle Prism

Open Cylinder

Correct answer:

Open Cylinder

Explanation:

In the three-dimensional world a coffee mug without a handle has a bottom and top in the shape of a circle. The three-dimensional connection between the top and the bottom creates a cylinder. Therefore, the geometric shape that best describes the coffee mug is an open cylinder since there is no top on the coffee mug.

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Example Question #11 : 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:

Rectangle

Circle

Sphere

None of the other answers

Cylinder

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.

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Example Question #12 : Modeling With Geometry

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:

$58398.16 \text{ kilograms}$

$58397.84 \text{ kilograms}$

$9513.03 \text{ kilograms}$

$358489.87 \text{ kilograms}$

$4756.52 \text{ kilograms}$

Correct answer:

$9513.03 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$0.1629 = \frac{m}{58398}$

m = 9513.03

Thus the mass of the balloon is 9513.03 kilograms.

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

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

Round your answer to decimal places.

Possible Answers:

$55879.66 \text{ kilograms}$

$53678.96 \text{ kilograms}$

$51563.09 \text{ kilograms}$

$53677.04 \text{ kilograms}$

$25781.54 \text{ kilograms}$

Correct answer:

$51563.09 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$\\0.9606 = \frac{m}{53678} \\\\m = 51563.09$

Thus the mass of the balloon is $51563.09 \text{ kilograms}$ .

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Example Question #3 : 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 decimal places

Possible Answers:

$4756.52 \text{ kilograms}$

$58398.16 \text{ kilograms}$

$9513.03 \text{ kilograms}$

$358489.87 \text{ kilograms}$

$58397.84 \text{ kilograms}$

Correct answer:

$9513.03 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$0.1629 = \frac{m}{58398}$

m = 9513.03

Thus the mass of the balloon is $9513.03 \text{ kilograms}$ .

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Example Question #4 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

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

Round your answer to decimal places.

Possible Answers:

$8324.02 \text{ kilograms}$

$58743.86 \text{ kilograms}$

$4162.01 \text{ kilograms}$

$414565.98 \text{ kilograms}$

$58744.14 \text{ kilograms}$

Correct answer:

$8324.02 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$\\0.1417 = \frac{m}{58744} \\\\m = 8324.02$

Thus the mass of the balloon is $8324.02 \text{ kilograms}$ .

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

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

Round your answer to decimal places.

Possible Answers:

$22177.88 \text{ kilograms}$

$86937.49 \text{ kilograms}$

$44355.77 \text{ kilograms}$

$86938.51 \text{ kilograms}$

$170399.84 \text{ kilograms}$

Correct answer:

$44355.77 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$\\0.5102 = \frac{m}{86938} \\\\m = 44355.77$

Thus the mass of the balloon is $44355.77 \text{ kilograms}$ .

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

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

Round your answer to decimal places.

Possible Answers:

$1319.15 \text{ kilograms}$

$72480.96 \text{ kilograms}$

$2638.31 \text{ kilograms}$

$1991236.26 \text{ kilograms }$

$72481.04 \text{ kilograms}$

Correct answer:

$2638.31 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

$\\0.0364 = \frac{m}{72481} \\\\m = 2638.31$

Thus the mass of the balloon is $2638.31 \text{ kilograms}$ .

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Example Question #7 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

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

Round your answer to decimal places.

Possible Answers:

$7132090.91 \text{ kilograms}$

$78453.01 \text{ kilograms}$

$78452.99 \text{ kilograms}$

$862.98 \text{ kilograms }$

$431.49 \text{ kilograms}$

Correct answer:

$862.98 \text{ kilograms }$

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.

$d = \frac{m}{v}$

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

$\\0.011 = \frac{m}{78453} \\\\m = 862.98$

Thus the mass of the balloon is $862.98 \text{ kilograms }$ .

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

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

Round your answer to decimal places.

Possible Answers:

$18053.35 \text{ kilograms}$

$77002.53 \text{ kilograms}$

$77003.47 \text{ kilograms}$

$36106.71 \text{ kilograms}$

$164220.52 \text{ kilograms}$

Correct answer:

$36106.71 \text{ kilograms}$

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.

$d = \frac{m}{v}$

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

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

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All Common Core: High School - Geometry Resources

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Common Core: High School - Geometry : Modeling with Geometry

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

All Common Core: High School - Geometry Resources

Example Questions

Example Question #431 : High School: Geometry

Example Question #11 : Modeling With Geometry

Example Question #12 : Modeling With Geometry

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

Example Question #3 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

Example Question #4 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

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

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

Example Question #7 : Apply Density Concepts To Area And Volume Situations: Ccss.Math.Content.Hsg Mg.A.2

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

All Common Core: High School - Geometry Resources

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