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quester [9]
3 years ago
14

Name the shortest segment from Point A to Line CD. Write and Solve an inequality for X.

Mathematics
1 answer:
Misha Larkins [42]3 years ago
5 0
Attached the solution with work shown.

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Solve 2x + 5y = −13 3x − 4y = −8
zlopas [31]
If you would like to solve 2x + 5y = -13 and 3x - 4y = -8, you can do this using the following steps:

2x + 5y = -13       /*4
3x - 4y = -8         /*5
_______________
8x + 20y = -52
15x - 20y = -40
_______________
8x + 15x + 20y - 20y = -52 - 40
23x = -92
x = -92 / 23
x = -4

<span>3x - 4y = -8 
</span>3 * (-4) - 4y = -8
-12 - 4y = -8
12 + 4y = 8
4y = 8 -12
4y = -4
y = -1

The correct result would be x = -4 and y = -1.
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3 years ago
A triangle can be formed with side lengths 3 in, 5 in, and 9 in. True or False?
Lelechka [254]

Answer:

no this cant form a right triangle but can form a regular triangle

Step-by-step explanation:

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4 years ago
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Eric Kweeder can paint a wall in
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The answer of the question is 12minutes
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What agency is in charge of printing money
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The Bureau of Engraving and Printing (BEP)  is a government agency within the United States Department of the Treasury that designs and produces a variety of security products for the United States government, most notable of which is Federal Reserve Notes (paper money) for the Federal Reserve, the nations central bank...
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3 years ago
In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
fredd [130]

Answer:

1) Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) Amount \ of \, water \ remaining \ in \, the \ tank \ is \  \frac{x^3(6-\pi) }{6}

Step-by-step explanation:

1) Here we have;

First tank A

Volume of tank = x³

The  volume of the sphere = \frac{4}{3} \pi r^3

However, the diameter of the sphere = x therefore;

r = x/2 and the volume of the sphere is thus;

volume of the sphere = \frac{4}{3} \pi \frac{x^3}{8}= \frac{1}{6} \pi x^3

For tank B

Volume of tank = x³

The  volume of the spheres = 8 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 2·D = x therefore;

r = x/4 and the volume of the sphere is thus;

volume of the spheres = 8 \times \frac{4}{3} \pi (\frac{x}{4})^3= \frac{x^3 \times \pi }{6}

For tank C

Volume of tank = x³

The  volume of the spheres = 64 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 4·D = x therefore;

r = x/8 and the volume of the sphere is thus;

volume of the spheres = 64 \times \frac{4}{3} \pi (\frac{x}{8})^3= \frac{x^3 \times \pi }{6}

Volume occupied by the spheres are equal therefore the three tanks contains the same volume of water

2) For the 4th tank, we have;

number of spheres on side of the tank, n is given thus;

n³ = 512

∴ n = ∛512 = 8

Hence we have;

Volume of tank = x³

The  volume of the spheres = 512 \times \frac{4}{3} \pi r^3

However, the diameter of the spheres 8·D = x therefore;

r = x/16 and the volume of the sphere is thus;

volume of the spheres = 512\times \frac{4}{3} \pi (\frac{x}{16})^3= \frac{x^3 \times \pi }{6}

Amount of water remaining in the tank is given by the following expression;

Amount of water remaining in the tank = Volume of tank - volume of spheres

Amount of water remaining in the tank = x^3 - \frac{x^3 \times \pi }{6} = \frac{x^3(6-\pi) }{6}

Amount \ of \ water \, remaining \, in \, the \ tank =  \frac{x^3(6-\pi) }{6}.

5 0
3 years ago
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