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Neporo4naja [7]
2 years ago
15

HELP HELP HELP HELP HELP

Mathematics
1 answer:
Deffense [45]2 years ago
7 0

Answer:

please help brainly.com/question/23055780

Step-by-step explanation:

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Angle XYZ is bisected by line YW the measure of angle XYW is 52 what is the measurement of angle WYZ what is the measurement of
Aliun [14]

Answer:

Step-by-step explanation:

2(2x+7) = 6x-8

4x + 14 = 6x - 8

2x = 22

x = 11

m<XYZ = 58

m<XYW = 29

7 0
3 years ago
Which expression is equal to (x+2)(−3x2+3x+1)?
padilas [110]
Multiply everything in the second parenthesis by x.

-3x^3 + 3x^2 + x

Multiply everything in the second parenthesis by 2.

-6x^2 + 6x + 2

Combine these two equations together.

-3x^3 + 3x^2 + x - 6x^2 + 6x + 2

Combine like terms.

-3x^3 - 3x^2 + 7x + 2

Hope this helps!
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7 0
3 years ago
Need answer ASAP ASAP please ASAP ASAP thank you:)
Paha777 [63]

Answer:

42°

Step-by-step explanation:

3x + 54° = 180° [Supplementary angles]

=> 3x = 180 - 54

=> 3x = 126

=  > x =  \frac{126}{3}

=> <u>x = 42</u><u>°</u><u> </u><u>(Ans)</u>

5 0
3 years ago
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If 60% of people said that they like
PilotLPTM [1.2K]

Answer:

40%

Step-by-step explanation:

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3 years ago
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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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