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Karo-lina-s [1.5K]
3 years ago
5

Write an equation that can be used to answer the question. Then solve. Round to

Mathematics
1 answer:
Pachacha [2.7K]3 years ago
8 0
Report this clown who put the first answer he’s trying to get your ip
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Double the difference of a number n and 3 is -10. Which equation models this "number riddle" correctly?
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7 0
3 years ago
A bottle of water is 2/3 full. You drink the given portion of
Citrus2011 [14]

Answer:

1/2

Step-by-step explanation:

Step 1: You would multiply 2/3 by 3/4.

(Hint: if it is ____ of ____, you would most likely multiply)

Step 2: 2×3/3×4

Step 3: 2×3=6

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Hope this helps :)

3 0
2 years ago
Read 2 more answers
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
11 5/9 + 10 1/3 plz help
Likurg_2 [28]
10 1/3= 31/3
31/3= 93/9

11 5/9= 104/9
93/9 + 104/9 = 197/9
197/9= 21 8/9



your answer is 21 & 8/9


Good Luck
XD
7 0
3 years ago
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