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salantis [7]
3 years ago
12

Find two integers whose product is 138 such that one of the integers is seven less than five times the other integer.

Mathematics
1 answer:
Oksana_A [137]3 years ago
4 0

Answer: the integers are 6 and 23

Step-by-step explanation:

Let x represent one of the integers.

Let y represent the other integer.

The product of the integers is 138. This means that

xy = 138 - - - - - - - - - -1

One of the integers is seven less than five times the other integer. This means that

x = 5y - 7

Substituting x = 5y - 7 into equation 1, it becomes

y(5y - 7) = 138

5y² - 7y = 138

5y² - 7y - 138 = 0

5y² + 23y - 30y - 138 = 0

y(5y + 23) - 6(5y + 23)

y - 6 = 0 or 5y + 23 = 0

y = 6 or y = - 23/5

The only possible value of y is 6

x = 138/y = 138/6

x = 23

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Answer:

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Step-by-step explanation:

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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}.

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3 years ago
An ice chest contains 9 cans of apple juice6 cans of grape juice5 cans of orange juiceand 4 cans of pineapple juice Suppose that
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Answer:

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Step-by-step explanation:

There are 24 total juice boxes.

18 of them are not grape

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The second time, 1 is removed, then another is removed. This creates:

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3 years ago
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