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Andrej [43]
3 years ago
9

Write the inequality!! Connie’s dog Fido weighs 35 pounds. Her vet placed Fido on a diet. What inequality can you write to find

the average number of pounds Fido must lose monthly to reach a healthier weight of 28 pounds within 6 months?
Mathematics
1 answer:
Butoxors [25]3 years ago
6 0

Answer:   \frac{35-28}{x} \leq 6

Step-by-step explanation:

Let x be the average number of pounds Fido must loss.

Since, the initial weight of Fido is 35 pounds.

And, After losing the weight, the new weight of Fido in pounds = 28 pounds.

Then the time taken for losing the weight

= \frac{\text{ The weight it losses}}{\text{ Average of losing weight per month}}

= \frac{35-28}{x}

According to the question,  it must lose weight within 6 months,

Thus,   \frac{35-28}{x}\leq 6

Which is the required inequality to find the average number of pounds per month.

By solving it we, get,    x\geq \frac{7}{6}

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In the illustration below, the three cube-shaped tanks are identical. The spheres in any given tank
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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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