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3 years ago

15

At the zoo on Monday, 65 percent of the people in attendance were children. If 520 children were in attendance, how many people

in total were visiting the zoo on Monday?

1 answer:

Anvisha [2.4K]3 years ago

8 0

A total of 800 people visited the zoo on Monday

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What is the definition of math

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3 years ago

a can complete a piece of work in 10 days B can do it in 12 days C can do it in 15 days in how many days will they finish the wo

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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

Explain why you cannot square each side of the equation when verifying a trigonometric identity.

Sergeu [11.5K]

This is because when we do verification of an identity, we must work separately on both sides, and to see in the end if we can get an equality. Because if we square both sides, that already means that we assume that the equality exist in the beginning, so no need to verify the identity.

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3 years ago

Read 2 more answers

BRAINLIESTTT ASAP! PLEASE HELP ME :)

11Alexandr11 [23.1K]

<h3>Answer: B) Only the first equation is an identity</h3>

========================

I'm using x in place of theta. For each equation, I'm only altering the left hand side.

Part 1

cos(270+x) = sin(x)

cos(270)cos(x) - sin(270)sin(x) = sin(x)

0*cos(x) - (-1)*sin(x) = sin(x)

0 + sin(x) = sin(x)

sin(x) = sin(x) ... equation is true

Identity is confirmed

---------------------------------

Part 2

sin(270+x) = -sin(x)

sin(270)cos(x) + cos(270)sin(x) = -sin(x)

-1*cos(x) + 0*sin(x) = -sin(x)

-cos(x) = -sin(x)

We don't have an identity. If the right hand side was -cos(x), instead of -sin(x), then we would have an identity.

7 0

3 years ago