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Fudgin [204]
1 year ago
11

i know this is probably simple but ive forgotten how to do it and i cant find it in my notes help appreciated!

Mathematics
1 answer:
Ronch [10]1 year ago
8 0

Answer:

This is pythagoras

A squared plus b squared equals c squared.

The unknown side is c as its the hypotenuse,

So the length of the first side is 6.4 this squared = 40.96

The other length is 12 as 6 is the radius so the diameter of the circle amd therfore the length of the triangle is 12.

12 squared is 144

144 + 40.96 = 184.96

The length of c squared is 184.96

To find the length of c alone we calculate the square root.

The square root of 184.96 = 13.6

? = 13.6

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41.7

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Circle X with a radius of 6 units and circle Y with a radius of 2 units are shown.
Natali [406]

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Translate the circles so they share a common center point, and dilate circle Y by a scale factor of 3.

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5 0
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}.

5 0
3 years ago
*please help and explain how you got the answer!*
stepan [7]

Answer:

0

Step-by-step explanation:

A= (4-1)^3

simplify A to be 3^3

Which gives us 27

B=(2*3)^2-9

simplify B

first multiply the 2 numbers in paranthesis which gives us 6. raise it to the power of 2 which is 39 and then subtract 9. Gives us 27.

C=15^3*4-12

Simplify the exponent first. 3*4 gives us 12 and 12-12 equals 0. Anything raised to the power of 0=1

If A-B^C is the equation we can write 27-27 raised to the power of 1 which is 0

7 0
2 years ago
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