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3 years ago

A baseball has a 45 cm diameter. what is the volume of the contents of the ball

2 answers:

5 0

First you need to find the radius. To do that you need to divide the diameter by two;

45 / 2 = 22.5

Then you plug the radius in the formula for a sphere. Which is...

V = (4 / 3)πr^3
First you need to find the volume with out pi

V = (4/3) π (45)^3
V= π (4/3) (91,125)
Your answer with out pi is...
V= (68,343.75) π cm^3
And with pi is...
V = 214,708.22 cm^3

Pepsi [2]3 years ago

3 0

Answer:

47732.14 cm³

Step-by-step explanation:

To find the volume of the sphere with a diameter of 45 cm, we simply use the formula for calculating volume of a sphere, since a base ball is also spherical in shape.

Volume of a sphere = 4/3 πr³

But in the question, we were given diameter, so we can find our radius using the diameter given.

radius = diameter / 2

radius = 45 / 2

radius(r) =22.5 cm

The constant value of π is always 22/7 or 3.142 unless stated otherwise.

We can now go ahead and insert our values into the formula;

Volume of baseball = 4/3 πr³

=4/3 × 22/7 × (22.5)³

= 1002375 / 21

=47,732.14 cm³

Volume of the baseball = 47,732.14 cm³

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3 years ago

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3 years ago

What is the value of y if 13/9 = y/18

Maslowich

Answer:

$y=26$

Step-by-step explanation:

1. Swap sides

$\frac{13}{9}=\frac{y}{18}$

Swap sides:

$\frac{y}{18}=\frac{13}{9}$

2. Isolate the y

$\frac{y}{18}=\frac{13}{9}$

Multiply to both sides by 18:

$\frac{y}{18}\cdot 18=\frac{13}{9}\cdot 18$

Group like terms:

$\frac{1}{18}\cdot 18y=\frac{13}{9}\cdot 18$

Simplify the fraction:

$y=\frac{13}{9}\cdot 18$

Multiply the fractions:

$y=\frac{13\cdot 18}{9}$

Simplify the arithmetic:

$y=26$

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Why learn this:

Linear equations cannot tell you the future, but they can give you a good idea of what to expect so you can plan ahead. How long will it take you to fill your swimming pool? How much money will you earn during summer break? What are the quantities you need for your favorite recipe to make enough for all your friends?
Linear equations explain some of the relationships between what we know and what we want to know and can help us solve a wide range of problems we might encounter in our everyday lives.

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Terms and topics

Linear equations with one unknown

The main application of linear equations is solving problems in which an unknown variable, usually (but not always) x, is dependent on a known constant.

We solve linear equations by isolating the unknown variable on one side of the equation and simplifying the rest of the equation. When simplifying, anything that is done to one side of the equation must also be done to the other.

An equation of:

$ax+b=0$

in which $a$ and $b$ are the constants and $x$ is the unknown variable, is a typical linear equation with one unknown. To solve for $x$ in this example, we would first isolate it by subtracting $b$ from both sides of the equation. We would then divide both sides of the equation by $a$ resulting in an answer of:

$x = -b/a$

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