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liraira [26]
2 years ago
14

Paul and Art are going to start a business selling fresh vegetables in their neighborhood. They

Mathematics
1 answer:
bearhunter [10]2 years ago
5 0

The system of 5x + 2y ≤ 25 is given by 20x + 15y ≤ 120 and 5x + 2y ≤ 25

<h3>Equation</h3>

An equation is an expression used to show the relationship between two or more variables and numbers.

Let x represent the cucumber and y represent the amount of carrots, hence:

20x + 15y = 2(60)

20x + 15y ≤ 120  (1)

Also:

5x + 2y ≤ 25 (2)

The system of 5x + 2y ≤ 25 is given by 20x + 15y ≤ 120 and 5x + 2y ≤ 25

find out more on Equation at: brainly.com/question/2972832

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Item 16 A sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. What is the difference of the vol
Zigmanuir [339]

Answer:

672\pi \text{ cm}^3.

Step-by-step explanation:

We have been given that a sphere has a radius of 8 centimeters. A second sphere has a radius of 2 centimeters. We are asked to find the difference of the volumes of the spheres.      

We will use volume formula of sphere to solve our given problem.

\text{Volume of sphere}=\frac{4}{3}\pi r^3, where r is radius of sphere.

The difference of volumes would be volume of larger sphere minus volume of smaller sphere.

\text{Difference of volumes}=\frac{4}{3}\pi(\text{8 cm})^3-\frac{4}{3}\pi(\text{2 cm})^3

\text{Difference of volumes}=\frac{4}{3}\pi(512)\text{ cm}^3-\frac{4}{3}\pi(8)\text{ cm}^3

\text{Difference of volumes}=\frac{4}{3}\pi(512-8)\text{ cm}^3

\text{Difference of volumes}=4\pi(168)\text{ cm}^3

\text{Difference of volumes}=672\pi\text{ cm}^3

Therefore, the difference between volumes of the spheres is 672\pi \text{ cm}^3.

3 0
3 years ago
Determine which relation is a function.
slamgirl [31]
The last one is the correct answer
5 0
3 years ago
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Solve this polynomial:
nadezda [96]

Answer:

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Step-by-step explanation:

Perhaps you want to simplify the expression.

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Solve the following equation. Then place the correct number in the box provided. 3x + 1 + 5x = 7 +15 + 7x find X
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