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Blababa [14]
3 years ago
12

1. clark has 5 times as many pennies as dimes. If he has a total of 4.29, which two linear equations best model this system?

Mathematics
1 answer:
Alisiya [41]3 years ago
4 0
I wouldn’t really know -_-
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I got 0.45

Good luck!
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Write an expression for the area of the shaded region in its simplest form. Show all of your steps.
patriot [66]
The area of the shaded region will be the area of the rectangle minus the area of the white square inside of it:

((x+10)(2x+5)) - ((x+1)(x+1))

First, FOIL both of the areas separately:

(2x^2 + 5x + 20x + 50) - (x^2 + x + x + 1)

Simplify within the parentheses by adding like terms:

(2x^2 + 25x + 50) - (x^2 + 2x + 1)

Now, subtract one equation from the other:

2x^2 + 25x + 50
-x^2  -    2x  -   1
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This will be the equation for the area.
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3 years ago
Educators are testing a new program designed to help children improve their reading skills. The null hypothesis of the test is t
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3 years ago
The perimeter of a rectangular field is 350m. If the length of the field is 99m, what is it’s width?
Juliette [100K]

Answer:

The width of the field is 3.535353535m

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3 years ago
Select the point that is a solution to the system of inequalities
Zarrin [17]
<h2>Hello!</h2>

The answer is:

The point that is a solution to the system of inequalities is C(4,2).

<h2>Why?</h2>

To find the point that is a solution to the system of inequalities, we need to evaluate it into the given inequalities. If the point is a solution to the system of inequalities, both inequalities will be satisfied.

We are given the inequalities:

y\leq 2x-2

y\leq x^{2} -3x

So, substituting the given points into the given inequalities, we have:

- A. (2,1)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\1\leq 2*(2)-2\\\\1\leq 4-2\\\\1\leq 2

Substituting into the second inequality, we have:

y\leq x^{2} -3x

1\leq (2)^{2} -3(2)

1\leq (2)^{2} -3(2)

1\leq 4 -6

1\leq -2

Therefore, since 1 is not less or equal to -2, the point A(2,1) is not a solution to the system of inequalities.

- B. (-2,-1)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\-1\leq (-2)*(2)-2\\\\-1\leq -4-2\\\\-1\leq -6

Substituting into the second inequality, we have:

y\leq x^{2} -3x

-1\leq (-2)^{2} -3(-2)

-1\leq 4 +6

-1\leq 10

Therefore, since -1 is not less or equal to -6, the point B(-2,-1) is not a solution to the system of inequalities.

- C. (4,2)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\2\leq (4)*(2)-2\\\\2\leq 8-2\\\\2\leq 6

Substituting into the second inequality, we have:

y\leq x^{2} -3x

2\leq (4)^{2} -3(4)

2\leq 16 -12

2\leq 4

Therefore, since 2 is less than 6, and 2 is less than 4, the point B(4,2) is a solution to the system of inequalities.

- D(1,3)

Substituting into the first inequality, we have:

y\leq 2x-2\\\\3\leq (1)*(2)-2\\\\3\leq 2-2\\\\3\leq 0

Substituting into the second inequality, we have:

y\leq x^{2} -3x

3\leq (1)^{2} -3(1)

3\leq 1 -3

3\leq -2

Therefore, since 3 is not less than 0, and 3 is not less than -2, the point D(1.3) is not a solution to the system of inequalities.

Hence, the point that is a solution to the system of inequalities is C(4,2).

Have a nice day!

6 0
3 years ago
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