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miss Akunina [59]
4 years ago
15

Haley used unit cubes to build a rectangular prism that is 5 units long, 3 units wide, and 4 units tall. Jeremiah used unit cube

s to build a rectangular prism that has twice the volume of Haley's prism. Jeremiah's prism is 4 units long and 3 units wide. How tall is Jeremiah's prism?
Mathematics
1 answer:
ad-work [718]4 years ago
8 0

Answer:

10 units

Step-by-step explanation:

Let us find the volume of Haley's prism and compare it with the volume of Jeremiah's.

The volume of Haley's prism is:

V = 5 * 3 * 4 = 60 cubic units

Jeremiah's prism has twice the volume of Haley's:

V(J) = 2 * 60 = 120 cubic units

This implies that:

120 = 4 * 3 * h

where h = height of prism

=> 120 = 12h

=> h = 120 / 12 = 10 units

Jeremiah's prism is 10 units tall.

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D. -2 ≤ y ≤ 3

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This is because range is based on y-values and so the lowest y-value here is -2 and the greatest one is 3, inclusive.

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Find MP.<br><br> MN=17<br> NP=3y<br> MP=5y+9
8_murik_8 [283]

Answer:

MP = 29

Step-by-step explanation:

Given:

MN = 17

NP = 3y

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MN + NP = MP (segment addition postulate)

17 + 3y = 5y + 9 (substitution)

3y = 5y + 9 - 17 (subtracting 17 from each side)

3y = 5y + 8

3y - 5y = 5y - 8 - 5y (Subtracting 5y from both sides)

-2y = 5y - 8 - 5y

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Plug in the value of y

MP = 5(4) + 9 = 20 + 9

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3 years ago
Prove that the left side equals to the right side:<br> (sin x/(1-cos x))+(sin x/(1+cos x))=2 csc x
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The sum of two positive integers, x and y, is not more than 40. The difference of the two integers is at least 20. Chaneece choo
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Answer:

The value of x lies between 20 and 40 and value of y lies between -20 and 10

Step-by-step explanation:

We are given that

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First we convert inequality equation into equality equation to find the solution of the given system of inequality equation.

Therefore, we can write as

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Substitute y=10 in equation (3) we get

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Substitute y=0 in equation (3)

x=40

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Substitute y=0 in equation (4)

x=20

Substitute x=0 and y=40 and in equation (1)

40\leq 40

Hence, the equation is true.Therefore, the shaded region below the line.

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