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zloy xaker [14]
3 years ago
8

The base of a prism has an area of 27 square inches and a perimeter of 36 inches. The surface area of this prism is 144 square i

nches. Which equation below could be used to find h, the height of this prism, in inches?
A 27h = 144
B 36h = 144
C 2(27) + 36h = 144
D 2(36) + 27h = 144
Mathematics
1 answer:
77julia77 [94]3 years ago
5 0

The base of a prism has an area of 27 square inches and a perimeter of 36 inches. The surface area of this prism is 144 square inches. Which equation below could be used to find h, the height of this prism, in inches?

The base of a prism has an area of 27 square inches and a perimeter of 36 inches. The surface area of this prism is 144 square inches. Which equation below could be used to find h, the height of this prism, in inches?

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Given

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Manny= 90

Jim =100

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Now their percentage

Manny percentage

\begin{gathered} manny=\frac{90}{300}\times100\text{ \%} \\  \\ Manny\text{ percentage=}\frac{90}{3}\text{ \%} \\  \\ Manny\text{ percentage =30  \%} \end{gathered}

Jim Percentage

\begin{gathered} Jim=\frac{100}{300}\times100\text{ \%} \\ Jim\text{ percentage=}\frac{1}{3}\times100\text{  \%} \\ Jim\text{ percentage=33.333\%} \end{gathered}

Bob percentage

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Find f(2) if f(x)= (x +1)^2
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How do I find the missing value
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Okay. Since the "y" value and 82* are on the same straight line, their values will always add up to make a sum of 180. So if you subtract 82 from 180 you get 98*. So your "y" value is equal to 98*.

Now, all of the degree values added together should give you a sum of 360, so now we must add the values we know, in order to find the "x" value. So, 112*+82*+98*=292. Now we know that the difference of 360 and 292 will give us the x value. So finally, 360-292=68. So your "x" value is equal to 68.

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What is the solution set for 2x+5y>-1 and 4x-3<-3?
bekas [8.4K]

PROBLEM ONE

•

Solving for x in 2x + 5y > -1.

•

Step 1 ) Subtract 5y from both sides.

2x + 5y > -1

2x + 5y - 5y > -1 - 5y

2x > -1 - 5y

Step 2 ) Divide both sides by 2.

2x > -1 - 5y

\displaystyle\frac{2x}{2} > \displaystyle\frac{-1 - 5y}{2}

\displaystyle\ x > \frac{-1 - 5y}{2}

So, the solution for x in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 - 5y}{2}

•

Solving for y in 2x + 5y > -1.

•

Step 1 ) Subtract 2x from both sides.

2x + 5y > -1

2x - 2x + 5y > -1 - 2x

5y > -1 - 1x

Step 2 ) Divide both sides by 5.

5y > -1 - 1x

\displaystyle\frac{5x}{5} > \frac{-1 -1x}{5}

\displaystyle\ x > \frac{-1 -1x}{5}

So, the solution for y in 2x + 5y > -1 is...

\displaystyle\ x > \frac{-1 -1x}{5}

•

PROBLEM TWO

•

Solving for x in 4x - 3 < -3.

•

Step 1 ) Subtract 3 from both sides.

4x - 3 < -3

4x -3 - 3 < -3 - 3

4x < 0

Step 2 ) Divide both sides by x.

4x < 0

\displaystyle\frac{4x}{4}

x < 0

So, the solution for x in 4x - 3 < -3 is...

x < 0

•

•

- <em>Marlon Nunez</em>

7 0
2 years ago
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