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Inessa [10]
3 years ago
8

Use a net to find the surface area of the right triangular prism shown below:

Mathematics
1 answer:
AnnZ [28]3 years ago
4 0
I don't exactly know what a net is, but I can figure out the surface area (maybe). When you put this pattern together, you have to make a couple of assumptions.
First of all the 2 triangles are connected to a base that's 8 by 13. 
Base
The area of the base = 8 * 13 = 104 square feet.

Two Triangles.
The area of any triangle is 
Area = 1/2 b * h
Area = 1/2 *12 * 5
Area = 30 square feet. There are 2 triangles so ...
The total area of both is 60 square feet.

Overlap
Now we come to the tough part.  how big is it from the first dotted line to the second one?
The answer is 5. But why? Well the 5 comes from the the length of the side that has to be covered by this pattern. The 5 is part of the triangle. 

How far across is the dotted line? The answer is 8. Why again? That's the width of the whole prism or pattern.
So the area between the two dotted lines is 5 * 8 = 40

Finally the area over the 12. 
The question is how far is it from the second dotted line to the last boundary? The answer is 12. And it's width is 8
So the area = 12 * 8 = 96

Total area
96 + 60 + 40 + 104 = 300 square feet. 
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3 years ago
Evaluate 2x2+y23 for x = 1.5 and y = 3. Show all of your work.
Lera25 [3.4K]

Answer:

73.5

Step-by-step explanation:

I assume you are trying to find 4x+23y. To do this, let's plug in 1.5 for x and 3 for y. When we do so, we get 4*1.5 + 23*3 = 6 + 69 = 75. Hope this helps!

6 0
2 years ago
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Vlad1618 [11]

Answer:

g(x) = 1: 0

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Step-by-step explanation:

Just by looking at the graph, at the origin, <em>x</em><em> </em>is 0 and g(x)<em> </em>is 1. Now, look at where <em>x</em><em> </em>is -2. Below that, you see where g(-2) is because -2 REPLACED <em>x</em><em>.</em>

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4 0
3 years ago
What is the quantity (7+7+7)/(-7-7-7) equal to?<br><br><br><br>please help ​
Readme [11.4K]

Answer:

- 1

Step-by-step explanation:

\frac{7+7+7}{-7-7-7}

= \frac{21}{-21}

= - 1

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2 years ago
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