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Serhud [2]
3 years ago
13

|25 POINTS| Suppose you have one pyramid with a base that is 10 meters long and 10 meters wide with a slant that is 15 meters. N

ow, if the length of the base is increased to 15 meters, what is the difference in the surface area between the new pyramid and the old pyramid?
125 meters squared.
525 meters squared.
600 meters squared.
400 meters squared.
Mathematics
2 answers:
ruslelena [56]3 years ago
3 0

Answer:

125 meters squared

Step-by-step explanation:

Surface area of initial pyramid:

  • A= 10²+4*1/2*10*15= 400 m²

Surface area of extended pyramid:

  • A= 10*15+2*1/2*15*15+2*1/2*10*15= 525 m²

The difference is:

  • 525- 400= 125 m²
inysia [295]3 years ago
3 0

Answer:

First pyramid is = 400 meters squared

A= 10²+4*1/2*10*15= 400 m²

second pyramid = 675 meters squared

A= 10*15+2*1/2*15*15+2*1/2*10*15= 525 m²

Subtract the two =

525- 400= 125 m²

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Answer:

15

Step-by-step explanation:

if you put x=-15

f(x) = 5*-5 +40

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7 0
3 years ago
What is the order of the set A={-1,20,3,17,5}
laiz [17]

Hey!

Hope this helps...

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3 0
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6 0
3 years ago
Convert the rectangular coordinates (-9, 3V3) into polar form. Express the angle
Whitepunk [10]

Answer:

(6\sqrt{3},\,\frac{5\pi}{6})

Step-by-step explanation:

The radius  r  can be found from the relationship

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The point is in Quadrant II (-, +), so use the inverse cosine function to find the angle.

\cos{\theta}=\frac{x}{r}=\frac{-9}{6\sqrt{3}}\\\cos{\theta}=-\frac{9}{6\sqrt{3}} \cdot \frac{\sqrt{3}}{\sqrt{3}}\\\cos{\theta}=-\frac{9\sqrt{3}}{6\cdot3}\\\cos{\theta}=-\frac{\sqrt{3}}{2}\\\\\cos^{-1}\frac{-\sqrt{3}}{2}}=\frac{5\pi}{6}

See the attached image.

7 0
3 years ago
A triangle has side lengths of 39, 80, and<br> 87. Is this triangle a right triangle?
iren2701 [21]

Answer:

Step-by-step explanation:

39²=1521

80²=6400

39²+80²=1521+6400=7921

87²=7569

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so it is not aright angled triangle.

4 0
2 years ago
Read 2 more answers
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