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Galina-37 [17]
4 years ago
10

Can someone please help me with these two questions, because I really don't understand them.

Mathematics
1 answer:
polet [3.4K]4 years ago
7 0
~Hello There!~

19. You can use Pythagoras to figure out the missing length of the triangle
√15² - 9² = 12
(12 x 9)/2 = 54
((20 + 32) / 2) x 9 = 234
234 + 54 = 288

That's what i got ^^ Not 100% sure if its correct

20. 8 + 3 + 6 + 3 = 20
20/4 = 5
Mean = 5

Hope This Helps You!
Good Luck :)
Have A Great Day ^_^

- Hannah ❤

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29. Write iſ as an improper fraction, and multiply the
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Answer: 29 = 5/6

Step-by-step explanation:

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Question Help
klasskru [66]

Answer:

16 times

Step-by-step explanation:

\textsf{Volume of a cone}=\sf \dfrac{1}{3} \pi r^2 h \quad\textsf{(where r is the radius and h is the height)}

If only the radius is changed, the change in volume will be proportionate to the multiplicative factor squared.

\sf \implies Volume =\dfrac{1}{3} \pi (ar)^2h=\dfrac{1}{3} \pi (a^2)r^2h

Therefore, if the cone is quadrupled (multiplied by 4), the volume of the larger cone will be 4² times greater than the volume of the smaller cone, so <u>16 times greater </u>than the smaller cone.

<h3><u>Proof</u></h3>

Given:

  • radius = 6
  • height = 9

Substituting the given values into the formula:

\sf \implies Volume =\dfrac{1}{3} \pi (6)^2(9)=108 \pi \: \:cubic\:units

If the radius is quadrupled:

  • radius = 6 × 4 = 24
  • height = 9

Substituting the new given values into the formula:

\sf \implies Volume =\dfrac{1}{3} \pi (24)^2(9)=1728 \pi \: \:cubic\:units

To find the number of times greater the volume of the large cone is than the volume of the smaller cone, divide their volumes:

\sf \implies \dfrac{V_{large}}{V_{small}}=\dfrac{1728\pi}{108\pi}=16

So the volume of the larger cone is <u>16 times greater</u> than the volume of the smaller cone.

4 0
2 years ago
Read 2 more answers
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