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Ainat [17]
3 years ago
6

Use Cavalieri’s Principle to calculate the exact volume of an oblique cylinder with a height of 10 meters and a circular base wi

th a radius of 8 meters
Mathematics
2 answers:
nirvana33 [79]3 years ago
7 0

Answer: Volume of an oblique cylinder is 2011.43 sq.m.

Step-by-step explanation:

Since we have given that

Radius of a cylinder = 8 meters

Height of a cylinder = 10 meters

Cavelieri's Principle states that volume of an oblique cylinder is same as the volume of right circular cylinder with equal radius and height.

As we know the formula for "Volume of cylinder":

Volume=\pi r^2h\\\\V=\frac{22}{7}\times 8\times 8\times10\\\\V=\frac{14080}{7}\\\\V=2011.43

Hence, Volume of an oblique cylinder is 2011.43 sq.m.

Aleksandr-060686 [28]3 years ago
4 0
I think it is 640 pi
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If a circle is equally divided into 10 angles, what is the measure of each angle
scoundrel [369]

Answer:

36°

Step-by-step explanation:

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4 0
3 years ago
The perimeter of a regular six-sided figure, also known as a regular hexagon, is 24 inches. What is the length of each side?
deff fn [24]
The length of each side is 4 inches.
3 0
3 years ago
M&lt;1=115<br> M&lt;2<br> Could y’all give answers for all though
Makovka662 [10]

Answer:

1) m<2 = 115°

2) m<5 = 83°

3) m<8 = 45°

4) m<4 = 137°

Step-by-step explanation:

When two lines cross over each other, creating four angles, the opposite angles are equal.

So in question one, <1 is equal to <2, the angle opposite it, and so it is 115°.

Similarly, in question two, <5 = <6 = 84°.

In question three, <8 = <9 = 45°.

and in question four, <3 = <4 =137°.

3 0
3 years ago
Read 2 more answers
P = √mx_ £x MX t make x the subject
vladimir2022 [97]

So, making x subject of the formula, x = [m - 2pt³ ±√(m²  - 4pt²m)]/{2t⁵}

<h3>How to make x subject of the formula?</h3>

Since p = √(mx/t) - t²x

So, p + t²x = √(mx/t)

Squaring both sides, we have

(p + t²x)² = [√(mx/t)]²

p² + 2pt²x + t⁴x² = mx/t

Multiplying through by t,we have

(p² + 2pt²x + t⁴x²)t = mx/t × t

p²t + 2pt³x + t⁵x² = mx

p²t + 2pt³x + t⁵x² - mx = 0

t⁵x² + 2pt³x - mx + p²t = 0

t⁵x² + (2pt³ - m)x + p²t = 0

Using the quadratic formula, we find x.

x = \frac{-b +/-\sqrt{b^{2} - 4ac} }{2a}

where

  • a = t⁵,
  • b = (2pt³ - m) and
  • c =  p²t

Substituting the values of the variables into the equation, we have

x = \frac{-(2pt^{3} - m) +/-\sqrt{(2pt^{3} - m)^{2} - 4(t^{5})(p^{2}t) } }{2t^{5} }\\= \frac{-(2pt^{3} - m) +/-\sqrt{4p^{2} t^{6} - 4pt^{2}m + m^{2} - 4p^{2}t^{6} } }{2t^{5}}\\= \frac{-(2pt^{3} - m) +/-\sqrt{m^{2}  - 4pt^{2}m } }{2t^{5}}\\= \frac{m - 2pt^{3}  +/-\sqrt{m^{2}  - 4pt^{2}m } }{2t^{5}}

So, making x subject of the formula,  x = \frac{m - 2pt^{3} +/-\sqrt{m^{2}  - 4pt^{2}m } }{2t^{5}}

Learn more about subject of formula here:

brainly.com/question/25334090

#SPJ1

6 0
1 year ago
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