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GenaCL600 [577]
1 year ago
11

15

Mathematics
1 answer:
alexandr1967 [171]1 year ago
7 0

By using parallel lines and transversal lines concept we can prove m∠1=m∠5.

Given that, a║b and both the lines are intersected by transversal t.

We need to prove that m∠1=m∠5.

<h3>What is a transversal?</h3>

In geometry, a transversal is a line that passes through two lines in the same plane at two distinct points.

m∠1+m∠3= 180° (Linear Pair Theorem)

m∠5+m∠6=180° (Linear Pair Theorem)

m∠1+m∠3=m∠5+m∠6

m∠3=m∠6

m∠1=m∠5 (Subtraction Property of Equality)

Hence, proved. By using parallel lines and transversal lines concept we can prove m∠1=m∠5.

To learn more about parallel lines visit:

brainly.com/question/16701300

#SPJ1

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The radius of a cone is decreasing at a constant rate of 7 inches per second, and the volume is decreasing at a rate of 948 cubi
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The height of cone is decreasing at a rate of 0.085131 inch per second.        

Step-by-step explanation:

We are given the following information in the question:

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\displaystyle\frac{dr}{dt} = -7\text{ inch per second}

The volume is decreasing at a constant rate.

\displaystyle\frac{dV}{dt} = -948\text{ cubic inch per second}

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We have to find the rate of change of height with respect to time.

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V = \displaystyle\frac{1}{3}\pi r^2 h

Instant volume =

525 = \displaystyle\frac{1}{3}\pi r^2h = \frac{1}{3}\pi (99)^2h\\\\\text{Instant heigth} = h = \frac{525\times 3}{\pi(99)^2}

Differentiating with respect to t,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)

Putting all the values, we get,

\displaystyle\frac{dV}{dt} = \frac{1}{3}\pi \bigg(2r\frac{dr}{dt}h + r^2\frac{dh}{dt}\bigg)\\\\-948 = \frac{1}{3}\pi\bigg(2(99)(-7)(\frac{525\times 3}{\pi(99)^2}) + (99)(99)\frac{dh}{dt}\bigg)\\\\\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi} = (99)^2\frac{dh}{dt}\\\\\frac{1}{(99)^2}\bigg(\frac{-948\times 3}{\pi} + \frac{2\times 7\times 525\times 3}{99\times \pi}\bigg) = \frac{dh}{dt}\\\\\frac{dh}{dt} = -0.085131

Thus, the height of cone is decreasing at a rate of 0.085131 inch per second.

3 0
2 years ago
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