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Anika [276]
3 years ago
13

∠A and \angle B∠B are vertical angles. If m\angle A=(5x-9)^{\circ}∠A=(5x−9) ∘ and m\angle B=(8x-30)^{\circ}∠B=(8x−30) ∘ , then f

ind the value of x
Mathematics
1 answer:
elena55 [62]3 years ago
5 0

9514 1404 393

Answer:

  x = 7

Step-by-step explanation:

Vertical angles have the same measure, so ...

  m∠A = m∠B

  (5x -9)° = (8x -30)°

  21 = 3x . . . . . . . . . divide by °, add 30-5x

  7 = x . . . . . . . . . . divide by 3

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Slope= 4

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12-8/6-5

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7 0
3 years ago
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
inessss [21]

Answer:

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:

The radius of a cone is decreasing at a constant rate.

\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}

Instant radius = 99 inch

Instant Volume = 525 cubic inches

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

Volume of cone =

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
3 years ago
If ON=8x*8, LM=7x+4, NM=x-5, and OL=3y-6, find the values of x and y for which LMNO must be a parallelogram. The diagram is not
lana66690 [7]
ON = 8x • 8
LM = 7x + 4
NM = x - 5
OL = 3y - 6

OL is congruent & parallel to NM
LM is congruent & parallel to ON

So,
8x \times 8 = 7x + 4
Simplify
64x = 7x + 4
subtract 7x from both sides
57x = 4
divide 57 from both sides
x =  \frac{4}{57}
Substitute x into equations
8x \times 8 = 7x + 4 =  4  \frac{28}{57}

3y - 6 = x - 5
3y - 6 = 4 \frac{28}{57}  - 5
3y - 6 =  -  \frac{28}{57}
NM & OL = -28/57
ON & LM = 4 + (28/57)


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Answer:

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8 0
3 years ago
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