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zzz [600]
2 years ago
11

A beacon is flashing on top of a 50 foot tower. A 6 foot tall man walks constantly away from the tower at 5 feet/sec. At the ins

tant the man is 30 feet away from the tower, what rate is the tip of his shadow moving away from the tower
Mathematics
1 answer:
Viefleur [7K]2 years ago
7 0

Answer:\frac{253}{44}

Step-by-step explanation:

ignore the "at the instant the man is 30 feet away" part, set it as X and the man's shadow as Y.

Similar triangles so we can do \frac{50}{x+y}  = \frac{6}{y}.

Solve for it we get 44y = 6x

Differentiate relative to time t, we get 44y' = 6x'.

change in x (x') is equal to 5. And we get the answer y' = \frac{33}{44}.

the \frac{33}{44} ft/sec is the rate of which the length of the shadow is changing. add 5 to it for the rate of the tip of his shadow moving away from the tower.

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Alla [95]

Answer: A. sqrt of 500.

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5 0
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uysha [10]

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7 0
3 years ago
A right circular cone is undergoing a transformation in such a way that the radius of the cone is increasing at a rate of 1/2 in
Ivenika [448]

Answer:

The volume is decreasing at the rate of 1.396 cubic inches per minute

Step-by-step explanation:

Given

Shape: Cone

\frac{dr}{dt} =\frac{1}{2} --- rate of the radius

\frac{dh}{dt} =-\frac{1}{3} --- rate of the height

r = 2

h = \frac{1}{3}

Required

Determine the rate of change of the cone volume

The volume of a cone is:

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

Differentiate with respect to time (t)

\frac{dV}{dt} = \frac{\pi}{3}(2rh \frac{dr}{dt} + r^2 \frac{dh}{dt})

Substitute values for the known variables

\frac{dV}{dt} = \frac{\pi}{3}(2*2*\frac{1}{3}* \frac{1}{2} - 2^2 *\frac{1}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}* \frac{1}{2} - \frac{4}{3})

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}(\frac{1}{2} - 1))

\frac{dV}{dt} = \frac{\pi}{3}(\frac{4}{3}*- 1)

\frac{dV}{dt} = -\frac{\pi}{3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{7*3}*\frac{4}{3}

\frac{dV}{dt} = -\frac{22}{21}*\frac{4}{3}

\frac{dV}{dt} = -\frac{88}{63}

\frac{dV}{dt} =-1.396in^3/min

The volume is decreasing at the rate of 1.396 cubic inches per minute

3 0
3 years ago
If the current in a circuit is 2 +j2 amps and the resistance is 3-j2 ohms, what is the voltage
victus00 [196]

Answer:

The voltage is 10 + 2j volts.

Step-by-step explanation:

Ohm's first law:

By Ohm's first law, the voltage is the multiplication of the current by the resistance, that is:

V = R*i

The voltage is measured in Volts, the current in amps and the resistance in ohms.

In this question:

Both the current and the resistance have complex parts. We have to know that:

j^{2} = -1

We have that:

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Then

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The voltage is 10 + 2j volts.

4 0
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Ilia_Sergeevich [38]

Answers are in bold:

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divide both sides by -5

-10÷-5=-5x÷-5

2=x

I hope this is good enough:

6 0
3 years ago
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