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crimeas [40]
3 years ago
11

A small radio transmitter broadcasts in a 29 mile radius. If you drive along a straight line from a city 32 miles north of the t

ransmitter to a second city 40 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter
Mathematics
1 answer:
hram777 [196]3 years ago
7 0

Answer: He will pick up a signal from the transmitter 29.43 miles of the drive.

Step-by-step explanation:

Since we have given that

Radius = 29 mile

So, equation of circle would be

x^2+y^2=29^2=841

There is a straight line from a city 32 miles north i.e. (0,32) to a second city 40 miles east of the transmitter i.e. (40,0)

So, equation of line would be

y-32=\dfrac{0-32}{40-0}(x-0)\\\\y-32=\dfrac{-32}{40}x\\\\y-32=-0.8x\\\\y+0.8x=32\\\\y=32-0.8x

So, we will put this value in the equation of circle.

x^2+(32-0.8x)^2=841\\\\x^2+1024+0.64x^2-51.2x=841\\\\1.64x^2-51.2x=841-1024\\\\1.64x^2-51.2x=-183\\\\1.64x^2-51.2x+183=0

So, x= 4.12 and x = 27.1

So, y = 28.7, y = 10.32

So, distance between them would be

\sqrt{(x_2-x_1)^2+(y_2-y_1)^2}\\\\=\sqrt{(27.1-4.12)^2+(28.7-10.32)^2}\\\\=29.43

Hence, he will pick up a signal from the transmitter 29.43 miles of the drive.

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5 0
3 years ago
A baseball diamond is a square with side 90 ft. A batter hits the ball and runs toward first base with a speed of 31 ft/s. (a) A
julsineya [31]

Answer:

a) -13.9 ft/s

b) 13.9 ft/s

Step-by-step explanation:

a) The rate of his distance from the second base when he is halfway to first base can be found by differentiating the following Pythagorean theorem equation respect t:

D^{2} = (90 - x)^{2} + 90^{2}   (1)

\frac{d(D^{2})}{dt} = \frac{d(90 - x)^{2} + 90^{2})}{dt}

2D\frac{d(D)}{dt} = \frac{d((90 - x)^{2})}{dt}  

D\frac{d(D)}{dt} = -(90 - x) \frac{dx}{dt}   (2)

Since:

D = \sqrt{(90 -x)^{2} + 90^{2}}

When x = 45 (the batter is halfway to first base), D is:

D = \sqrt{(90 - 45)^{2} + 90^{2}} = 100. 62

Now, by introducing D = 100.62, x = 45 and dx/dt = 31 into equation (2) we have:

100.62 \frac{d(D)}{dt} = -(90 - 45)*31          

\frac{d(D)}{dt} = -\frac{(90 - 45)*31}{100.62} = -13.9 ft/s

Hence, the rate of his distance from second base decreasing when he is halfway to first base is -13.9 ft/s.

b) The rate of his distance from third base increasing at the same moment is given by differentiating the folowing Pythagorean theorem equation respect t:

D^{2} = 90^{2} + x^{2}  

\frac{d(D^{2})}{dt} = \frac{d(90^{2} + x^{2})}{dt}

D\frac{dD}{dt} = x\frac{dx}{dt}   (3)

We have that D is:

D = \sqrt{x^{2} + 90^{2}} = \sqrt{(45)^{2} + 90^{2}} = 100.63

By entering x = 45, dx/dt = 31 and D = 100.63 into equation (3) we have:

\frac{dD}{dt} = \frac{45*31}{100.63} = 13.9 ft/s

Therefore, the rate of the batter when he is from third base increasing at the same moment is 13.9 ft/s.

I hope it helps you!

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3 years ago
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~I hope this is correct and has helped you, have a great day/night~

Step-by-step explanation:

6 0
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