A Small Radio Transmitter Broadcasts
+38 A small-scale radio transmitter broadcasts in a 27 mile radius. If you lot drive along a direct line from a metropolis 35 miles north of the transmitter to a second city 31 miles east of the transmitter, during how much of the bulldoze will you pick up a signal from the transmitter?
+124599 Let the transmitter be located at (0,0)
The equation for its effective coverage area will be a circumvolve centered at (0,0) with a radius of 27
And then....the equation is
x^2 + y^ii = 27^2
x^2 + y^2 = 729
Let the signal 35 miles north of the transmitter be (0, 35)
Let the point 31 miles east of the trasmitter be (31,0)
The line connecting these (the line of travel) will take a slope of [ 35 - 0] / [ 0 - 31] = -35/31
The equation of this line will exist
y = (-35/31)10 + 35
We could solve this algebraically, but a graph seems easier to bargain with
See here : https://world wide web.desmos.com/estimator/ksd2p5jdwl
The signal will exist received betwen the points (8.221 , 25.718) and (26.523, 5.055)
The altitude betwixt these points is
√ [ ( 26.523 - 8.221)^2 + ( 25.718 - v.055)^2 ] ≈ 27.6 miles (1)
The distance between (0, 35) and (31,0) = √ [ ( 31)^2 + ( 35)^2 ] ≈ 46.eight miles (2)
And so....you will receive the bespeak nigh (1)/ (ii) = 27.6 / 46.8 ≈ 59% of the drive
+38 Sad for the bother simply is there whatsoever gamble you could evidence me how to do it algebraically?
It'south review for a examination and I won't have acess to a graphing calculator.
+124599 Yeah....I can work it out for you.....in just a few....
+124599 We demand to notice the x intersection points of
x^ii + y^two = 729 and
y = (-35/31)ten + 35
So....subbing the second equation into the first for y we have that
x^2 + [ (-35/31)x + 35]^2 = 729
x^ii + 1225x^two/961 -2450x/31 + 1225 = 729
[961 + 1225]x^2 / 961 - 2450x/31 + 496 = 0 multiply through by 961
2186 x^2 - 75950x + 476656 = 0
Using the quadratic formula we accept that
75950 ±√ [ (75950^2 - 4(2186)(476656) ]
_____________________________________
2 * 2186
Evaluating this we get that 10 = eight.221 and 10 = 26.523
Putiing the commencement value into the linear equation nosotros have that
y = (-35/31)(viii.221) + 35 ≈ 25.718
Putting the second vallue into the linear eqation we have that
y = (-35/31)(26.523) + 35 ≈ 5.055
So....the intersection points of the circumvolve and line of travel are ( 8.221, 25.718) and (25.523, five.055)
And I just used the distance formula twice to solve the balance of the problem
A Small Radio Transmitter Broadcasts,
Source: https://web2.0calc.com/questions/been-stuck-on-this-one-help-would-be-appreciated
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