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A small radio transmitter broadcasts in a 23 mile radius. If you drive along a straight line from a city 31 miles north of the transmitter to a second city 32 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

 

 

This is what I have so far:

 

y= (-31/32)x+31

 

x^2+y^2=23^2

 

x^2+(-31/32 x+31)^2 = 23^2

 

(1985 x^2 - 61504 x + 984064)/1024 = 529

 

x= 30752/1985 - (32 sqrt(66001))/1985 = 11.350624115856267752682526003638252741963852552082890846307116452

 

x= 30752/1985 + (32 sqrt(66001))/1985 = 19.633758755680256176788506741953686804635643669579577667546787829

 

However, I need help with getting the right answer by finding the distance between the two x points basically I'm looking for a distance that represents part of the drive from A to B.:

 

 Sep 30, 2021
 #1
avatar+35059 
+1

The total 'x' distance of the drive goes from x = 0    to  x = 32      for 'x' distance 32   

 

the part of the drive where you can hear the radio goes from x = 11.35   to  19.64     for x distance  8.29 

 

percentage of distance you can hear the station is then       8.29/32 = 25.9%

 

 

 

 

You COULD find the distance between the starting and end points    and then the points of intersection...BUT that is a much more difficult method than is necessary...    

distance formula   d^2 = (x1-x2)^2  + (y1-y2) ^2    (recognize this as modified pythagorean theorem)    .....   so for the two intersection points you would have to calculate the 'y' coordinates (you have the x's)

    for A to B      d^2 =  (0-32)^2  + ( 31-0)^2      resulting in d =  44.55 miles

 

  I would use a quadratic formula calculator like this one:

https://www.calculatorsoup.com/calculators/algebra/quadratic-formula-calculator.php

 Sep 30, 2021
edited by ElectricPavlov  Sep 30, 2021
 #2
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+1

What are you on about guest.

 

That is a direct extract copy from one of my recent answers.

 

You are one very lazy rude person!

 Sep 30, 2021

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