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.:

Guest Sep 30, 2021

#1**+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

ElectricPavlov Sep 30, 2021