Question

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Answer 1

Given the radius of 50 miles and the line joining the cities at (0, 56) and (58, 0), the transmitter signal can be picked during 59.24 miles of the drive.

How can the duration of signal reception be found?

Radius of broadcast of the transmitter = 50 miles

Location of starting point = 56 miles north of the transmitter

Location of destination city = 58 miles east of the transmitter

Therefore we have;

Slope of the line joining the two cities

= 56 ÷ (-58) = -0.966

Which gives the equation of the line as follows;

y = -0.966•x + 56

The equation of the circle is;

${x}^{2} + {y}^{2} = {50}^{2}$

${x}^{2} + { ( - 0.966x + 56)}^{2} = {50}^{2}$

1.933156•x^2 - 108.192•x + 636 = 0

Which gives;

• x = 6.67 or x = 49.29

Therefore;

When x = 6.67, we have;

• y = -0.966 × 6.67 + 56 = 49.56

When x = 49.29, we have;

• y = -0.966 × 49.29 + 56 = 8.4

The length of the drive, during which the driver can pick the signal, l, is therefore;

l = √((49.56-8.4)^2 + (49.29-6.67)^2) = 59.24 miles

• The length of the drive during which the signal is received is 59.24 miles

Learn more about the equation of a circle here:

brainly.com/question/502872

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