A small radio transmitter broadcasts in a 69 mile radius. If you drive along a straight line from a city 93 miles north of the transmitter to a second city 78 miles east of the transmitter, during how much of the drive will you pick up a signal from the transmitter?

See Explanation

Step-by-step explanation:

According to the Question,

Thus,

The distance that you get reception is the length of the chord created by the intersection of the circle defining the edge of transmission and the line defining the car trip.

And,

The Transmitter at the origin

City to the north at (0,93)  & City to the east at (78,0)  

the Slope is M=(-93/78)

Intercept is B= y - mx ⇒  93 - (-93/78)(0) = 93

The equation of the line between the cities is y = (-93/78)x + 93

Now, Solve the above two Equations

The intersection is gotten from the picture or solving:

x^2 + [(-93/78)*x + 93]^2 = 69^2

Now,

From the Pythagorean theorem the total distance of the trip is:  

d1 = √(93^2 + 78^2) ≈ 121.37miles  

And the distance when the signal is picked up is:

d2 =√ [(67.952-23.627)^2 + (64.82 - 11.98)^2] ≈ 68.96 miles

You will pick up a signal from the transmitter in (d2/d1)*100 = 56% of the drive.

what do u need to know

step-by-step explanation:

5=812a+9+3

step 1: simplify both sides of the equation.

5=812a+9+3

5=(812a)+(9+3)(combine like terms)

5=812a+12

5=812a+12

step 2: flip the equation.

812a+12=5

step 3: subtract 12 from both sides.

812a+12−12=5−12

812a=−7

step 4: divide both sides by 812.

812a /812 = -7/812

a= -1/116

so the answer is

a=   -1/116