Asmall radio transmitter broadcasts in a 44 mile radius. if you drive along a straight line from a city 56 miles south of the transmitter to a second city 53 miles west of the transmitter, during how much of the drive will you pick up a signal from the transmitter? (hint: you will need to find the equations of the circle and line, the intersection and the differnence of the distances.) also show work for finding the points of intersection.

the transmitter signal is picked up for 42.63m of the drive.

Ans = 42.63m

Step-by-step explanation:

Solution

Let triangle ABC formed by line from 56 miles south of transmitter, to the transmitter itself then to 53 miles west of the transmitter with sides

AB BC AC

Where AB = 56 miles

and BC = 53 miles

Therefore AC = Sqr((56m)^2 +(53m)^2) = 77.10m

The point of intersection of the radius of the radio transmitter signal and the triangle formed by the path of travel of the traveller and the lines AB and AC 

To find the perpendicular line that can be drawn from C to AB we have from trigonometric relations

56 × sin(t) = 53 × sin (90 - t) because the traveller moves from directly south of the transmitter to directly west of the transmitter

Hence we have

56×sin(t) = 53×cos(t) because sin(90-t) = cos(t)

Rearranging 56×sin(t) = 53×cos(t) we have

1=(56×sin(t))/ (53×cos(t))

or (sin(t)/ cos(t))=1/(56/53)=53/56

That is tan(t)=53/56 and ACTAN(t) = 43.42°

Angle (t)

Drawing a perpendicular line from the point of the radio transmitter C to the travel path of the traveller AB   and calling the point of Intersection E we have EC = 53×sin(43.42)=38.49m

It is seen that the distance from the point of intersection of the radius of the radio transmitter and intersection of the line CE and AB  is EI1 where I1 is the first point of intersection of the radius of the radio transmitter and the line AB

EI1=Sqr((44m)^2-(38.49m)^2)

= 21.31m

Also since the triangles CEI1 and CEI2 are identical, it follows that EI1 = EI2 = 21.32m

The distance over which the traveller will be able to receive the signal from the radio transmitter while travelling from point A to point B is the distance I1 to I2 which is equal to 2*I

=2×21.31=42.63m

Ans = 42.63m

5x + 5

step-by-step explanation:

you should always add the diagrams and graphs we cannot see. this particular question is kind of borderline because we can make a graph for you.

(f + g)x = -x^2 +3x + 5 + x^2 + 2x

(f + g)x = 5x + 5

the graph of z = 5x + 5 is shown below. always use desmos when in doubt.