Acar and a motorcycle leave at noon from the same location, heading in the same direction. the average speed of the car is 30 mph slower than twice the speed of the motorcycle. in two hours, the car is 20 miles ahead of the motorcycle. find the speed of both the car and the motorcycle, in miles per hour.

Motorcycle  : 40 miles per hour

Car: 50 miles per hour

Step-by-step explanation:

Hi, we have to write a system of equations.

The average speed of the car is 30 mph slower than twice the speed of the motorcycle.

Where

C = speed rate of the car

M = speed rate of the motorcycle

since distance = speed x time

In two hours(time), the car is 20 miles(distance) ahead of the motorcycle.  

Since the car is ahead of the motorcycle , the speed is the difference of both.(C-M)

Replacing the value of C by 2 M -30  , in the second equation:

20 = [(2M-30)-M ]2

20 = [M-30]2

20 = 2M-30(2)

20= 2M -60

20+60 =2M

80 = 2M

80/2 =M

M =40 km per hour

Replacing M in the first equation:

C = 2 (40)-30 = 80-30

C = 50 km per hour

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3