Suppose a blue car is 1/2 mile south of an intersection, driving north towards the intersection at 40 mph (miles per hour). At the same time, a red car is 1/2 mile east of the intersection, driving east (away from the intersection) at an unknown speed. The the straight-line distance between the blue and red car is increasing at a rate of 15 mph. What is the speed of the red car?

dx/dt  = - 18,79 mph

Step-by-step explanation:

The two cars with the intersection point and the straight-line distance between the cars make up a right triangle. In that right triangle, the legs are the distance between each car and the intersection point, and the distance between cars is the hypothenuse

If we call  x  and  y  distances between blue car and red car respectively  and L the hypothenuse by Pythagoras theorem we have:

L²  =  x²  + y²      (1)

Tacking derivatives on both sides of the equation

2*L*dL/dt  = 2*x*dx/dt + 2*y*dy/dt

And from equation (1)

L²  = (0,5)² + (0,5)²    ⇒  L = √(0,5)²  + (0,5)²   ⇒ L = 0,5*√2

By subtitution in equation (2)

2*(0,5*√2)*15 = 2*0,5*dx/dt + 2*0,5*40

(15*√2 - 40 ) / 1  = dx/dt  [mph]

dx/dt  = - 18,79 mph

Note the( - ) sign is equivalent to say that the car is driving away from the intersection point

answer: the value of x is 7cm

step-by-step explanation:

answer rather than having to add and subtract the measurements of the semi-circle, notice that the semicircle to the left of ad is actually identical in area to the blank semicircle-shaped space to the left of bc. so if we transferred the red semicircle into the blank space, we would end up with a square of side length 24 cm. the area of a square is s^2 = 24^2 = 676 cm^2.

u also answered this so i think u would know but i am here to

put in or use mine