A police car is approaching a right-angled intersection from the north at 35 mph, and a speeding car is heading east at 30 mph. When the police car is 0.75 miles north of the intersection and the speeding car is 1 mile to the east, what is the rate that the distance between the police car and car is changing?

The rate that the distance between the police car and car is changing = 3 m/s

Step-by-step explanation:

The given speed of the police car = 35 mph

The speed of the speeding car = 30 mph

The position of the police car = 0.75 miles north

The position of the speeding car = 1 miles east

The distance of the police car from the speeding car, d = √(0.75² + 1²) = 1.25

d² = x² + y²

2d·dd/dt = 2x·dx/dt + 2·y·dy/dt

Where;

dx/dt = 30 mph

dy/dt = -35 mph

x = 1

y = 0.75

Substituting gives;

2×1.25 ×dd/dt = 2×1×30 - 2×0.75×35

2×1.25 ×dd/dt = 7.5

dd/dt = 7.5/2.5 = 3

dd/dt = 3 m/s

The rate that the distance between the police car and car is changing = dd/dt = 3 m/s

i think it is everything has to equal 180 so, if i correct 4=c. 6=d. 8=a. 7=b. and the last question would be c. because they are parallel lines and they are congruent.

step-by-step explanation:

the reason everything has to equal to 180 is because that is how it works. i'm guessing that is its rule. 4 and 8 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 6 and 7 are congruent and parallel so, that would be 30+30 because they have the same number. it would equal 60. 60+60 is 180. so, now you have your answer