Two cars start moving from the same point. One travels south at 24 mi/h and the other travels west at 18 mi/h. At what rate (in mi/h) is the distance between the cars increasing two hours later

The rate at which the distance between the two cars is increasing is 30 mi/h

Step-by-step explanation:

Given;

speed of the first car, v₁ = 24 mi/h

speed of the second car, v₂ = 18 mi/h

Two hours later, the position of the cars is calculated as;

position of the first car, d₁ = 24 mi/h  x 2 h  = 48 mi

position of the second car, d₂ = 18 mi/h  x 2 h = 36 mi

The displacement of the two car is calculated as;

displacement, d² = 48² + 36²

                        d² = 3600

                        d = √3600

                        d = 60 mi

The rate at which this displacement is changing = (60 mi) / (2h)

                                                                                 = 30 mi/h

answer: 2530

hope this

9) i can't see the angle so whatever it is just do this: cos(angle) x hypotenuse and then use that answer and they hypotenuse and plug it into this: (answer you just got)^2 + b^2 = (hypotenuse)^2. the first answer will be the answer of the adjacent side and the second answer after this equation will be the opposite side.

10) y = 14 x = √147 or 12.12435565

11) x = 4.8

12) x = 4.2

13) x = 9.3

14) pq: 13.2

qr : 17.6

angle p: 53º

15) angle c: 54.3º

angle a: 35.7º

16) h = 21040

v = 1841

step-by-step explanation:

10) 7/sin(30) = 14 = y

7^2 + b^2 = 14^2

49 + b^2 = 196

b^2 = 196 - 49

√b^2 = √147

x = √147

11) cos(70) x 14 = 4.788282007

x = 4.8

12) tan(25) x 9 = 4.196768923

x = 4.2

13) sin(31) x 18 = 9.270685348

x = 9.3

14) pq: sin(37) x 22 = 13.23993051

pq: 13.2

qr: cos(37) x 22 = 17.56998122

qr : 17.6

angle p: 180 - (37 + 90) = 53º

15) cos-1(7/12) = 54.31466529

angle c: 54.3º

angle a: 180 - (90 + 54.31466529) = 35.68533471

angle a: 35.7º

16) 4 x 5280 = 21120

cos(5) x 21120

h = 21039.63202

h = 21040

sin(5) x 21120

v = 1840.729287

v = 1841