Two jet planes travelling in opposite directions were 3800 miles apart at the end of 4 hours. if one traveled 80 miles per hour faster than the other, find the rate of each plane.

Their total speed is (3800 mi)/(4 h) = 950 mi/h. This lets you write two equations for the speeds (x, y) of the two planes:

... x + y = 950

... x - y = 80 . . . . . . one travels 80 mi/h faster than the other

Adding the two equations gives

... 2x = 1030

... x = 515

... y = 515 -80 = 435

The speeds of the planes are 515 miles per hour and 435 miles per hour.

Comment on the problem

This is a problem of the generic type "sum and difference problem." You are given the total of the two speeds and the difference between them. The higher speed is the average of those two numbers: (950+80)/2 = 515.

This is the generic solution for any "sum and difference problem:" the larger value is the average of the sum and difference.

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