Aplane leaves jfk international airport and travels due west at 570 mi/hr. another plane leaves 20 minutes later and travels 22° west of north at the rate of 585 mi/h. to the nearest ten miles, how far apart are they 40 minutes after the second plane leaves. attach a picture if convenient.

557 miles

Step-by-step explanation:

Plane 1

It travels for 60 minutes (20+40) = 1 hour

Distance = 1 hour* 570 mph =  570 miles  in a negative x direction

(-570,0)

Plane 2

travels 40 minutes = 2/3 hour

Distance = 2/3 * 585 mph = 390 miles

-x direction = 390 sin 22

y direction = 390 cos 22

(-390 sin22, 390 cos22)

distance = sqrt ((x2-x1)^ + (y2-y1)^2)

              = sqrt( (-390 sin22 - -570)^2 + (390cos22)^2)

              = sqrt((-390 sin22 +570)^2 + (390cos22)^2)

              = sqrt( 423.9034286^2 +361.6017033^2)

              =sqrt(31044909086)

               =557.18031961590

the f

last graph i believe

step-by-step explanation:

worked it out i believe

(f - g)(x) = 5x^3 - 3

step-by-step explanation:

rearrange these functions vertically, with like terms in separate columns:

f(x) = 5x^3 - 2

g(x) =         +1

now subtract, column by column:

(f - g)(x) = 5x^3 - 3