Answer:
r = 225 Mil/h speed of the airplane in still air
Step-by-step explanation:
Then:
d is traveled distance and r the speed of the airplane in still air
so the first equation is for a 4 hours trip
as d = v*t
d = 4 * ( r + 25) (1) the speed of tail wind (25 mil/h)
Second equation the trip back in 5 hours
d = 5 * ( r - 25 ) (2)
So we got a system of two equation and two unknown variables d and
r
We solve it by subtitution
from equation (1) d = 4r + 100
plugging in equation 2
4r + 100 = 5r - 125 ⇒ -r = -225 ⇒ r = 225 Mil/h
And distance is :
d = 4*r + 100 ⇒ d = 4 * ( 225) + 100
d = 900 + 100
d = 1000 miles
Answer:
3x+12y+4 -- is the simplified version, variables would be x and y
WXY = XYW ( = 41 ) ⇒ XYW is a isosceles triangle
⇒ XW = YW
⇒ 54 = 6x + 6
⇒ 6( x + 1 ) = 54
⇒ x + 1 = 9
⇒ x = 8
ok done. Thank to me :>
Use the chain rule.
Let u = 25sin²(x), such that dy/dx = dy/du · du/dx
