You can
put this solution on YOUR website!Distance(d)=Rate(r)times Time(t) or d=rt; r=d/t and t=d/r
let r=rate (speed) of plane in still air
rate out with wind=r+45
rate returning against wind=r-45
also
let t=time it takes returning against the wind
then t-1=time it takes going out with the wind
Now we know that:
time going out with the wind=time returning against the wind minus 1 hour
So----
450/(r+45)=(450/(r-45))-1 multiply both sides by (r+45)(r-45) to get rid of fractions
450(r-45)=450(r+45)-(r+45)(r-45) get rid of parens
450r-20250=450r+20250-r^2+2025 subtract 20250, 2025 and also 450r from both sides
450r-450r-20250-20250-2025=450r-450r+20250-20250-r^2+2025-2025 collect like terms
-42525=-r^2 0r
r^2=42525 take square root of both sides
r= plus or minus 206.2
r=206.2 mph-------------------------speed in still air
CK
time out=450/(206.2+45)=450/(251.2)=1.791 hours
time returning=450(206.2-45)=450/(161.2)=2.791 hr
2.791-1.791=1 hour
Hope this helps----ptaylor
