can some set the equation up for me.
Science and medicine. A plane flies 720 mi against a steady 30-mi/h headwind and then returns to the same point with the wind. If the entire trip takes 10 h, what is the plane’s speed in still air?
i thought it was this set up but its not:
(720)/(30)=(x)/(10)
The correct equation that you need to solve is:
720/(V-30) + 720/(V+30) = 10
V-30 is the planes ground speed flying inti the wind, and V+30 is the ground speed with the 30 mph tail wind. What I wrote was the total time required for both trips.
DRAWLS ,
YOU JUST LOST ME....HOW AM I SUPPOST TO SOLVE FOR THIS
720/(V-30) + 720/(V+30) = 10
Get a common denominator of
(v-30)(v+30)
then multiplying both sides by that, then
720(V+30) + 720(V-30)= 10(v-30)(v+30)
then multily out, and solve for v.
During rush hour, Fernando can drive 35 miles using the side roads in the same time that it takes to travel 30 miles on the freeway. If Fernando’s rate on the side roads is 9 mi/h faster than his rate on the freeway, find his rate on the side roads.
To set up the equation for this problem, let's assume Fernando's rate on the freeway is x mi/h. Since his rate on the side roads is 9 mi/h faster, his rate on the side roads would be x + 9 mi/h.
We know that the time it takes for Fernando to drive 35 miles on the side roads is the same as the time it takes to travel 30 miles on the freeway.
The formula to calculate time is distance divided by rate.
So, the time taken to drive 35 miles on the side roads would be 35/(x + 9) hours, and the time taken to travel 30 miles on the freeway would be 30/x hours.
Since these two times are equal, we can set up the equation:
35/(x + 9) = 30/x
Now, we can solve for x. To do that, we can cross-multiply:
35x = 30(x + 9)
Now expand the equation:
35x = 30x + 270
Simplify further:
5x = 270
Now divide both sides by 5:
x = 54
Therefore, Fernando's rate on the freeway is 54 mi/h.
To find his rate on the side roads, we can add 9 to the freeway rate:
x + 9 = 54 + 9 = 63 mi/h.
So, Fernando's rate on the side roads is 63 mi/h.