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yaroslaw [1]
2 years ago
11

An airplane flying into a headwind travels the 1890-mile flying distance between two cities in 3 hours. On the return flight, th

e airplane travels this distance in 2 hours and 30 minutes. Find the airspeed of the plane and the speed of the wind, assuming that both remain constant.
Mathematics
1 answer:
aksik [14]2 years ago
4 0

Answer:

Velocity (against wind) = 1,890 / 3 = 630 mph

Velocity (with the wind) = 1,890 / 2.5 = 756 mph

Plane speed (still air) = 630 + wind speed

Plane speed (still air) = 756 - wind speed

Adding both equations equals

2 * Plane speed (still air) = 1,386

Plane Speed = 693 mph

Source: https://www.1728.org/veloccal.htm


Step-by-step explanation:


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Only truck 1 can pass under the bridge.

Step-by-step explanation:

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y^{2}=144(\frac{361-x^{2}}{361})

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y=\frac{12}{19}\sqrt{361-x^{2}}

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y=\frac{12}{19}\sqrt{361-(8)^{2}}

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