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Lady bird [3.3K]
3 years ago
15

You want to swim across a stream that is 150 feet wide. However as you swim the current carries you downstream and you end up 75

feet from where you
wanted to be. How far did you actually swim? (Round your answer to the nearest tenth)

No links!
Thanks
Mathematics
1 answer:
alekssr [168]3 years ago
8 0
Lol I just need points sorry
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3 years ago
An airplane leaves an airport at 9:00 p.M. With a heading of 270 and a speed of 610 mph. At 10:00 p.M. The pilot changes the he
timama [110]

Answer:

2330.51 miles

Step-by-step explanation:

Given that the speed of the airplane = 610 mph.

The airplane leaves an airport at 9:00 P.M. with a heading of 270 degrees and at 10:00 P.M., the pilot changes the heading to 310 degrees.

So, for 1 hour the plane is heading at 270 degrees and for 3 hours, from 10:00 p.m to 1:00 a.m, the plane is heading at 310 degrees as shown in the figure.

As, distance = time x speed, so

The distance covered at 270 degrees, d_1 = 1\times610=610 miles.

The distance covered at 310 degrees, d_2 = 3\times610=1830 miles.

Total distance covered, d, is the magnitude of the sum of vectors \vec{d_1} and \vec{d_2} as shown in the figure.

The angle between the vectors \vec{d_1} and \vec{d_2}, \theta=310-270=40 degree.

Magnitude of sum of the vectors \vec{d_1} and \vec{d_2},

d= \sqrt{|\vec{d_1}|^2+|\vec{d_2}|^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\theta} \\\\\Rightarrow d=\sqrt{610^2+1830^2+2|\vec{d_1}|\;|\vec{d_2}|\cos\(40^{\circ})}

\Rightarrow d=2330.51 miles

Hence, at 1:00 a.m, the airplane is at a distance of 2330.51 miles from the airport.

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