Two airplanes leave an airport at the same time. the velocity of the first airplane is 730 m/h at a heading of 44.3 ◦ . the velocity of the second is 590 m/h at a heading of 83◦ . how far apart are they after 2.9 h?

So airplane will be 1324.9453 m apart after 2.9 hour

Explanation:

So if we draw the vectors of a 2d graph we see that the difference in angles is  = 83 - 44.3 =

Distance traveled by first plane = 730×2.9 = 2117 m

And distance traveled by second plane = 590×2.9 = 1711 m

We represent these distances as two sides of the triangle, and the distance between the planes as the side opposing the angle 38.7.

Using the law of cosine, representing the distance between the planes, we see that:

d = 1324.9453 m

find the surface area of the sphere to the nearest square unit. use a calculator.

a. 254 in.2

b. 64 in.2

c. 127 in.2

d. 1,018 in.2

inverse v=i*r

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