Two airplanes leave an airport at the same time. The velocity of the first airplane is
680 m/h at a heading of 39º. The velocity of
the second is 560 m/h at a heading of 129º.
How far apart are they after 2.5 h?
Answer in units of m.

y > 1/3x

y < = -x + 4

hope this : )

d) by the cross product property, ab^2 = bc multiplied by ad

step-by-step explanation:

we are given

in the given triangle abc

angle a is 90° and segment ad is perpendicular to segment bc

we can see that

triangle(abc) and triangle(abd) are similar because they have common side and same angle (90 degree)

so, triangles(abc) and abd are similar

since, they are similar

so, the ratio of their sides must be equal

we get

now, we can cross multiply

and we get

so, this could be step to prove it

step-by-step explanation:

bob the banana

step-by-step explanation:

659