A long jumper leaves the ground at an angle of 20° above the horizontal, at a speed of 9 m/s. The height of the jumper can be modeled by h(x) = -0.051x2 +0.364.x, where h is the jumper's height in meters and x is the horizontal distance from the point of
launch.
Part: 0/3
Part 1 of 3
(a) At what horizontal distance from the point of launch does the maximum height occur? Round to 2 decimal places.
The long jumper reaches a maximum height when the horizontal distance from the point of
launch is approximately 0 meters.

intensity of the vector is v= √37 ≈ 6.08 units and make angle ∡α ≈ 9.46° with east direction

step-by-step explanation:

required vector is consists of the two componentsvx= 2+4=6 units   and vy= 1 unit and vx ⊥ vywe will use pythagorian theorem to find intensity of the vector vv∧2 = vx∧2 + vy∧2 => v = √vx∧2 + vy∧2 = √6∧2 + 1∧2 = √36+1 = √37 ≈ 6.08 unitsthe angle ∡α between vector and east direction we wil find with tanαtanα = 1/6 => α = arc tanα = arc 1/6 => α ≈ 9.46°

hope this

687.09 dollars       happy now

step-by-step explanation: 6870

1857 times .63 equals 1169.91 subtract that from 1857 you get 687.09

which expression represents the approximate length of bc?

step-by-step explanation:

i dont know sorry

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