Please help me A rocket is launched from a tower. The height of the rocket, y in feet, is related to the time after launch, x in seconds, by the given equation. Using this equation, find the maximum height reached by the rocket, to the nearest tenth of a foot.

y=-16x^2+212x+139

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2. dual - aspect

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the answer is a.

step-by-step explanation:

in three games of golf, alex shot two under par four times.

hello from mrbilldoesmath!

answer:

(7 x + 6 y) (49 x^2 - 42 x y + 36 y^2)

discussion:

the trick is to recognize that 343 = 7^3 and 216 = 6^3 and, in general,

a^3 + b^3 =   (a + b) (a^2 - a b + b^2).

in our case

343x^3+216y^3   =   (7x)^3 + (6y)^3

so take a = 7x and b = 6y in the above equation to get the answer.

you,

mrb

yx = 36

step-by-step explanation:

cos f = opposite side/ hypotenuse

cos f = yw / yx

we know cos f = 4/9 and yw =16

4/9 = 16/yx

multiply each side by yx

4/9 yx = 16/ yx * yx

4/9 yx = 16

multiply each side by 9/4

9/4 * 4/9 yx = 16 *9/4

yx = 36