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 time that the rocket will hit the ground, to the nearest 100th of second. y=-16x^2+189x+79

make sure you have version 2.50

try looking in the very back of the book : )

step-by-step explanation:

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тgє qυєѕтισи αѕкє∂: ιи тнє ѕєяιєѕ σя яєα¢тισиѕ тgαт σ¢¢υя ιи ρнσтσѕуитнєѕιѕ, ωнι¢н fσямѕ ωнєи ℓιgнт ѕтяιкєѕ α мσℓє¢υℓє σf ¢нℓσяσρнуℓℓ.

α. мσℓє¢υℓє σf ωαтєя

в.тнєє-¢αявσи ѕυgαя

¢. "єχ¢ιтє∂" єℓє¢тяσи

∂. иα∂ρн

му αиѕωєя: ιт ιѕ ¢. "єχ¢ιтє∂" єℓє¢тяσи

нσρє ι ¢συℓ∂ нєℓρ υ

alright so i think for this one it would to draw it out. this may be hard for me to try and draw, but i will do my best on here:

        r

  /       |     \

notice how it forms 2 triangles inside one bigger triangle.

so if we know the angle spr is the same as psr that is saying the measure of angle p is the same as angle s. this means the big triangle psr is either isosceles or equilateral. in both cases, the triangle is perfectly symmetrical when split down the middle, which is in this case rq. because we know it is perfectly symmetrical on both sides, we know that the two triangles inside the bigger one are equal.