A model rocket is launched with an initial upward velocity of 183 ft/s. The rocket's height h (in feet) after t seconds is given by the following. h=183t-16t^2
Find all values of t for which the rocket's height is 95 feet.

0.55 seconds and 10.89 seconds

Step-by-step explanation:

h = 183t − 16t²

95 = 183t − 16t²

16t² − 183t + 95 = 0

Solve with quadratic formula:

t = [ -(-183) ± √((-183)² − 4(16)(95)) ] / 2(16)

t = (183 ± √27409) / 32

t = (183 ± √27409) / 32

t ≈ 0.55 or 10.89

answer: it's d

step-by-step explanation:

hello from mrbilldoesmath!

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

discussion:

in general, the factorization of a^3 + b^3 is

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

in our case note that 343 = 7^3 and 216 = 6^3 so the equation becomes:

343x^3 + 216y^3 = (7x)^3 + (6y)^3

if you think of a (above) as 7x and b as 6y the factorization becomes

343x^3 + 216y^3 =

((7x + 6y) (   (7x)^2 - 7x*6y + (6y)^2)   =

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

you,

mrb

(a + b) (a^2 - a b + b^2)