Aball is thrown in the air from a ledge. its height in feet is represented by f(x) = 16(x2 – 5x – 6), where x is the number of seconds since the ball has been thrown. the height of the ball is 0 feet when it hits the ground. how many seconds does it take the ball to reach the ground?

we are given

equation in height as

Since, the height of the ball is 0 feet when it hits the ground

so, we can set f(x)=0

and then we can solve for x

now, we can factor it and then we can solve for x

So, it takes 6 seconds the ball to reach the ground...........Answer

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

it should be a.

hope it