Which compound inequality is equivalent to ax-b>c for all real numbers a, b, and c, where c>=0?

(2,6)

step-by-step explanation:

cause i just did this and got the answer

you're welcome

answer: integers are just whole numbers: positive, negative, and zero.

i = {-∞, -2, -1, 0, 1, 2, 3, ∞}

let x = 1st integer

let y = 2nd integer

x = 3y + 17   {equation 1}

xy = -24         {equation 2}

substitute 3y+17 in place of x n equation 2

and solve for y.

(3y+17)y = -24

3y2 + 17y = -24

3y2 + 17y + 24 = 0

from quadratic equation:

y = [-17±√(172-4(3)(24))]/(2(3))

y = (-17 ±√1)/6

y = (-17+1)/6   or     y = (-17-1)/6

y = -16/6           or     y = -18/6

y = -2 2/3         or   y = -3

since we are dealing with

y = -3

x = 3y+17 = -9+17 = 8

the integers are -3, 8

step-by-step explanation:

answer: -10

step-by-step explanation:

s(t) = 1 - 10t

s(x) = 1 - 10x

s(x + h) = 1 - 10(x + h)

            = 1 - 10x - 10h

s(x + h) - s(x) = 1 - 10x - 10h - (1 - 10x)

                      = 1 - 10x - 10h - 1 + 10x

                      = -10h

= -10