On a coordinate plane, a parabola opens up. It goes through (negative 6, 0), has a vertex at (negative 2, negative 16), has a y-intercept at (0, negative 12), and goes through (2, 0). What are the x-intercepts of the graph of the function f(x) = x2 + 4x – 12?

answer:   n ≤ -3 or n < -5

step-by-step explanation:

-5 + 2n ≥ -11

+5             +5

2n ≥ -6

2/2 ≤ -6/2

n ≤ -3

-6 - 10n > 44

+6           +6

-10n > 50

-10/-10 50/-10

n < -5

it’s supposed to be -4+45=41

step-by-step explanation:

x ≥ 18 is your answer

step-by-step explanation:

let "a number" = x

(2/3)x + 17 ≥ 29

isolate the variable (x). do the opposite of pemdas (parenthesis, exponent (& root), multiplication, division, addition, subtraction).

first, subtract 17 from both sides.

(2/3)x + 17 (-17) ≥ 29 (-17)

(2/3)x ≥ 29 - 17

(2/3)x ≥ 12

next, multiply 3/2 to both sides to isolate the x

(3/2)(2/3)x ≥ 12(3/2)

x ≥ 12(3/2)

simplify. multiply, then divide

x ≥ (12 * 3)/2

x ≥ (36)/2

x ≥ 18

x ≥ 18 is your answer

~

rotation point l

step-by-step explanation: