What are the zeros of the function? f(x)=x3+x2−6x

x = - 3, x = 0, x = 2

Step-by-step explanation:

To find the zeros equate f(x) to zero, that is

x³ + x² - 6x = 0 ← factor out x from each term

x(x² + x - 6) = 0

x(x + 3)(x - 2) = 0

Equate each factor to zero and solve for x

x = 0

x + 3 = 0 ⇒ x = - 3

x - 2 = 0 ⇒ x = 2

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

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