The variable $x$ varies directly as the square of $y$, and $y$ varies directly as the cube of $z$. If $x$ equals $-16$ when $z$ equals 2, what is the value of $x$ when $z$ equals $\frac{1}{2}$?

If varies directly as , then there is some constant for which

Similarly, there is some constant such that

Given that when , we have

Now when , we get

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

~

these sign is used to tell weather the number is greater or less than the number   <   or >     but this sign means its equal =    

hope this you and

step-by-step explanation: