Miguel stated that any monomial can be a cube root. Sylvia disagreed and said that a monomial cube root must have exponents divisible by 3. Who is correct, and why? - Sylvia is correct. Any variable term, when it is cubed, always has an exponent divisible by 3.

- Miguel is correct. Any monomial can be a perfect cube root because, when it is cubed, the variables will have exponents divisible by 3.

- Sylvia is correct. In order for the cube to have exponents that are divisible by 3, the cube root has to be divisible by 3.

- Miguel is correct. Sylvia confused the perfect square root with the perfect cube root.

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hello umm i don't understand your question can you explain it more pls

step-by-step explanation:

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:

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