Teresa stated that the heights of the students in her class were not a function of their ages. which reasoning could justify teresa’s statement? two students are the same age but have different heights. two students have the same height but different ages. no two students are the same age or height.

The answer is Two students are the same age but have different heights. Hope this helps!

A) Two students are the same age but have different heights.

(2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))

step-by-step explanation:

simplify the following:

x/x^2 + 3 x + 2 - (x (x + 1))/(x + 2)

combine powers. x/x^2 = x^(1 - 2):

x^(1 - 2) + 3 x + 2 - (x (x + 1))/(x + 2)

1 - 2 = -1:

x^(-1) + 3 x + 2 - (x (x + 1))/(x + 2)

put each term in 1/x + 3 x + 2 - (x (x + 1))/(x + 2) over the common denominator x (x + 2): 1/x + 3 x + 2 - (x (x + 1))/(x + 2) = (x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)):

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2))

(x + 2)/(x (x + 2)) + (3 x^2 (x + 2))/(x (x + 2)) + (2 x (x + 2))/(x (x + 2)) - (x^2 (x + 1))/(x (x + 2)) = ((x + 2) + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2)):

(2 + x + 3 x^2 (x + 2) + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

3 x^2 (x + 2) = 3 x^3 + 6 x^2:

(2 + x + 3 x^3 + 6 x^2 + 2 x (x + 2) - x^2 (x + 1))/(x (x + 2))

2 x (x + 2) = 2 x^2 + 4 x:

(2 + x + 6 x^2 + 3 x^3 + 2 x^2 + 4 x - x^2 (x + 1))/(x (x + 2))

-x^2 (x + 1) = -x^2 - x^3:

(-x^2 - x^3 + 3 x^3 + 6 x^2 + 2 x^2 + 4 x + x + 2)/(x (x + 2))

grouping like terms, 3 x^3 - x^3 + 6 x^2 + 2 x^2 - x^2 + 4 x + x + 2 = (3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2:

((3 x^3 - x^3) + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

3 x^3 - x^3 = 2 x^3:

(2 x^3 + (6 x^2 + 2 x^2 - x^2) + (x + 4 x) + 2)/(x (x + 2))

6 x^2 + 2 x^2 - x^2 = 7 x^2:

(2 x^3 + 7 x^2 + (x + 4 x) + 2)/(x (x + 2))

x + 4 x = 5 x:

answer:   (2 x^3 + 7 x^2 + 5 x + 2)/(x (x + 2))