Based on the solid line segment, which of the following could be the predicted width, in
millimeters, of a rhododendron leaf when the temperature is T°F, where 0< T <12?

oof tough question

step-by-step explanation:

sorry i can't ya : c gl tho

(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))

ca = ec

cpctc makes it so that corresponding parts of the triangle you proved to be equal are equal, the other options aren't part of the part of the triangles you proved to be equal.

answer: 0.1,0.85

step-by-step explanation: