Arhombus is inscribed in a rectangle that is w meters wide with a perimeter of 56 m. each vertex of the rhombus is a midpoint of a side of the rectangle. express the area of the rhombus as a function of the rectangle's width

let the rectangle l meters long

perimeter = 2l + 2w

56 = 2l + 2w

28 = l + w

l = 28 - w

area of rhombus = ½•l•w

f(w) = ½•(28-w)•w

= 14w - ½w²

step-by-step explanation:

always plug in x. in this case, x equals 4.

i'm going to first solve 4(2+3x)

1) 4(2+3(4)) use pemdas to solve.

i'm going to multiply whats in the parenthesis first, so 3 times 4=12.

2)4 (2+12) i'm then going to multiply 4 by both numbers in the parenthesis (because multiplying comes before adding)

4 times 2= 8.   4 times 12= 48

4) (8+48)= 56 (answer)

i'm then going to solve 8+12x and plug in 4 for x.

1) 8 +12(4)

multiplying comes before adding, so i'm gonna multiply 12 and 4.

12 times 4= 48.

2) 8+48=56

both equal the same!