A rectangle has a perimeter of 80. if the length is four more than twice the width, what is the length of the rectangle

28

Step-by-step explanation:

l = length

w = width

p = perimeter

w = w

l + l + w + w = p (adding all the rectangle's sides together gets us our perimeter)

l = 2w + 4 (four more than twice the width)

Substitute 2w + 4 for 'l' in the original equation

(2w + 4) + (2w + 4) + w + w = 80

Solve equation normally:

6w + 8 = 80

6w = 72

w = 72/6 = 12

(w = 12)

Length = 2w + 4

Substitute 12 for 'w':

2(12) + 4 = l

24 + 4 = l

28 = l

The length is 28

x=30

step-by-step explanation:

answer: (c) m∠qpo + (2x + 16)° = 180°

step-by-step explanation:

the opposite angles of a quadrilateral are supplementary.

so, ∠o + ∠q = 180°   and   ∠p + ∠r = 180°

since ∠r = 2x + 16°, we can use substitution as follows:

∠p + ∠r = 180°

∠p + (2x + 16)° = 180°