Arectangle with area 80 square inches has a length that is four less than four times the width

length=16 inches

width= 5 inches

Step-by-step explanation:

Area=l*w=80 square inches

l*w=80    (equation 1)

l=4w-4     (equation 2)

Putting value of l from equation 2 in equation 1

(4w-4)w=80

4w^2-4w-80=0   (equation 3)

Taking 4 common from equation

4(w^2-w-20)=0

w^2-w-20=0

w^2-5w+4w-20=0

w(w-5)+4(w-5)=0

(w+4)(w-5)=0

w=-4, w=5

ignoring the negtive width

we'll solve with w=5 by putting it in equation 2

l=(4*5)-4

l=20-4

l=16

what is the approximate angle of acb if abc is 25cm and the length of segment bc is 15

step-by-step explanation: