Suppose that the width of a certain rectangle is 1 inch more than one-fourth of its length. the perimeter of the rectangle is 62 inches. find the length and width of the rectangle.

Step-by-step explanation:

Let:

x=length

(x/4)+1=width

P=2l+2w

62=2x + 2[(x/4)+1]

62=2x + 1/2x + 2

multiply both sides by 2

124=4x+x+4

5×=120

x=24

length: 24 inches

width: 7 inches

hope this helps!!

-3

step-by-step explanation:

a drawing really does here:

so i draw the horizontal line y=2 and the vertical line x=-2 at this intersection i put b (this is where my right angle is going to be).   so point b is at (-2,2).   point a is 5 units long and is also on y=2 so this point has to be (3,2) or (-7,2) -doesn't matter i will go with (3,2).

now we know c is at (-2,y) we don't know the y yet bet we know that c is on x=-2.

we have the area of the triangle is .5bh=12.5

we know the base length is 5 (ab)

we have the height is (2-y) if you look at your

plug in

.5(5)(2-y)=12.5

solve for y to get the y-coordinate for c

2.5(2-y)=12.5

divide both sides by 2.5 giving:

2-y=5

now subtract 2 on both sides

-y=3

y=-3

i need a picture

step-by-step explanation: