he length of a rectangular picture is 5 inches greater than twice the width. if the perimerter is 112 inches,?

Question

he length of a rectangular picture is 5 inches greater than twice the width. if the perimerter is 112 inches, find the dimensions of the frame

Answer 1

Let: L=length, W=width, P=perimeter

L=5+2W

P=2L+2W, P=112
2L+2W=112
Substituting L into the equation, we get:

2(5+2W) +2W = 112
10 + 4W + 2W = 112
6W = 112-10
6W = 102
W = 17

L = 5+2W
L = 5+ (2*17)
L = 39

So, the dimensions of the frame is 39 x 17 inches.

Answer 2

let the length be l and width be w

given : l = 5 + 2w

perimeter of a rectangle = 2l + 2 w = 112 (given)

therefore 2(5+2w) +2w = 112
10 + 6w =112
6w = 102
w = 17 inches

therefore l = 5 + 2(17) = 39 inches

Answer 3

The formulation for perimeter is 2L+2w For this actual case: 2L+2w=28 that's suggested that L = 2w-a million by ability of substitution, 2(2w-a million)+2w=28 4w-2+2w=28 6w-2=28 6w=30 w=5 L=2w-a million Substitution: L=2(5)-a million L=10-a million L=9 The rectangle has a length of 9 inches and a width of 5 inches.

Answer 4

Let l = length
w = width

Given
P = 112

P = 2l + 2w so
2l + 2w = 112

Or
l + w = 56

Also (length is 5 in. greater than twice the width)
l = 2w + 5

Subs. the l value to the first equation
2w + 5 + w = 56

combine and transpose
3w = 51

Divide
w = 17

subs. to l
l = 2w + 5
l = 2(17) + 5
l = 34 + 5
l = 39

Answers:
w = 17 inches
l = 39 inches

^_^

Answer 5

The length is 39 inches and the width is 17 inches

Answer 6

Width= 17 inches
Length = 39 inches
Nye

Answer 7

l = 2w + 5
P = 2(l + w)

112 = 2((2w + 5) + w)
112 = 2(2w + 5 + w)
112 = 2(3w + 5)
56 = 3w + 5
3w = 51
w = 17

l = 2w + 5
l = 2(17) + 5
l = 34 + 5
l = 39

Length = 39 inches
Width = 17 inches

Answer 8

width is 17"
length is 39" ( 2(17) + 5)

Answer 9

breadth = x
length = 2*x+5

p=112inch

6*x+10=112
x=17:breadth
length=39