the length of a rectangle is 5 ft more than twice the width.
a)if x represents the width of the rctangle, represent the perimeter of the rectangle in terms of x.
b) if the perimeter if the rectangle is 2 ft more than eight times the width of the rectangle, find the dimensions of the rectangle.
the length of a rectangle is 5 ft more than twice the width(x)
would be written as: length = 2x+5
Now you can proceed to calculate the perimeter=2(W+L), and attempt part (b).
13
13,5
13,4 Sorry typo
To solve this problem, let's break it down step by step:
a) If x represents the width of the rectangle, we are told that the length of the rectangle is 5 ft more than twice the width. So the length can be expressed as (2x + 5).
The formula for the perimeter of a rectangle is given by: P = 2(length + width)
Substituting the given expressions for length and width, we have:
P = 2((2x + 5) + x)
Simplifying the expression inside the parentheses:
P = 2(3x + 5)
Expanding the expression:
P = 6x + 10
Therefore, the perimeter of the rectangle, in terms of x, is 6x + 10.
b) We are given that the perimeter of the rectangle is 2 ft more than eight times the width. So we can set up the equation:
Perimeter = 8(width) + 2
Substituting the formula for the perimeter from part a:
6x + 10 = 8x + 2
Now, let's solve for x:
6x - 8x = 2 - 10
-2x = -8
x = -8 / -2
x = 4
Plugging the value of x back into the expression for the length, we find:
Length = 2x + 5 = 2(4) + 5 = 8 + 5 = 13
Therefore, the dimensions of the rectangle are width = 4 ft and length = 13 ft.