Answer:
Lenth=15, width=4/5
Step-by-step explanation:
SO first we wanna make a equation about ths
The length of a rectangle is 11 centimeters less than five times its width.
L = lenth
W= width
L=5w+11
so w(5w+11)
and you wanna solve it
skip cuz im bad
w = 4/5 or w = -3
-3 is not possible so w = 4/5
L=15
check:
15*4/5=12
The dimensions of the rectangle are:
Length = 4 cm
Width = 3 cm
To solve this problem, we will represent each of the given dimensions using algebraic expressions as follows:
Let:
Width of the rectangle = w cm
Length of the rectangle = [tex](5w - 11)[/tex] cm
Area of the rectangle = 12 sq cm
Formula for area of a rectangle = length [tex]\times[/tex] width
Therefore:
[tex](5w - 11) \times w = 12\\5w^2 - 11w = 12\\5w^2 - 11w - 12 = 0[/tex]
Factorize:
[tex]5w^2 + 4w - 15w - 12 = 0\\w(5w + 4) -3(5w + 4) = 0\\(w - 3)(5w + 4)\\(w - 3) = 0\\w = 3[/tex]
or
[tex]5w + 4 = 0\\5w = -4\\w = -\frac{4}{5}[/tex]
Width of the rectangle cannot be a negative number, so we would go with:
[tex]w = 3[/tex] cm
The width of the rectangle is therefore 3 cm
Find the length by plugging in the value of w into [tex]5w - 11[/tex]
Length of the rectangle = [tex]5(3) - 11[/tex]
= [tex]15 - 11\\[/tex]
= 4
Length of the rectangle is 4 cm
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