183k views
0 votes
Help need answer asap #struggling

Help need answer asap #struggling-example-1

1 Answer

3 votes

Answer:

b) Dimensions of garden:
length = 10 feet
width = 3 feet

Explanation:

I am only doing the part b. Part a is just a sketch of a rectangle with the dimensions computed

b) Find the dimensions of Amelia's garden

Let us use the variable L to represent the length and the variable W to represent the width of the garden

We are given that the length = 4 plus twice the width
In algebraic equation terms this would be
L = 4 + 2W


We are given that the area is 30 ft²
Area of a rectangle = LW
So
LW = 30

Substitute for L in terms of W:
(4 + 2W)W = 30
4W + 2W² = 30

Move 30 to the left and rearrange terms on the left
2W² + 4W - 30 = 0'

Divide by 2:
W² + 2W - 15 = 0

This is a quadratic equation that can be solved using factoring
Find the factors of -15 and see which of them when added will give a value of 2 which is the coefficient of W in the quadratic equation

Factors of -15 are
-15 1 => sum = 15 + 1 = - 14 X
-5 3 => sum = -5 + 3 = -2 X
5 -3 => sum = 5 + (-3) = 2 √

Given the correct factors we can rewrite the equation as
(W + 5)(W - 3) = 0

So either W + 5 = 0 or W - 3 =0

(If you multiply W + 5 by W - 3 you will get W² + 2W - 15)

Therefore the solutions to the quadratic equation are
W + 5 = 0 ==> W = -5 ; not possible, dimensions have to be positive
W - 3 = 0 ==> W = 3 ; this is the solution

So we have W, the width of the garden as 3 feet

Substitute W = 3 in the equation for length:
L = 4 + 2W

L = 4 + 2 x 3

L = 10

So length = 10 feet

User Mikiqex
by
7.9k points
Welcome to QAmmunity.org, where you can ask questions and receive answers from other members of our community.