CURRENTLY 6 LESSON DAYS BEHIND AND VERY DEPRESSED/STRESSED!! You are making a rectangular table. The area of the table should be 10 ft^2. You want the length of the table to be 1 ft shorter than twice its width. What should the dimensions of the table be?
Okay, so, I'm not sure how to put this into an equation. The example in the book is very different to this problem. Please help!!
How do I solve that using a quadratic formula?
w(2w-1)=10
2w^2-w-10 = 0
(2w-5)(w+2) = 0
w = 5/2
so, the length is 4
I'm sorry to hear that you're feeling stressed and depressed, but I'm here to help you with this math problem. Let's break it down step by step.
We're asked to find the dimensions of a rectangular table that has an area of 10 ft^2. Let's call the width of the table "w" and the length "l."
According to the problem, we're given two conditions:
1. The area of the table is 10 ft^2, so we can write the equation:
w * l = 10
2. The length of the table is 1 ft shorter than twice its width, so we can write the equation:
l = 2w - 1
Now, we have a system of two equations. We can solve it by substituting one equation into the other.
Let's start by substituting the value of "l" from the second equation into the first equation:
w * (2w - 1) = 10
Now, we simplify and solve the quadratic equation:
2w^2 - w - 10 = 0
We can factor this quadratic equation or use the quadratic formula to find the values of "w":
w = (-b ± √(b^2 - 4ac)) / (2a)
In this case, a = 2, b = -1, and c = -10. Plugging these values into the formula, we get:
w = (-(-1) ± √((-1)^2 - 4 * 2 * -10)) / (2 * 2)
Simplifying further, we have:
w = (1 ± √(1 + 80)) / 4
w = (1 ± √81) / 4
w = (1 ± 9) / 4
This gives us two possible values for "w":
w1 = (1 + 9) / 4 = 10 / 4 = 2.5
w2 = (1 - 9) / 4 = -8 / 4 = -2
Since the width cannot be negative, we can discard the negative value. Therefore, the width of the table is 2.5 ft.
Now, we can substitute this value back into the second equation to find the length:
l = 2w - 1
l = 2(2.5) - 1
l = 5 - 1
l = 4
Therefore, the dimensions of the table should be a width of 2.5 ft and a length of 4 ft.