SOLUTION: The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in
Algebra ->
Expressions-with-variables
-> SOLUTION: The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in
Log On
Question 1031851: The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit. Found 3 solutions by stanbon, mananth, MathTherapy:Answer by stanbon(75887) (Show Source):
You can put this solution on YOUR website! The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
Area = L*W so 225 = L*W so W = 225/L
-------------------------
P = 2(L+W)
P = 2L + 2W
0 = 2L + 2W - 60
or
0 = 2L + 225/L - 60
Quadratic:: 2L^2 - 60L + 225 = 0
-------------------------------
B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit.
Comment:: Not sure what methods were discussed.
----------
Cheers,
Stan H.
------------
You can put this solution on YOUR website! The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
You can put this solution on YOUR website!
The perimeter of a rectangular garden is 60 m and its area is 225 m^. Find the length of the garden. Use L = length and W = width.
A. Write a quadratic equation in terms of L in the form 0 = aL + bL + c that represents the area of the the flower bed.
B. Solve the quadratic equation you wrote in part A using one of the methods discussed in this unit.
With L and W being the length and width, respectively, we get: 2(L + W) = 60______2(L + W) = 2(30)_______L + W = 30_______W = 30 - L
As area is , we can say that: L(30 - L) = 225
Solve by any of the 4 methods to get dimensions: