The like terms are 16x and 2x
Well it would be (x+3)(x+3)
using the FOIL method you end up with:
x^2 + 6x + 9
:D
Answer:
x = 10 cm, y = 5 cm gives a minimum area of 300 cm^2.
Step-by-step explanation:
V= x^2y = 500
Surface area A = x^2 + 4xy.
From the first equation y = 500/x^2
So substituting for y in the equation for the surface area:
A = x^2 + 4x * 500/x^2
A = x^2 + 2000/x
Finding the derivative:
dA/dx = 2x - 2000x^-2
dA/dx = 2x - 2000/x^2
This = 0 for a minimum/maximum value of A, so
2x - 2000/x^2 = 0
2x^3 - 2000 = 0
x^3 = 2000/ 2 = 1000
x = 10
Second derivative is 2 + 4000/x^3
when x = 10 this is positive so x = 10 gives a minimum value of A.
So y = 500/x^2
= 500/100
= 5.
First, we need to get the total area of the rectangular pool.
Area = l * w
Area = 20 ft * 50 ft
Area = 1000 ft^2
Then the deck is 456 ft^2 with the width of 20ft, the same as the rectangular pool.
Area = l * w
456 = l * 20
l = 456 / 20
l = 22.8 ft.
So the walkway is 22.8ft wide.
Answer:
rise over run
Step-by-step explanation:
In this case it is undefined
