Arectangular area is to be enclosed by a wall on one side and fencing on the other three sides. if 18 meters of fencing is are used, what is the maximum area that can be enclosed?
a: 9/2 m^2

b: 81/4 m^2

c: 27 m^2

d: 40 m ^2

e: 81/2 m^2

Explanation:

Let L be the length and W be the width.

We have only 2 sides are fenced

         Fencing = 2L + W

Fencing = 18 m

          2L + W = 18

           W = 18 - 2L

We need to find what is the largest area that can be enclosed.

     Area = Length x Width

      A = LW

       A = L x (18-2L) = 18 L - 2L²

For maximum area differential is zero

So we have

      dA = 0

      18 - 4 L = 0

        L = 4.5 m

      W = 18 - 2 x 4.5 = 9 m

Area = 9 x 4.5 = 40.5 m² = 81/2 m²

Option E is the correct answer.

the slope of the line is 60

one point on the line is (40,10)

step-by-step explanation:

in the equation outside the parenthesis the equation always tells you the slope. in this case it is 60.

in the parenthesis is also says 40 which is x because it is near the x and 10 is y because it is near y.

b, i believe, but not sure.