Robert has available 400 yards of fencing and wishes to enclose a rectangular area. Express the areaAof the rectangle as a function of the widthwof the rectangle. For what value ofwis the arealargest? What is the maximum area?

The answer is below

Step-by-step explanation:

a)Given that the length of fencing available is 400 yards. This means that the perimeter of the rectangle is 400 yards.

the perimeter of a rectangle is given as:

Perimeter = 2(length + width) = 2(l + w)

Hence;

400 = 2(l + w)

200 = l + w

l = 200 - w

The area of a rectangle is given as:

Area = length × width

Area = (200 - w) × w

Area = 200w - w²

b) For a quadratic equation y = ax² + bx + c. it has a maximum at x = -b/2a

Hence, for the area = 200w - w² a=-1, b = 200, the maximum width is at:

w = -b/2a = -200/2(-1) = -200/-2 = 100

A width of 100 yard has the largest area

c) l = 200 - w = 200 - 100 = 100 yards

Area = l × w = 100 × 100 = 10000 yd²

The maximum area is 10000 yd²

10.416 meters per second

step-by-step explanation:

distance=1 00 meters

time= 9.6 seconds in decimal form or 9 3/5 as a mixed number

average speed= 100/9.6

=10.416 meters per second or 10 52/125 as a mixed number.

what are the statments

step-by-step explanation: