we have 342 students, subtract 12 because we need only kids riding the bus, now we have 330 kids, we divide the 330 kids between 6 buses 330/6, our answer is 55.

4 0

3 years ago

Three of the sides will require fencing and the fourth wall already exists. If the farmer has 116 feet of fencing, what are the

dolphi86 [110]

Answer:

29 ft x 58 ft

Step-by-step explanation:

Let x be the length of each side perpendicular to the wall, and y be the length of the side parallel to the wall.

The amount of wire available is:

$116 = 2x+y\\y=116-2x$

The area of the region is:

$A=xy=x(116-2x)\\A(x)=116x-2x^2$

The value of 'x' for which the derivate of the area function is zero will yield the maximum area:

$A(x)=116x-2x^2\\A'(x) = 116-4x=0\\x=29\ ft$

The value of y is:

$y=116-2*29\\y=58\ ft$

The dimensions of the region with the largest area are 29 ft x 58 ft.

7 0

3 years ago

Solve V = 10t +7f for f​

Solve V = 10t +7f for f