David has available 480 yards of fencing and wishes to enclose a rectangular area.

Step-by-step explanation:

Total fencing David has = 480 yards

Means He can surround all 4 sides of the rectangular area using this fencing.

We can say that the total fencing he has in terms of length L and width W of the rectangle will be L+L+W+W = 480

So , 2(L+W) =480

Divide both sides by 2 to get

L+W = 240

Subtract W from both sides so we get L in terms of W

L= 240 - W

(a) We know that the area of a rectangle is Length times width (L × W)

So ,  Area  = (240-W) W

        Area  = 240W - W^2

(b) By comparing the area equation with ax^2+bx+c (standard quadratic equation)

we get a=-1 and b =240

Area is maximum when

So for W = 120 yd the area is largest .

(c) Maximum area of the rectangle can be found by plugging this W = 120 in the Area equation that we got in part (a)

So the maximum area ,

A =

A =28800-14400

A = 14400

Hence Maximum area = 14400 sq yd

m = M (u-V)/(v-u)

Step-by-step explanation:

We know that:

If we sum -mu to both sides of the equation we get:

Now let's sum -MV  to both sides:

Now we can factor the m and M terms:

Finally we can divide by v-u ***

Since (v-u)/(v-u) =1

m = M (u-V)/(v-u)

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If u=v it will imply that V=u and tehrefore the two mases will travel with the same velocity to the same direction and no collision would take place

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