T.
a rectangular playground space is to be fenced in using the wall of a daycare building
for one side and 200 meters of fencing for the other three sides. the area a(x) in
square meters of the playground space is a function of the length x in meters of each of
the sides perpendicular to the wall of the daycare building.
a.
what is the area of the playground when x = 50?
b.
write an expression for a(x).
c.
what is a reasonable domain for a in this context?

We know that for a rectangle of length L and width W the area is just given by:

A = L*W.

And the perimeter is:

P  = 2*(L + W)

The solutions are:

a) 500 square meters.

b) A(x) = x*(200m - 2*x)

c) 0m < x < 100m

Here we have a rectangle of a perimeter equal to 200m + L

Where L is the side that is along the wall, and we call "x" as the length of the side perpendicular to the wall.

Then we also have:

200m = 2*x + L

a) If x = 50, we can use the above equation to find the length of the other side.

200m = 2*50m + L

200m = 100m + L

200m - 100m = L

100m = L

Then the two measures of the rectangle are:

L = 100m

x = 50m

Then the area is:

A = 100m*50m = 500m^2

b) Now we want a general expression for A(x).

We have:

A(x) = x*L

And we know that:

200m = 2*x + L

Then we can isolate L to get:

200m - 2*x  =L

Now we can replace that in the area equation:

A(x) = x*(200m - 2*x)

Now the equation only depends on the variable x, as wanted.

c) The domain of a function is the set of possible inputs that we can use.

Here, x represents the length of one of the sides of a rectangle, so we should start with:

x > 0m

So x is a positive number.

Now let's go to the equation:

200m - 2*x  =L

L is also a measure, so L also must be larger than zero meters, then we can write:

200m - 2*x > 0m

Now we can solve this for x:

200m > 2*x

200m/2 > x

100m > x

Writing together the two inequalities, we get:

0m < x < 100m

This is the domain of the area function.

If you want to learn more, you can read:

link

  a) 5000 m²

  b) A(x) = x(200 -2x)

  c) 0 < x < 100

Step-by-step explanation:

b) The remaining fence, after the two sides of length x are fenced, is 200-2x. That is the length of the side parallel to the building. The product of the lengths parallel and perpendicular to the building is the area of the playground:

  A(x) = x(200 -2x)

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a) A(50) = 50(200 -2·50) = 50·100 = 5000 . . . . m²

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c) The equation makes no sense if either length (x or 200-2x) is negative, so a reasonable domain is (0, 100). For x=0 or x=100, the playground area is zero, so we're not concerned with those cases, either. Those endpoints could be included in the domain if you like.

i get you, what you should do is think of the positive things that would come out of it. it will make yours parents proud and raise your grade. you should always try and if you can't focus tell the teacher.