Here we want to find an area equation as a function of the width for a given rectangle, and evaluate it in different values.
The image of the rectangle is missing, I think that I found the correct image online, and I will post it at the end of the answer.
We will get the table:
[tex]\left[\begin{array}{ccc}w&A\\1m&9m^2\\2m&16m^2\\3m&21m^2\\4m&24m^2\\5m&25m^2\\6m&24m^2\\7m&21m^2\\8m&16m^2\\9m&9m^2\end{array}\right][/tex]
Let's see how to find this.
We know that for a rectangle of length L and width W, the perimeter is given by:
P = 2*(W + L)
In this case, we have that the width is w, and the other side measures 10m - w
Then the perimeter will be:
P = 2*(w + 10m - w) = 2*10m = 20m
Which is the amount of fencing that you have, 20m.
Now that we saw that, we can write the area as the product of the two measures, so we have:
A(w) = w*(10m - w) = 10*w - w^2
What we want to do, is evaluate the function in different values of w, and complete a table of w vs A(w).
Where evaluate means, for example for w = 1m, we get:
A(1m) = 10m*1m - (1m)^2 = 10m^2 -1m^2 = 9m^2
And we want to do that for w from 1m to 9m.
Then we just compute:
A(2m) = 10m*2m - (2m)^2 = 20m^2 - 4m^2 = 16m^2
A(3m) = 10m*3m - (3m)^2 = 30m^2 - 9m^2 = 21m^2
A(4m) = 10m*4m - (4m)^2 = 40m^2 - 16m^2 = 24m^2
A(5m) = 10m*5m - (5m)^2 = 50m^2 - 25m^2 = 25m^2
A(6m) = 10m*6m - (6m)^2 = 60m^2 - 36m^2 = 24m^2
A(7m) = 10m*7m - (7m)^2 = 70m^2 - 48m^2 = 21m^2
A(8m) = 10m*8m - (8m)^2 = 80m^2 - 64m^2 = 16m^2
A(9m) = 10m*9m - (9m)^2 = 90m^2 - 81m^2 = 9m^2
Now that we got all the values, we put them in the table:
[tex]\left[\begin{array}{ccc}w&A\\1m&9m^2\\2m&16m^2\\3m&21m^2\\4m&24m^2\\5m&25m^2\\6m&24m^2\\7m&21m^2\\8m&16m^2\\9m&9m^2\end{array}\right][/tex]
If you want to learn more, you can read:
https://brainly.com/question/8629807
