Arectangular field has an area of 1,764 m2. the width of the field is 13 m more than the length. what is the perimeter of the field? 121 m 144 m 168 m 170 m

There's an interesting way to work this that you might not think of at first. Consider the length that is the average of length and width. Call it x. Then the short side of the field is x-6.5 m, and the long side is x+6.5 m, so that the difference between those side lengths is 13 m = (x+6.5 m) -(x-6.5 m).

The area is

... (x +6.5)(x -6.5) = x² -6.5² = 1764

Solving for x gives

... x = √(1764 +6.5²) = √1806.25 = 42.5

At this point, you can solve for the side lengths, then add them to give perimeter. However, you already know that the sum of the short and long sides is 2x = 85 m, so the perimeter is 2×85 m = 170 m.

to plot points   for x just start at the origin (middle) and go to the left however many times it say(if the number is a positive) if the number is a negative then go right however many times it says if it is 0 go nowhere

now keep you pencil there and go up however many times it says on y(if it is positive) if it is negative then go down however many times it says. if it says 0 go nowhere.

you have just plotted a point! now when you finish plotting all of them connect them with a ruler

plot each point as (-3, 7), (0,4), (3,1), (8,-4)   in the parenthesis on the left of the comma is x and the right is y.

the correct option is b

step-by-step explanation: