A rectangular field is nine times as long as it is wide. If the perimeter of the field is 1700 feet, what are the dimensions of the field?

A) Write an equation you can use to answer the given question. Let w be the width of the field. Do not
solve the equation yet.

The equation is
X (Make sure you use the correct variable.)

B) Use your equation to find the dimensions of the field.

The width of the field is
x feet.
The length of the field is
x feet.

I. L = 10.35 feet

II. W = 93.15 feet.

Step-by-step explanation:

Let the length of the rectangle be L.

Let the width of the rectangle be W.

Given the following data;

Perimeter of rectangular field = 1700 feet

Translating the word problem into an algebraic expression, we have;

W = 9L

Mathematically, the formula for the perimeter of a rectangle is;

P = 2(L + W)

A. To write an equation;

X = P = 2(L + W)

B. To find the dimensions of the field;

207 = 2(L + 9L)

207 = 2L + 18L

207 = 20L

L = 207/20

L = 10.35 feet

To find the weight;

W = 9L

W = 9 * 10.35

W = 93.15 feet.

Therefore, the width of the field is

93.15 feet and the length of the field is

10.35 feet.

c

step-by-step explanation:

it isn't d, because the interquartile range is from 27-57. and the interquartile range is basically half the data, it's pretty obvious.

it isn't b, because like i said earlier, the interquartile range is from 27-57. 57-27 is 30. the real range has to be longer

not a either. although it looks like it, 3/4 of the data is not greater than 27. but if c is wrong than this should be correct

the correct answer is c, because 13 is the point in which the plot starts

6.9

step-by-step explanation:

simple math really

6 1/2 times 0.6 = 3.9

but if you are just doing 6 and not 6 1/2 then the answer is 3.6