A rectangular field whose length is 33 less than four times its width needs to be enclosed with exactly 104 meters of fencing.

What are the dimensions of the field?

2(x + 4x-33) = 104

104=2L+2W=2(4W-33)+2W

104=10W-66
10W=170
W=17m
L=68-33=35

To find the dimensions of the field, we can set up equations based on the given information and solve them simultaneously.

Let's assume the width of the rectangular field is "w" meters. Therefore, the length of the field would be (4w - 33) meters.

The perimeter of a rectangle is given by the formula: P = 2(length + width)

In this case, the perimeter is given as 104 meters. So, we can set up the equation:

104 = 2((4w - 33) + w)

Now, let's solve this equation to find the value of "w" (the width):

104 = 2(5w - 33)
104 = 10w - 66
10w = 104 + 66
10w = 170
w = 170/10
w = 17

Therefore, the width of the rectangular field is 17 meters.

To find the length, substitute the value of "w" back into the expression: length = 4w - 33

Length = 4(17) - 33
Length = 68 - 33
Length = 35

Therefore, the length of the rectangular field is 35 meters.

Hence, the dimensions of the field are 17 meters (width) and 35 meters (length).