Answer:
The width of the football field is 160 feet.
The length of the football field is 360 feet.
Step-by-step explanation:
Let w represent width of the football field.
We have been given that the length is 200 ft more than the width, so the length of the field would be
.
We are also told that the perimeter is 1,040 ft. We know that football field is in form of rectangle, so perimeter of field would be 1 times the sum of length and width. We can represent this information in an equation as:

Let us solve for w.






Therefore, the width of the football field is 160 feet.
Upon substituting
in expression
, we will get length of field as:

Therefore, the length of the football field is 360 feet.
So basically you are going to line up your equations (already done). Next, you will just add each term. So x + 3x = 4x, then 2y - 2y = 0 (we wouldn't put anything) then 7 - 3 = 4
So then we have all of our terms figured out and will have 4x = 4 then divide 4x by 4 to get x alone and then you will also divide the four that is by itself by four to equal one. So, your answer is x = 1. I hope that helps!
So, I came up with something like this. I didn't find the final equation algebraically, but simply "figured it out". And I'm not sure how much "correct" this solution is, but it seems to work.
![f(x)=\sin(\omega(x))\\\\f(\pi^n)=\sin(\omega(\pi^n))=0, n\in\mathbb{N}\\\\\\\sin x=0 \implies x=k\pi,k\in\mathbb{Z}\\\Downarrow\\\omega(\pi^n)=k\pi\\\\\boxed{\omega(x)=k\sqrt[\log_{\pi} x]{x},k\in\mathbb{Z}}](https://tex.z-dn.net/?f=f%28x%29%3D%5Csin%28%5Comega%28x%29%29%5C%5C%5C%5Cf%28%5Cpi%5En%29%3D%5Csin%28%5Comega%28%5Cpi%5En%29%29%3D0%2C%20n%5Cin%5Cmathbb%7BN%7D%5C%5C%5C%5C%5C%5C%5Csin%20x%3D0%20%5Cimplies%20x%3Dk%5Cpi%2Ck%5Cin%5Cmathbb%7BZ%7D%5C%5C%5CDownarrow%5C%5C%5Comega%28%5Cpi%5En%29%3Dk%5Cpi%5C%5C%5C%5C%5Cboxed%7B%5Comega%28x%29%3Dk%5Csqrt%5B%5Clog_%7B%5Cpi%7D%20x%5D%7Bx%7D%2Ck%5Cin%5Cmathbb%7BZ%7D%7D)