There’s not enough context could you maybe post a picture?
Answer:
23
Step-by-step explanation:
<span>First, we write an equation to represent that the fencing lengths add up to 568 feet. we call the side of the fence that has three segments of its length x and the side with only two segments y. We write 3x + 2y = 568. We also know that the area of the rectangle is equal to xy, so area = xy. We put y in terms of x using our first equation and find that y = (568 - 3x)/2. We plug this into our area equation and find that area = (568x - 3x^2)/2. To find the maximum we set the derivative equal to 0 and end up with 0 = 284 - 3x. We solve for x and get 94 and 2/3. We then put that into our first equation to find y = 142. So the dimensions that maximize the area are 94 2/3 x 142.</span>
-50m^4 n^7 = 10m^2n^7(-5m^2)<span>
and
40m^2 n^10 = 10</span>m^2 n^7(4n^3)
Answer: GCF = 10m^2 n^7
