Answer: 
Step-by-step explanation:
GCF = Greatest common factor.
For example: 6 is the GCF of 12 and 18.
The given expression: 
It can be written as

Taking out common factor
, we get
![5y(x^2-6xy-7y^2)\\\\ =5y(x^2+xy-7xy-7y^2)\ \ [\because-6xy=xy-7xy]\\\\=5y(x(x+y)-7y(x+y))\\\\= 5y((x+y)(x-7y))\\\\=5y(x+y)(x-7y)](https://tex.z-dn.net/?f=5y%28x%5E2-6xy-7y%5E2%29%5C%5C%5C%5C%20%3D5y%28x%5E2%2Bxy-7xy-7y%5E2%29%5C%20%20%5C%20%5B%5Cbecause-6xy%3Dxy-7xy%5D%5C%5C%5C%5C%3D5y%28x%28x%2By%29-7y%28x%2By%29%29%5C%5C%5C%5C%3D%205y%28%28x%2By%29%28x-7y%29%29%5C%5C%5C%5C%3D5y%28x%2By%29%28x-7y%29)
Hence, 
= 2025
When you are told to find the smallest length possible, you perform L.C.M(Least common multiples)
For this, you divide the given lengths using the numbers that divides all through.
I have added an image to this answer. Go through it for more explanation
Not sure if you wanted answers or explanation on what to do. Ill just do the explanation and if you want the answers just lmk C:
R=1/2 d, and C=pi*2r. So you basically take the given number and plug it into these formulas.
We use the given data above to calculate the volume of gasoline that is being burned per minute by commercial airplanes.
Amount burned of 1 commercial airplane = <span>3.9 × 10³ ml of gasoline per second
Number of airplanes = </span><span>5.1 × 10³ airplanes
We calculate as follows:
</span> 3.9 × 10³ ml of gasoline per second / 1 airplane (5.1 × 10³ airplanes)(60 second / 1 min ) = <span>1.2 x 10^9 mL / min</span>