Answer:
Options C. and D. are correct.
Step-by-step explanation:
Let 
Equation of a line is given by 

Put 

Put 

Put 

Put 

Put 

So, points
satisfy the equation 
Therefore,
points
lie on the line through
and 
Options C. and D. are correct.
Let's first say that L=W+44
and then remember that perimeter is P=2L+2W
replace the L with W+44
we then get P=2(W+44)+2W, now I'll solve it
P=2W+88+2W
P=4W+88
substitute 288 for P
288=4W+88
200=4W
50=W
so now we now how wide the court is. add 44 to find the length which gives you L=94
as always plug the numbers back into your perimeter equation to ensure L and W are correct
Answer:
0.12
Step-by-step explanation:
15 divided by 125 is equal to 0.12