A rectangle with sides 15 and w has perimeter

We want this quantity to be at most 50, so it must be less than or equal to 50:

For the record, this implies that

So, the width can be at most 10.
Answer:
x= 3/7 i.d.k if this is correct I did it in my brain
Answer:
y ≥−3
Step-by-step explanation:
1 By inspecting the graph, the range is:
y\ge -3y≥−3
Done
Answer:
5
Step-by-step explanation:
Use the distance formula 
insert the numbers into the formula 
2+1=3 and 0-4=-4
=9 and
= 1
9+16=25
the square root of 25 is 5