Answer:
slope=6/5
Step-by-step explanation:
The formula for finding the slope is y1-y2/x1-x2=slope. So if we substitute in the numbers from you question, it would be (-3-3)/(-2-3)=slope
(-3-3)/(-2-3)=slope
-6/-5=slope
6/5=slope
That does not make any since...
Its 0.343 because it cannot be expressed as a fraction or whole number
The second choice is correct, but I believe that either its +1/3 ( in the question) or -3y as that will give you the answer of: y=1/3x+6
Answer:
Below
I hope its not too complicated

Step-by-step explanation:



