anonymous
  • anonymous
I don't understand Linear Functions. This is what it states write the rule for the linear function. x y -3 -17 -1 -7 1 3 3 13 using (1,3) and (3,13): m=______=_______=_______ How do I figure this out?
Mathematics
jamiebookeater
  • jamiebookeater
See more answers at brainly.com
At vero eos et accusamus et iusto odio dignissimos ducimus qui blanditiis praesentium voluptatum deleniti atque corrupti quos dolores et quas molestias excepturi sint occaecati cupiditate non provident, similique sunt in culpa qui officia deserunt mollitia animi, id est laborum et dolorum fuga. Et harum quidem rerum facilis est et expedita distinctio. Nam libero tempore, cum soluta nobis est eligendi optio cumque nihil impedit quo minus id quod maxime placeat facere possimus, omnis voluptas assumenda est, omnis dolor repellendus. Itaque earum rerum hic tenetur a sapiente delectus, ut aut reiciendis voluptatibus maiores alias consequatur aut perferendis doloribus asperiores repellat.

Get this expert

answer on brainly

SEE EXPERT ANSWER

Get your free account and access expert answers to this
and thousands of other questions

anonymous
  • anonymous
y=mx+c where c is the y-intercept and m is the gradient. \[m=\frac{y_2-y_1}{x_2-x_1}\]
anonymous
  • anonymous
how did you get that? I'm a big dummy lol
anonymous
  • anonymous
These are the rule to find out the equation of a straight line. All you need to do to find the gradient is plug in the values for the two points.

Looking for something else?

Not the answer you are looking for? Search for more explanations.

More answers

anonymous
  • anonymous
ok so if I'm looking at the graphand I'm using 1,3 and 3,13 then the m= what? see I don't get it. When I look at the graph I see where the 1,3 is but I don't get how to make the m= anything
anonymous
  • anonymous
(1,3) --> (x1,y1) and (3,13) ---> (x2,y2) So \[m=\frac{y_2-y_1}{x_2-x_1}\]\[=\frac{13-3}{3-1}\]\[=\frac{10}{2}\]\[=5\]

Looking for something else?

Not the answer you are looking for? Search for more explanations.