Question

Assistance is required.

Answer 1

m=3, b=-6

Answer 2

That's what I think it is.

Answer 3

u ar right

Answer 4

When I checked, it said I was wrong.

Answer 5

m=+3 b=-6

Answer 6

Incorrect

Answer 7

hold on

Answer 8

You can apply the slope formula
\(\Large m = \frac{y_2-y_1}{x_2-x_1}\)
This says you subtract the y values together (y2-y1) over the difference of the x values (x2-x1). Then divide those two differences.

Answer 9

@jhonyy9 @XioGonz

Answer 10

assistance you will not get

Answer 11

\(\color{#0cbb34}{\text{Originally Posted by}}\) @imnotsmartlol
assistance you will not get
\(\color{#0cbb34}{\text{End of Quote}}\)
rude lol

Answer 12

@jhonyy9

Answer 13

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
You can apply the slope formula
\(\Large m = \frac{y_2-y_1}{x_2-x_1}\)
This says you subtract the y values together (y2-y1) over the difference of the x values (x2-x1). Then divide those two differences.
\(\color{#0cbb34}{\text{End of Quote}}\)
^^

Answer 14

Actually, I recalculated, and it's actually m=-2, b=5

Answer 15

Thank you anyways

Answer 16

Bruh, right when I get the answer smh

Answer 17

\(\color{#0cbb34}{\text{Originally Posted by}}\) @dontsaymyname
Bruh, right when I get the answer smh
\(\color{#0cbb34}{\text{End of Quote}}\)
Lol rip your track record

Answer 18

\(\color{#0cbb34}{\text{Originally Posted by}}\) @Downpour17
\(\color{#0cbb34}{\text{Originally Posted by}}\) @dontsaymyname
Bruh, right when I get the answer smh
\(\color{#0cbb34}{\text{End of Quote}}\)
Lol rip your track record
\(\color{#0cbb34}{\text{End of Quote}}\)
shush >.> did u get the answer?

Answer 19

Yes that's correct @Downpour17

since,
\(\Large m = \frac{y_2-y_1}{x_2-x_1}\)

\(\Large m = \frac{5-11}{0-(-3)}\)

\(\Large m = \frac{5-11}{0+3}\)

\(\Large m = \frac{-6}{3}\)

\(\Large m = -2\)

Answer 20

Yep it was m=-2, b=5. I was dumb enough to use a 3rd grade method

Answer 21

Anyways, the question's answerd. Thank you everyone. Goodbye!

Answer 22

\(\color{#0cbb34}{\text{Originally Posted by}}\) @jimthompson5910
Yes that's correct @Downpour17

since,
\(\Large m = \frac{y_2-y_1}{x_2-x_1}\)

\(\Large m = \frac{5-11}{0-(-3)}\)

\(\Large m = \frac{5-11}{0+3}\)

\(\Large m = \frac{-6}{3}\)

\(\Large m = -2\)
\(\color{#0cbb34}{\text{End of Quote}}\)
good job !