find the slope of a line that passes through (-2, -3) and (1, 1) 1/3 1 2 4/3
Use the slope formula\[m = \frac{ y_2 - y_1 }{ x_2-x_1 }\]where x_2 and y_2 represents one coordinate and the x_1 and y_1 represents the 2nd coordinate. Slope = rise over run right?
yes its rise over run
since your point (1,1) is higher than your point (-2,-3) on a graph. use the (1,1) as your x_2, and y_2 and the other point (-2,-3) as your x_1 and y_1
sooo |dw:1385526239897:dw|
I don't understand what I do next
@agent0smith
\[\large m = \frac{ y_2 - y_1 }{ x_2-x_1 }\] \[\large m = \frac{ 1 - (-3) }{ 1 - (-2) }\] \[\large m = ???\]
so I do -3 x -3 first I guess which = 9 then 1 - 9 = -8 -2 x -2 = 4 1 - 4 = -3 righttttttttttttttttt????
then 8 divided by 3?
See jim's post, follow it slowly and carefully. Not sure why you're doing -3*-3 and -2*-2, there's no multiplication involved.
the (-3) in it doesn't that mean your supposed to multiply
1-(-3) means 1 minus negative 3
also, 1 - (-3) is really the same as 1 + 3
What makes you think (-3) means you multiply...? Make sure you realize what mistake you are making there...
ok so 1 - -3 = 4
Correct
1 - -2 = 3
the reason I thought it was multiplication was because commonly with pemdas and sorts the parenthesis can mean you multiply the things inside
so 4/3
But still... what would you multiply it by (-3).. there is nothing other than a -3 in there. There's nothing to multiply it by.
(-3) does not mean (-3)^2
I thought it was sorta like -3^ with the exponents and I though oh it must multiply by itself
Make sure you don't continue thinking that... you will make a lot of mistakes if you do. If there's no exponent, don't assume there's one supposed to be there.
ya sorry math confuses me haha :) thank you for helping me though I really appreciate it
And yes, 4/3 is correct :)
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