graph y=6x-3

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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.

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Do you know the slope and y-intercept of the equation?
nope
|dw:1441861628631:dw|

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so the slope would be -6/1 and the y-intercept would be -3?
The y-intercept is -3. Correct! The slope is basically the number before \(x\). Is that -6?
yes
Wrong.
\(\color{#0cbb34}{\text{Originally Posted by}}\) @ludvic y=6x-3 \(\color{#0cbb34}{\text{End of Quote}}\) Is the slope -6? Is the number before \(x\), -6?
its 1
Do you know where \(x\) is?
yes
Where is it? What's number before x?
-6
Where did you get -6?
|dw:1441862025788:dw|
The problem already came like that in the book
What is the equation again?
my bad its positive 6/1
I'm sorry I saw it wrong
\(\color{#0cbb34}{\text{Originally Posted by}}\) @ludvic The problem already came like that in the book \(\color{#0cbb34}{\text{End of Quote}}\) I never say your problem was wrong. I said the slope you told me is wrong.
Okay. That's correct! So the slope is 6/1 or 6. Meaning you "rise" 6 times then "run" 1 time.
thank you
Since the y-intercept is -3, then plot (0, -3). |dw:1441862193136:dw|
Can you use the "Reply using Drawing" button to show me where (0, -3) is? :)
lol she thinks she's done with the problem... oh well
:'(
It is rise OVER run. :D
|dw:1441863363747:dw| Don't pay attention to this ^
let me try again
oops sorz 0_0. I'm pretty sure she left already though.
yup and when she logs in, she'll know the answer easily :(
I just sign up for the website and someone helped me with a problem so I wanted to help someone else on here, because it took like an hour.
When you're giving out direct answers, that's not helping. Besides, it's against the code of conduct of openstudy.
ok thanks for telling me :D
I did thank you. I just wanna help someone that I can but i'm not the best in math. I just started Alg 2 0_o. But I guess it's better to not answer at all unless I know exactly how to do it. I don't want to teach someone the wrong way.

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