Find the slope of the line graphed below. A. 5/3 B. 3/5 C. -5/3 D. -3/5
\( \color{Black}{\Rightarrow Slope = \Large{\Delta y \over \Delta x}}\)
Simply find the change in \(y\) for every unit change in \(x\).
Right. Can you just see my explanation and then find the slope?
slope = tan\(\theta\) .. = \(\large{\frac{opposite}{adjacent}}\) calculare opposite by using pythagoras theorem ..
Or you could just note down two points \((-3,0) \text{ and} (0,5)\)
yes .. both r correct
put opposite = 5 and adjacent = -3
May i know what r u getting? @dasDaName ?
@mathslover are you trying to find \(\theta\)?
\[Slope = {y_2 - y_1 \over x _2 - x_1} \Longrightarrow {5 - 0 \over 0 - (-3) }\]
no ..@lgbasallote .. yes @ParthKohli adjacent's place = 0-(-3)=3 ..
then why propose tan theta?
Why make it so complicated?
haha, im not sure how to start this.do i just randomly pick to points on the graph thats on the line?
Yes, the points that you can manage.
weren't you always keen on going "by the book" and being very complicated @ParthKohli ? suddenly had a change of views? lol
Two random points. Yep. It'd be better if the coordinates are integers.
ok, (−3,0) and(0,5), then the next step would be? nah il try to manage it
The slope formula is the next step.
\[Slope = {y_2 - y_1 \over x_2 - x _1} \]
\[(5-0)(0-(negative)3) \]
Yes.
There's a division sign in between, so careful.
correct ?
Yeah. What's 5 - 0?
*5 and 3
?
Yeah
alright, i really appreciate it man, thanks
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