I REALLY REALLY NEED HELP! WILL MEDAL!
@sweetburger
@k_lynn
part A is the slope
The rate of changes is essentially the slope of the line.
For part B, all you have to do is find the slope of the information in the table and subtract it from the slope you got from the graph.
part B is easy once you know the slope you basically just subtract the slope from that year to get the previous one :)
You're going to need to use the slope formula.
rise over run :)
What would the slope be?
use the rise over run formula so the rise would be 14 because that is the cost per bushel
\(slope=\Large\frac{y_2-y_1}{x_2-x_1}\) Where, \((x_1,y_1)\) and \((x_2,y_2)\) are both on the line.
So the slope is 14?
no the slope would be 14/2 so what does 14/2 simplify to be?
Ohh yeea 7
Part B is 2 then, because 7 is 2 more dollars than 5 from the prev one
@SydtheKid913 @StudyGurl14 Am I right?
you got it! great job :D \(\cal\Huge\color{red}{V}\color{orangered}{I}\color{darkorange}{C}\color{orange}{T}\color{gold}{O}\color{yellow}{R}\color{greenyellow}{Y}\color{yellowgreen}{!}\color{olivedrab}{!}\color{green}{!}\color{#088A4B}{!}\color{#088A68}{!}\color{#088A85}{!}\color{#086A87}{!}\color{#084B8A}{!}\color{#08298A}{!}\color{#08088A}{!}\color{#29088A}{!}\color{#4B088A}{!}\color{#6A0888}{!}\color{#8A0886}{!}\color{#8A0868}{!}\color{#8A084B}{!}\color{#8A0829}{!}\)
Yesss! thanks you guys, that rly helped me :D
:D no problem that's what I'm here for \(\def\Syd{\bf\Huge\color{lime}{\Huge\circledS}_\color{blue}{Y_\color{red}{D}\color{yellow}{T}\color{aqua}{H}_\color{darkmagenta}{E}\color{darkorange}{K}^\color{chartreuse}{I}_\color{crimson}{D}\color{blueviolet}{9}\color{blue}{1}\color{fuchsia}{3}^\ddot{\smile}}\color{lime}{♫}}\Syd\)
Nice job helping @SydtheKid913
I might have some more questions in a little while, ill tag you once I post them if you dont mind. :)
And you are correct @AgentLocke
thanks! @StudyGurl14 I've been working on my helping skills :D
Yes! wow didnt know how easy this question was, haha
Thanks again guys :D
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