Ask your own question, for FREE!
Mathematics 13 Online
OpenStudy (anonymous):

**MEDAL AND FAN** This may be odd, but can someone help me please? I wrote an essay on why using algebraic methods over graphing is easier when solving non-linear equations. These were the points that I used: 1. You get a more accurate answer when using algebraic methods. 2. It can be difficult to graph complex curves on a graph. 3. Sometimes, you run out of room on your graph paper for your graph. I must now provide an graph of a system of non-linear equations that supports my viewpoint. Can someone please draw or find a graph that proves one of my points above?

OpenStudy (anonymous):

@djspeed Can you help?

OpenStudy (anonymous):

sure sorry

OpenStudy (anonymous):

Thank you :) it's fine! Or at least provide me with a non-linear equation that would prove one of my points above and then I can draw the graph.

OpenStudy (anonymous):

hold on @Jaynator495 @jeffyblood @freegirl112

OpenStudy (anonymous):

@SyedMohammed98

OpenStudy (anonymous):

um im trying to get some freands to come

OpenStudy (anonymous):

Thank you :)

OpenStudy (anonymous):

not going to stop tell get you some help

OpenStudy (anonymous):

Thank you! I keep looking for graphs and trying to think of equations and I just cant think of any!

OpenStudy (anonymous):

sweet ok @SyedMohammed98 this person needs help

OpenStudy (anonymous):

Please! :)

OpenStudy (anonymous):

darn

OpenStudy (anonymous):

@triciaal

OpenStudy (anonymous):

:(

OpenStudy (anonymous):

should be coming soon

OpenStudy (anonymous):

Thank you! Hopefully someone can come to the rescue! lol

OpenStudy (anonymous):

lol you funny

OpenStudy (anonymous):

@endlesslight254 last person of my freands on then I start saying random names

OpenStudy (anonymous):

yes

OpenStudy (anonymous):

That's what I am about to start doing!

OpenStudy (anonymous):

i think ur question killed my brain

OpenStudy (anonymous):

Lol, dang it!

OpenStudy (anonymous):

oh OH GOSH ITS GOING TO BLOW exploshin

OpenStudy (anonymous):

it seems to complicated O.O

OpenStudy (anonymous):

._.

OpenStudy (anonymous):

yes another freand came on

OpenStudy (anonymous):

@Unanimoose

OpenStudy (anonymous):

@hartnn , @Nurali

OpenStudy (anonymous):

@DullJackel09 @ryanvarghese12790 @Quxxn @NeonStrawsForever

OpenStudy (anonymous):

know one is coming:-(

OpenStudy (anonymous):

YES

OpenStudy (triciaal):

don't every agree with not having enough room on paper because you need to choose the right scale to get it on paper.

OpenStudy (anonymous):

You do or do not agree with this?

OpenStudy (anonymous):

well this is all you guys im going

OpenStudy (anonymous):

need any more help tag me

OpenStudy (anonymous):

Thank you for your help @djspeed :) I really appreciate it!!!

OpenStudy (anonymous):

NP anytime

OpenStudy (triciaal):

thought provoking!! @jim_thompson5910, @mathmath333 @hartnn ?

hartnn (hartnn):

if the solution is large, then the scale of the graph can be adjusted accordingly. to accomodate the solution point.

hartnn (hartnn):

though i agree with the 1st 2 points :)

OpenStudy (anonymous):

Thank you !!! I believe I am going to go with this graph to prove my 1st point. The answer to this equation is (-6.5, 2.4) and (-2.4, 6.5).This would be very difficult to come up with by using the graph.

hartnn (hartnn):

1. Line is linear, don't you need 2 non linear equations? 2. solution points... 6.5, 2.4 .. are very easy to get from graph even if we have a scale of 0.5 units per cm

hartnn (hartnn):

how about these to prove your point x^2 = 22 y, y^2 = 19x :)

hartnn (hartnn):

look at the 2nd solution! no way you can get it accurate from a graph :P

OpenStudy (anonymous):

Yes! I see what you mean :) Yours makes more sense. I will use that one. Thank you so so so much!

hartnn (hartnn):

welcome ^_^

Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!
Can't find your answer? Make a FREE account and ask your own questions, OR help others and earn volunteer hours!

Join our real-time social learning platform and learn together with your friends!