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BloomLocke367
 one year ago
Help with PreCal?
BloomLocke367
 one year ago
Help with PreCal?

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anonymous
 one year ago
Best ResponseYou've already chosen the best response.0posting specifics helps

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1It says to solve algebraically and confirm graphically, and the first problem is \(v^25=82v^2\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1I was getting there, lol XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf v^25=82v^2\implies v^2+2v^2=8+5\implies v=?\)

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0put the variables on one side constants on the other keep the balance of the equation

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1When I solved aglebraically I got \(v=\sqrt{\frac{13}{3}}\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1But when I graphed it, I don't think it's right... but I may have made an error. That's why I'm asking for help.

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1oops, I meant to say when I graphed it, it didn't match my algebraic solution.

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0notice, is a quadratic equation, 2nd degree polynomial, thus 2 answers \(\bf v^25=82v^2\implies v^2+2v^2=8+5\implies v=\pm\sqrt{\cfrac{13}{3}}\)

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1I forgot to put that, but I know that. lol

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0hmmm ok.... so.. whathmm how did you do the graph anyway? just a table of values?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1I set the equations equal to zero

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1and used a calculator

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0so.... I'd think... the graph is ok...no?

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1obviously, I got a parabola with the yintercept 13. And my zeroes were \(\pm\)2.082

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1Is that correct?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0\(\bf \pm\sqrt{\cfrac{13}{3}}=\pm2.08166599946613273528\) so, the zeros look fine

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1ohhhh... so I was right, sorry. I was thinking of something else when I graphed it and that was my problem lol

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1Thanks *facepalm*

triciaal
 one year ago
Best ResponseYou've already chosen the best response.0dw:1440630977887:dw

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0http://fooplot.com/#W3sidHlwZSI6MCwiZXEiOiIzeF4yLTEzIiwiY29sb3IiOiIjMDAwMDAwIn0seyJ0eXBlIjoxMDAwLCJ3aW5kb3ciOlsiLTguMzQ5OTk5OTk5OTk5OTk4IiwiNy45IiwiLTEzLjU1MDAwMDAwMDAwMDAwMiIsIi0zLjU0OTk5OTk5OTk5OTk5OSJdfV0 drag the graph down, to see the intercepts anyhow, you can zoom in/out by using the mouse middlebutton so those are the intercepts

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1yeah, I have the graph right in front of me. I was just thinking I was looking for something else, I don't know why. XD

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0In every work of genius we recognize our own rejected thoughts: they come back to us with a certain alienated majesty. ~~ Ralph Waldo Emerson ~~

BloomLocke367
 one year ago
Best ResponseYou've already chosen the best response.1Very wise. I have another one, but I think I need to use completing the square to do it.. Can you maybe help?

anonymous
 one year ago
Best ResponseYou've already chosen the best response.0sure, post anew, more eyes :)
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