kiamonstaa
Y= 16x^2+190+0
A=
B=
C=
graph when done.



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ilikephysics2
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I think you have to do the quadratic formula

kiamonstaa
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umm i forgot what that is.... the b= something right?

uzumakhi
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in quadratic equation
term with x^2 is A
term with x is B
term with constant is C

ilikephysics2
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yes b + or  sqrt(b^24ac) all over at 2a

kiamonstaa
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okay thx

ilikephysics2
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in this case you can solve for your variables a b c

jazy
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\[b = b \pm \frac{ \sqrt{b^2  4ac} }{ 2 }\]

ilikephysics2
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jazy its 2a

ilikephysics2
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@jazy

jazy
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I thought I put that. But yes, over 2a. (:

jazy
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\[b = b \pm \frac{\sqrt{b^2  4ac}} { 2a }\]

kiamonstaa
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uughh this is sooo harrddd

jazy
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ax^2 + b^2 + c
16x^2 + 190 +0
what is a, b, and c?

kiamonstaa
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16
190
0

kiamonstaa
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oh. i got 380

uzumakhi
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how do you get 380

ilikephysics2
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you have the steps down, i think you know what you are doing

hartnn
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also 'x' term is missing..

kiamonstaa
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oh snap i didnt do the bottom hold on

kiamonstaa
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so... id plug in the answer i got from the eqation as x?

kiamonstaa
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i got 0.95

sudharsa
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it right @jazy

jazy
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I think you had it right up to the 380. You have to divide by 2(16) or 32.
What is 380 /32?

ilikephysics2
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you have it

kiamonstaa
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yea thats when i got 11.87

kiamonstaa
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then i pluged it in to the equation

jazy
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right(: what was the 0.95 for though?

kiamonstaa
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y= 16(11.87^2).............
the 11.87 is substituted for the x