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

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

Mathematics
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I think you have to do the quadratic formula
umm i forgot what that is.... the b= something right?
in quadratic equation term with x^2 is A term with x is B term with constant is C

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yes -b + or - sqrt(b^2-4ac) all over at 2a
okay thx
in this case you can solve for your variables a b c
\[b = -b \pm \frac{ \sqrt{b^2 - 4ac} }{ 2 }\]
jazy its 2a
I thought I put that. But yes, over 2a. (:
\[b = -b \pm \frac{\sqrt{b^2 - 4ac}} { 2a }\]
uughh this is sooo harrddd
ax^2 + b^2 + c -16x^2 + 190 +0 what is a, b, and c?
-16 190 0
oh. i got -380
how do you get -380
you have the steps down, i think you know what you are doing
also 'x' term is missing..
oh snap i didnt do the bottom hold on
so... id plug in the answer i got from the eqation as x?
i got 0.95
it right @jazy
I think you had it right up to the -380. You have to divide by 2(-16) or -32. What is -380 /-32?
you have it
yea thats when i got 11.87
then i pluged it in to the equation
right(: what was the 0.95 for though?
y= -16(11.87^2)............. the 11.87 is substituted for the x

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