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OpenStudy (haleyelizabeth2017):
Are you supposed to solve for w?
OpenStudy (anonymous):
i think...idk
OpenStudy (anonymous):
its w to the power of 2 where it has the /
OpenStudy (haleyelizabeth2017):
oh! so \[w^2+36=12w\]?
OpenStudy (haleyelizabeth2017):
Are you familiar with the quadratic formula?
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OpenStudy (haleyelizabeth2017):
Or completing the square?
OpenStudy (anonymous):
yeah! and no...
OpenStudy (haleyelizabeth2017):
Okay, well it's a good day to learn something new!
OpenStudy (haleyelizabeth2017):
So the quadratic formula is just this...\[\frac{-b\pm \sqrt{b^2-4ac}}{2a}\]
OpenStudy (anonymous):
What is that lol...
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OpenStudy (haleyelizabeth2017):
Is this Algebra 2?
OpenStudy (anonymous):
i belive its pre algebra...
OpenStudy (haleyelizabeth2017):
Really? never mind the quadratic equation then.
OpenStudy (haleyelizabeth2017):
I don't think I ever had powers in pre algebra lol...
OpenStudy (anonymous):
lol its okay ill move on to the next problem...
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OpenStudy (haleyelizabeth2017):
But, an easy (sort of cheating way) is just to plug in all the values for w and see which one works.
OpenStudy (anonymous):
ok
OpenStudy (trojanpoem):
Are you sure of this question ?
OpenStudy (trojanpoem):
0.5 w + 36 = 12 w ?
OpenStudy (trojanpoem):
w power 2 = w^2 not w/2
/ = division
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OpenStudy (anonymous):
sorry its, w to the power of 2 plus 36 = 12
OpenStudy (haleyelizabeth2017):
oh, not 12x?
OpenStudy (haleyelizabeth2017):
12w*
OpenStudy (anonymous):
it s12w
OpenStudy (anonymous):
any guesses on whats the answer?
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OpenStudy (haleyelizabeth2017):
If this is pre algebra, you won't know the quadratic equation so that's out of the question. All we can do in this case, is plug in each of the values and see which one deems true.\[(0)^2+36=12(0)\]\[(2)^2+36=12(2)\]\[(4)^2+36=12(4)\]\[(6)^2+36=12(6)\]
Can you simplify each of these for me?