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Solve for W: 2w^3-w^2+6w-3=0
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factor by grouping and you'll get ^_^:\[= w^2(2w-1) + 3(2w-1) = (2w-1)(w^2+3)\]
Thanks!
np ^_^
you can also use horner's method of factoring polynomial
not to be a stickler in the mud; but there is no "W" in the problem :)
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lol, he meant w :)
meant, schmeant.... lol
lol, alright ^_^
Not to be a stickler, but the phrase is, "a stick in the mud"
\[\log_{17} 17^{y}\]
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Remember that \[k = log_{b}x \iff b^k = x\]
so let \[ k = log_{17} 17^y \rightarrow 17^? = 17^y\]
canadian eh....
\[\log_{17} 17^{y}\]
... are you repeating the question for any particular reason?
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Nah, i got it. typo. Thanks anyway
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