Mathematics
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OpenStudy (zenmo):
How to simplify the point-slope form?
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OpenStudy (zenmo):
\[16+\frac{ 1 }{ 6 }(x-1)\]
OpenStudy (misty1212):
lol is this from the last problem ? just asking because that is not right if it is
OpenStudy (zenmo):
yea, i have the answer, but it doesn't count as the solution
OpenStudy (misty1212):
the point was not \((1,8)\) it was \((64,8)\)
OpenStudy (zenmo):
where did u get 8 from?
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OpenStudy (misty1212):
is it the last problem? the point was \((64,8)\) you got i t
OpenStudy (zenmo):
sec
OpenStudy (misty1212):
should be \[y-8=\frac{1}{6}(x-64)\]
OpenStudy (misty1212):
maybe i am remembering wrong
OpenStudy (zenmo):
yea but I'm suppose to use linearization form \[f(a)+f'(a)(x-a) \to find the equation\]
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OpenStudy (zenmo):
OpenStudy (misty1212):
ok right
so \(x_1=64\)
OpenStudy (misty1212):
and \(y_1=f(64)=8\)
OpenStudy (misty1212):
point slope gives \[y-8=\frac{1}{6}(x-64)\]
OpenStudy (zenmo):
isnt 64^2/3 = 16? cube root of 64 is 4, so 4^2=16
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OpenStudy (misty1212):
not sure where the 16 or the 1 came from
OpenStudy (misty1212):
ooh sorry
OpenStudy (misty1212):
you are right i am wrong
OpenStudy (misty1212):
but there is still no 1` in it, \(x_1=64\)
OpenStudy (misty1212):
\[y-16=\frac{1}{6}(x-64)\] that's better
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OpenStudy (zenmo):
Oh, I found the error, I used 1/6 for a in (x-a) instead of 64. Oops!
OpenStudy (zenmo):
/derp
OpenStudy (misty1212):
ooh