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OpenStudy (anonymous):
find the slope of the curve y = x^2 (x+2)^2 at point (1,2)
a. 81
b. 48
c. 64
d. 54
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OpenStudy (anonymous):
take the derivative, replace x by 1
OpenStudy (anonymous):
my answer is 64 i got wrong... i just wanted to know the answer...
OpenStudy (anonymous):
\[x^2(x+2)^2=x^2(x^2+4x+4)=x^4+4x^3+4x\]
\[y'=4x^3+12x^2+4\]
i get 20
OpenStudy (anonymous):
the answer is in the letter i just forgot..and i need it
OpenStudy (anonymous):
oh i made a mistak
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OpenStudy (anonymous):
\[x^2(x+2)^2=x^2(x^2+4x+4)=x^4+4x^3+4x^2\]
\[y'=4x^3+12x+8\] now i get 24
OpenStudy (anonymous):
i get 24 omg
OpenStudy (jlastino):
wth
OpenStudy (amistre64):
there is no point 1,2 on this curve
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OpenStudy (anonymous):
OMG MY CLASSMATE KING
OpenStudy (anonymous):
KINGkos is review his quiz and exam too hahahahha
OpenStudy (jlastino):
so the truth is revealed haha
OpenStudy (anonymous):
on tuesday is final exam thats y i memorized all the answer in quiz and homework.this question is my X in the quiz
OpenStudy (amistre64):
your question makes no sense ... chk it for typos
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