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Annette [7]
3 years ago
10

I am the key that is the hardest to turn What am I ??

Advanced Placement (AP)
2 answers:
alekssr [168]3 years ago
7 0
The key that is hardest to turn is an answer key hi hi hi hi hi hi 
laila [671]3 years ago
4 0
Donkey. They eat paper BTW.
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Please help me with this question, I don't know where to exactly start
Anarel [89]
3.

A.
find the area under the curve first
\int\limits^{\sqrt[3]{\pi}}_0 {x sin(x^3)} \, dx =0.61157809233184
then solve c∛π=0.61157809233184
c≈0.41757577488026
round if necicary



4.
alright, solve where they itnersect

x²=y=mx
x²=mx
x²-mx=0
x(x-m)=0
set to zero

x=0

x-m=0
x=m

they intersect at x=0 and x=m

which one is on top?
hmm
y=mx will always be on top (I don't know how to prove that but it's on top)

so we do
\int\limits^m_0 {mx-x^2} \, dx =8
solve
\int\limits^m_0 {mx-x^2} \, dx =[\frac{mx^2}{2}-\frac{x^3}{3}]^m_0  (\frac{m(m)^2}{2}-\frac{m^3}{3})-(0)=\frac{m^3}{2}-\frac{m^3}{3}=\frac{m^3}{6}=8

so
\frac{m^3}{6}=8
m³=48
m=2∛6
5 0
3 years ago
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