Given that a= 3i+7j and b= 5i-12j find: (1) 12a+3b
2 answers:
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3.
A.
find the area under the curve first
![\int\limits^{\sqrt[3]{\pi}}_0 {x sin(x^3)} \, dx =0.61157809233184](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5E%7B%5Csqrt%5B3%5D%7B%5Cpi%7D%7D_0%20%7Bx%20sin%28x%5E3%29%7D%20%5C%2C%20dx%20%3D0.61157809233184)
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
solve
![\int\limits^m_0 {mx-x^2} \, dx =[\frac{mx^2}{2}-\frac{x^3}{3}]^m_0](https://tex.z-dn.net/?f=%20%5Cint%5Climits%5Em_0%20%7Bmx-x%5E2%7D%20%5C%2C%20dx%20%3D%5B%5Cfrac%7Bmx%5E2%7D%7B2%7D-%5Cfrac%7Bx%5E3%7D%7B3%7D%5D%5Em_0)
so
m³=48
m=2∛6
A.
find the area under the curve first
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
solve
so
m³=48
m=2∛6