8 in0.4 in=20
This shows you how to inscribe an equilateral triangle into a given circle.
This shows you how to inscribe a circle inside a given triangle.
That should do it.
you mean with compass and straight edge?
#3
|z−3|=|z+i|=|z−3i|let z=x+iy(x−3)2+y2=x2+(y−3)2x2−6x+9+y2=x2+y2−6y+9−6x=−6yx=y
|x−3+xi|=|x+(1+x)i|(x−3)2+x2=x2+(x+1)2x2−6x+9=x2+2x+18=8xx=1z=1+i
#2
z2=ˉzit should be intuitively clear that |z|=1 so let z=eiθei2θ=e−iθ+2πk, k∈Z2θ=2π−θθ=2π3z=ei2π/3=−12+i√32
you can set up a ratio such that both numerator and denominator both go to infinity
yet the ratio has any finite value you like.
well your answer agrees with sim!
what's the extra factor of 4! for?
oh nm, it's because you are having order matter.
(sorry I edited accidentally, I think it is back to how it was now)
Above is wrong. Simulation is showing p is approximately 0.14
Still working on this.
y=2x−13y=−3x+922x−13=−3x+925x=105x=21
3x+4y=76x+4y=c−4y 6x+8y=cletting c=14 makes equation 2 twice equation 1 and thus the system has infinite solutions
ab4=12a5b5=7776a5b20=125b15=1257776=32b3=2b=21/324/3a=12a=3⋅22/3
Add the two equations2a=8bb=a4a+a(a4)2=40(a4)a316=9aa2=144a=±12b=±3a=0,b=0 is also a solution