\(x=\sqrt{12 + \sqrt{12 + \sqrt{12 + \dotsb}}}\\ x=\sqrt{12+x}\)
etc
Solve it using matrices.
Shouldn't be too hard
Here is one of many videos available that teaches this.
https://www.youtube.com/watch?v=RgnWMBpQPXk
\(x = \frac{1}{3-\frac{1}{3-\frac{1}{3-\ldots}}}\\~\\ x=\frac{1}{3-x}\)
Now you can solve it.
Your question makes no sense to me.
You say that theta has the terminal point of P
then you say P is the terminal point of acos(sin(theta))
It can't be both.
Yes it is :)
Thanks Logarythm
Matrices are cool for working out systems of equations:
hint:
\(f^{-1}(5)=13\)
Yes that looks right to me
If p(x + 2) = 3x^2 -7x + 1, then what is p(x)?
p(x)=3(x-2)^2-7(x-2)+1
I'll explain it back to front
if p(x)=3(x-2)^2-7(x-2)+1
then
p(x+2)=3(x+2-2)^2-7(x+2-2)+1
so
p(x+2)=3(x)^2-7(x)+1
\(a(a+2b)+b(b+2c)+c(c+2a)=\frac{104}{3}+\frac{7}{9}+-7\\ a^2+2ab+b^2+2bc+c^2+2ac=\frac{256}{9}\\ a^2+2ab+2ac+b^2+2bc+c^2=\frac{256}{9}\\ a(a+2b+2c)+b(b+2c)+c^2=\frac{256}{9}\\ a(a+b+c)+a(b+c)+b(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+bc+c^2=\frac{256}{9}\\ a(a+b+c)+(a+b)(b+c)+c(b+c)=\frac{256}{9}\\ a(a+b+c)+(a+b+c)(b+c)=\frac{256}{9}\\ (a+b+c)(a+b+c)=\frac{256}{9}\\ |a+b+c|=\frac{16}{3}\)