we'll use the same log cancellation rule, since this is pretty much the same thing as the other, just recall that ln = logₑ.
![\bf \textit{Logarithm Cancellation Rules} \\\\ log_a a^x = x\qquad \qquad \stackrel{\stackrel{\textit{we'll use this one}}{\downarrow }}{a^{log_a x}=x} \\\\\\ \textit{logarithm of factors} \\\\ log_a(xy)\implies log_a(x)+log_a(y) \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)+ln(5)=2\implies ln(x\cdot 5)=2\implies log_e(5x)=2 \\\\\\ e^{log_e(5x)}=e^2\implies 5x=e^2\implies x=\cfrac{e^2}{5}](https://tex.z-dn.net/?f=%20%5Cbf%20%5Ctextit%7BLogarithm%20Cancellation%20Rules%7D%0A%5C%5C%5C%5C%0Alog_a%20a%5Ex%20%3D%20x%5Cqquad%20%5Cqquad%20%5Cstackrel%7B%5Cstackrel%7B%5Ctextit%7Bwe%27ll%20use%20this%20one%7D%7D%7B%5Cdownarrow%20%7D%7D%7Ba%5E%7Blog_a%20x%7D%3Dx%7D%0A%5C%5C%5C%5C%5C%5C%0A%5Ctextit%7Blogarithm%20of%20factors%7D%0A%5C%5C%5C%5C%0Alog_a%28xy%29%5Cimplies%20log_a%28x%29%2Blog_a%28y%29%0A%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29%2Bln%285%29%3D2%5Cimplies%20ln%28x%5Ccdot%205%29%3D2%5Cimplies%20log_e%285x%29%3D2%0A%5C%5C%5C%5C%5C%5C%0Ae%5E%7Blog_e%285x%29%7D%3De%5E2%5Cimplies%205x%3De%5E2%5Cimplies%20x%3D%5Ccfrac%7Be%5E2%7D%7B5%7D%20)
![\bf \\\\[-0.35em] \rule{34em}{0.25pt}\\\\ ln(x)-ln(5)=2\implies ln\left( \cfrac{x}{5} \right)=2\implies log_e\left( \cfrac{x}{5} \right)=2\implies e^{log_e\left( \frac{x}{5} \right)}=e^2 \\\\\\ \cfrac{x}{5}=e^2\implies x=5e^2](https://tex.z-dn.net/?f=%20%5Cbf%20%5C%5C%5C%5C%5B-0.35em%5D%0A%5Crule%7B34em%7D%7B0.25pt%7D%5C%5C%5C%5C%0Aln%28x%29-ln%285%29%3D2%5Cimplies%20ln%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20log_e%5Cleft%28%20%5Ccfrac%7Bx%7D%7B5%7D%20%5Cright%29%3D2%5Cimplies%20e%5E%7Blog_e%5Cleft%28%20%5Cfrac%7Bx%7D%7B5%7D%20%5Cright%29%7D%3De%5E2%0A%5C%5C%5C%5C%5C%5C%0A%5Ccfrac%7Bx%7D%7B5%7D%3De%5E2%5Cimplies%20x%3D5e%5E2%20)
3x +2=180//////////////////
I would say 8.55, regardless you're talking about the median or not.
The second option. 10
Hope this helps<3
Answer:
D. Pythagorean
Step-by-step explanation:
Given the identity
cos²x - sin²x = 2 cos²x - 1.
To show that the identity is true, we need to show that the left hand side is equal to right hand side or vice versa.
Starting from the left hand side
cos²x - sin²x ... 1
According to Pythagoras theorem, we know that x²+y² = r² in a right angled triangle. Coverting this to polar form, we have:
x = rcostheta
y = rsintheta
Substituting into the Pythagoras firnuka we have
(rcostheta)²+(rsintheta)² = r²
r²cos²theta+r²sin²theta = r²
r²(cos²theta+sin²theta) = r²
(cos²theta+sin²theta) = 1
sin²theta = 1 - cos²theta
sin²x = 1-cos²x ... 2
Substituting equation 2 into 1 we have;
= cos²x-(1-cos²x)
= cos²x-1+cos²x
= 2cos²x-1 (RHS)
This shows that cos²x -sin²x = 2cos²x-1 with the aid of PYTHAGORAS THEOREM
