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seen Dec 10 at 20:31

Dec
10
accepted Prove the median and the altitude drawn to the hypotenuse make an angle congruent to the difference of the acute angles of a right triangle.
Dec
7
asked Prove the median and the altitude drawn to the hypotenuse make an angle congruent to the difference of the acute angles of a right triangle.
Dec
6
accepted Find a point C on an infinite line AB which, when connecting two other points M and N, would form congruent angles
Dec
6
asked Find a point C on an infinite line AB which, when connecting two other points M and N, would form congruent angles
Dec
1
accepted Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$
Dec
1
revised Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$
deleted 2 characters in body
Dec
1
comment Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$
@AndréNicolas Thanks. Would you mind explicitly pointing out where I went wrong in an answer? I've checked several times and these errors aren't immediately obvious to me
Dec
1
asked Prove by induction that $\frac{1}{2}+\frac{2}{2^2}+\frac{3}{2^3}+\cdots+\frac{n}{2^n}=2-\frac{n+2}{2^n}$
Nov
22
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Ah okay. And I suppose if $iz=\log(4\pm\sqrt{15})$, the $i$ becomes interchangeable on both sides?
Nov
22
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
The quadratic I got is $e^{2iz}-8e^{iz}+1$. Completing the square gives $e^{iz}=4\pm i\sqrt{15}$ right? The answer lists $\sqrt{15}$ not $i\sqrt{15}$
Nov
22
accepted Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Nov
22
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
@Integrator huh?
Nov
22
comment Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Thanks. When I complete the square I get $iz=\ln(4\pm i\sqrt{15})$, but the answer listed is $i\ln(4\pm\sqrt{15})$. Is it acceptable to extract out the $i$ like that?
Nov
22
asked Find 2 imaginary numbers that have a cosine of 4, using $\cos z =\frac{e^{iz}+e^{-iz}}{2}$
Nov
21
accepted Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
A few people seem to have substituted $\pi - \theta$ for $\theta$ - which I can kind of understand (it seems kind of arbitrary). I'm not sure how to plug in your substitution, though...
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Shouldn't it be $\pi \operatorname{cosec} \theta \space d\theta$?
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
Ah okay. Finally, why does $\int f(a + b-t)dt = - \int f(a + b -t)dt$?
Nov
21
comment Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$
How does $\frac{dx}{dt}=1$? And why does $\int f(x)dx=-\int f(a+b-t)dt$? Could you add a few more steps if that's okay? :)
Nov
21
asked Prove that $\int^b_af(x)dx=\int^b_af(a+b-t)dt$