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lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
are you allowed to use trigonometry?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
it was a cosh function so yes
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
is there no such thing as arc cosh?
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
If there is I have not been taught it yet
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
i'm sure there is though
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
I found the arc cosh function on my calculator, but how do I do this by hand?
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
I think if we make it into a quadratic equation with \(e^x\), you could make some progress here.
 one year ago

lgbasallote Group TitleBest ResponseYou've already chosen the best response.0
"progress"
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
so if we multiply everything by e^x to make it a quadratic
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
Well, we know the definition of \(\cosh x\) is: \[ \cosh x=\frac{e^x+e^{x}}{2} \]So: \[ 2\cosh x=2.5\implies\\ x=\pm\ln(2) \]
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
(Where \(x\in\mathbb{R}\))
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
where did the ln come from?
 one year ago

LolWolf Group TitleBest ResponseYou've already chosen the best response.1
We have: \[ \cosh^{1}z=\ln\left(z+\sqrt{(z+1)(z1)}\right) \]So: \[ \cosh^{1}1.25=\ln\left(1.25+\sqrt{(2.25)(.25)}\right)=\ln(2) \]
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
Yeah, for that quadratic method, move everything to left side and multiply by \(e^x\): \((e^x)^2  2.5 e^x + 1 = 0\) You can use quadratic formula: \( x = \frac{b \pm \sqrt{b^2  4ac}}{2a}\) \[ e^x = \frac{2.5 \pm \sqrt{(2.5)^2  4}}{2} \\ e^x = \frac{2.5 \pm \sqrt{2.25}}{2} \\ e^x = \frac{2.5 \pm 1.5}{2} \\ e^x = \frac{2.5 + 1.5}{2} \text{ or } e^x = \frac{2.5  1.5}{2} \\ e^x = 2 \text{ or } e^x = \frac{1}{2} \] Then we can just take natural logarithms for the same as above answer.
 one year ago

baldymcgee6 Group TitleBest ResponseYou've already chosen the best response.0
k thanks
 one year ago

AccessDenied Group TitleBest ResponseYou've already chosen the best response.3
You're welcome! :)
 one year ago

Hero Group TitleBest ResponseYou've already chosen the best response.0
I was wondering if something like that could be done. Now I have my answer.
 one year ago
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