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3

There is no universally-agreed-upon definition of median for a data set with an even number of values. You have $10$ values, of which five are $\le 15$ and the others $\ge 17$. The median, defined as "a value separating the lower half of the data from the upper half", could in principle be anything between $15$ and $17$, but it's most common to take the ...


0

As $\tan x = \tan (x + \pi n)$, where $n\in\mathbb{Z}$, if you have $\tan d = \frac{z}{a + b\sec c}$ then all values $d$ satisfies this equation are gives by $$ d = \tan^{-1}\left(\frac{z}{a + b\sec c}\right) + \pi n, \quad n\in \mathbb{Z}. $$


0

Is $d$ an angle between $-\frac{\pi}{2}$ and $\frac{\pi}{2}$? If so, you don't need the $n$. Divide both sides by $a+b\sec c$: $$\frac{z}{a+b\sec c} = \tan d$$ And apply the inverse tangent function. It satisfies $\tan^{-1}( \tan d) = d$ for all $-\frac{\pi}{2}$ and $\frac{\pi}{2}$: $$\tan^{-1}(\frac{z}{a+b\sec c}) = d$$ Given the values $z,a,b,c$ ...


3

The formulas given by Wolfram Alpha are not wrong, they're "generic". (Which is a very specific form of wrong which means "only wrong on a set so small that you shouldn't ever care if you're picking random numbers, but always contains the point your boss randomly asks about". Okay. Not really, but some days...) The not generic versions can be generated ...


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Let $g(x)=x^4+bx^3+f$, in case $b=0$ and $f\in \mathbb{R}$, we always have a complex zeros for $g$. Simply, putting $f=\pm A^4$, then $x^4+ A^4=(x^2+A^2i)(x^2-A^2i)$ and $ x^4- A^4=(x^2-A^2 )(x^2+A^2)$. In case $b\ne0$, it is possible to have real roots for $g$, e.g., let $f>0$ and let $b=-2-2f$ define $g$ on $[0,1]$, then $g(0)=f>0$ and $f(1)=1+b+f=...


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There's a real root if and only if $256f \leq 27b^4$. I derived this inequality from the derivative to locate the local minimums. The case is trivial if $b = 0$, so assume $b \neq 0$. Check that the end behavior of the function $g(x) = x^4 + bx^3 + f$ satisfies $g(x) \rightarrow \infty$ as $x \rightarrow \pm \infty$. Thus, there is no guarantee that there ...



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