Daily math problems for October 9, 2026 — Combinatorics, Trigonometry, Elementary Functions & more

$5!=5\cdot 4\cdot 3\cdot 2\cdot 1=120$.

$\tan {\displaystyle \frac{\pi }{4}}=1$, so $\arctan 1={\displaystyle \frac{\pi }{4}}$.

$x^2-4\ge 0\iff x\le -2$ or $x\ge 2$. Domain: $(-\mathrm{\infty },-2]\cup [2,+\mathrm{\infty })$.

$x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1{)}^2(x+2)$. So ${\displaystyle \frac{(x-1{)}^2(x+2)}{x-1}}=(x-1)(x+2)\to 0$ as $x\to 1$.

$\sigma$: $1\to 2\to 3\to 4\to 1$. So ${\sigma }^2$: $1\to 3$, $3\to 1$, $2\to 4$, $4\to 2$. That's $(13)(24)$.

$\bullet$ Take reciprocals of each equation, using $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$:

$$\frac{1}{x}+\frac{1}{y}=\frac{3}{5},\frac{1}{x}+\frac{1}{z}=\frac{2}{3},\frac{1}{y}+\frac{1}{z}=\frac{1}{4}$$

$\bullet$ Let $a=1\mathrm{/}x$, $b=1\mathrm{/}y$, $c=1\mathrm{/}z$. Sum all three:

$$2(a+b+c)=\frac{3}{5}+\frac{2}{3}+\frac{1}{4}=\frac{91}{60}$$

$$a+b+c=\frac{91}{120}$$

$\bullet$ Subtract each pair-sum from $a+b+c$:

$$a=\frac{61}{120},b=\frac{11}{120},c=\frac{19}{120}$$

$\bullet$ Recover $x,y,z$ and add:

$$x+y+z=\frac{120}{61}+\frac{120}{11}+\frac{120}{19}$$

$A={\displaystyle \frac{36\sqrt{3}}{4}}=9\sqrt{3}$.

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.

The slope is $8$ score points per additional hour. For a $0.5$-hour increase, the predicted change is $0.5\times 8=4$.

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.