Daily math problems for July 21, 2026 — Combinatorics, Calculus, Trigonometry & more

${\displaystyle \frac{n(n-1)}{2}}=15\Rightarrow n^2-n-30=0\Rightarrow n\in \{6,-5\}$. Only $n=6$ is admissible.

By the squeeze theorem, $-{\displaystyle \frac{1}{n}}\le {\displaystyle \frac{\sin n}{n}}\le {\displaystyle \frac{1}{n}}$ and both bounds tend to $0$, so the limit is $0$.

$\cos x=0⟺x=\pi \mathrm{/}2+k\pi$. On $[0,2\pi )$: $\pi \mathrm{/}2$ and $3\pi \mathrm{/}2$.

The cancellation law states: $a\ast b=a\ast c\Rightarrow b=c$. (Even without inverses, this property may or may not hold; in a group it always does.)

$\mathrm{\Delta }=16-4=12$, $a=-1$, so $f_{\max }=-{\displaystyle \frac{\mathrm{\Delta }}{4a}}=-{\displaystyle \frac{12}{-4}}=3$. The image is $(-\mathrm{\infty },3]$.

Outer: ${\sec }^2(x^2)$. Inner derivative: $2x$. So $f^{\mathrm{\prime }}(x)=2x{\sec }^2(x^2)$.

$\mathrm{\mid }z-1\mathrm{\mid }=\mathrm{\mid }z-(-1)\mathrm{\mid }$ describes the set of points equidistant from $1$ and $-1$. That's the perpendicular bisector of the segment between them, i.e. the imaginary axis $\mathrm{Re}(z)=0$.

$\bullet$ Multiply numerator and denominator by $2\sin 20\mathrm{°}$ and use $2\sin \theta \cos \theta =\sin 2\theta$:

$$\cos 20\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}={\displaystyle \frac{\sin 40\mathrm{°}\cos 40\mathrm{°}\cos 80\mathrm{°}}{2\sin 20\mathrm{°}}}$$

$\bullet$ Apply the identity again:

$$={\displaystyle \frac{\sin 80\mathrm{°}\cos 80\mathrm{°}}{4\sin 20\mathrm{°}}}$$

$\bullet$ And once more:

$$={\displaystyle \frac{\sin 160\mathrm{°}}{8\sin 20\mathrm{°}}}$$

$\bullet$ Use $\sin 160\mathrm{°}=\sin (180\mathrm{°}-20\mathrm{°})=\sin 20\mathrm{°}$:

$$={\displaystyle \frac{\sin 20\mathrm{°}}{8\sin 20\mathrm{°}}}={\displaystyle \frac{1}{8}}$$

On $[0,1]$: $1\le 1+x\le 2$, hence ${\displaystyle \frac{1}{2}}\le {\displaystyle \frac{1}{1+x}}\le 1$. So ${\displaystyle \frac{x^n}{1+x}}\le x^n$, giving $I_n\le {\int }_0^1{x}^{n}dx={\displaystyle \frac{1}{n+1}}$.

$2x-1=-x+5\Rightarrow 3x=6\Rightarrow x=2$, $y=3$. Intersection: $(2,3)$.