Daily math problems for September 21, 2026 — Sequences and series, Determinants, Trigonometry & more

$a_{20}=a_1+19d=2+19\cdot 3=2+57=59$.

A matrix with a zero row has determinant $0$.

${\displaystyle \frac{2\pi }{3}}\in [0,\pi ]$, so $\arccos \text{\hspace{0.17em}⁣}(\cos {\displaystyle \frac{2\pi }{3}})={\displaystyle \frac{2\pi }{3}}$.

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

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.)

$\mathrm{\mid }z-1\mathrm{\mid }=\mathrm{\mid }z-(-1)\mathrm{\mid }$ 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 $\mathrm{Re}(z)=0$.

Number of sequences with exactly two heads: $\left(\genfrac{}{}{0px}{}{3}{2}\right)=3$. Total sequences: $8$. Probability $=\frac{3}{8}$.

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

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: $(2,3)$.