Daily math problems for September 28, 2026 — Logs, Sequences, Trigonometry & more

$\bullet$ Rewrite $50$ using $100$:

$$50={\displaystyle \frac{100}{2}}$$

$\bullet$ Apply $\lg (M\mathrm{/}N)=\lg M-\lg N$:

$$\lg 50=\lg 100-\lg 2$$

$\bullet$ Since $\lg 100=2$ and $\lg 2=a$:

$$\lg 50=2-a$$

$S_{12}={\displaystyle \frac{12(2\cdot 50+11\cdot 10)}{2}}=6\cdot 210=1260$.

$\sin \text{\hspace{0.17em}⁣}(\pi +{\displaystyle \frac{\pi }{6}})=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$.

$4^x=2^{2x}=(2^x{)}^2=5^2=25$.

${\lim }_{x\to \mathrm{\infty }}{e}^{-x}=0$, so $y=0$ is a horizontal asymptote at $+\mathrm{\infty }$.

The polynomial $z^n-1$ has zero coefficient for $z^{n-1}$, so by Viète's formula the sum of its roots is $0$. Equivalently, the geometric sum ${\sum }_{k=0}^{n-1}{e}^{2\pi ik\mathrm{/}n}={\displaystyle \frac{1-1}{1-e^{2\pi i\mathrm{/}n}}}=0$ (for $n\ge 2$).

$\mathrm{\Delta }=36-32=4$, $a=1$, so $f_{\min }=-{\displaystyle \frac{4}{4}}=-1$ (attained at $x=3$).

$A\mathrm{△}\mathrm{\varnothing }=A$, so $\mathrm{\varnothing }$ is the identity.

$P(2)=8-2m+4=12-2m=0\Rightarrow m=6$.

A linear system is consistent (has at least one solution) iff $\mathrm{rank}(A)=\mathrm{rank}([A\mathrm{\mid }b])$ — the rank of the coefficient matrix equals the rank of the augmented matrix.