Daily math problems for July 28, 2026 — Elementary Functions, Algebra, Calculus & more

$f(1)=1$, $f(2)={\displaystyle \frac{1}{2}}$. As $x$ grows, ${\displaystyle \frac{1}{x}}$ shrinks — $f$ is strictly decreasing on $(0,+\mathrm{\infty })$.

$P(1)=1-2+1=0$. So $X-1$ divides $P$ exactly.

${\displaystyle \frac{3n^2+5}{2n^2-1}}={\displaystyle \frac{3+5\mathrm{/}{n}^2}{2-1\mathrm{/}{n}^2}}\to {\displaystyle \frac{3}{2}}$.

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

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

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 $n\ge 2$).

$A^2=A\cdot A=\left(\begin{array}{cc}3& 1\\ 5& 4\end{array}\right)\left(\begin{array}{cc}3& 1\\ 5& 4\end{array}\right)=\left(\begin{array}{cc}3\cdot 3+1\cdot 5& 3\cdot 1+1\cdot 4\\ 5\cdot 3+4\cdot 5& 5\cdot 1+4\cdot 4\end{array}\right)=\left(\begin{array}{cc}14& 7\\ 35& 21\end{array}\right)$.

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

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

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.