Daily math problems for October 8, 2026 — Calculus, Matrices, Algebra & more

$f^{\mathrm{\prime }}(x)=2x$, so $f^{\mathrm{\prime }}(3)=6$.

Add componentwise: $\left(\begin{array}{cc}1& 1\\ 8& 6\end{array}\right)$.

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

$\gcd (2,6)=2\mathrm{\ne }1$, so $\widehat{2}$ has no multiplicative inverse in ${\mathbb{Z}}_6$. ($\widehat{1}$ and $\widehat{5}$ are coprime to $6$ and invertible.)

$\bullet$ Evaluate each entry:

$$i^{2024}=(i^4{)}^{506}=1$$

$$\underset{x\to \mathrm{\infty }}{\lim }{(1+\frac{1}{x})}^{2x}=e^2\Rightarrow \ln (e^2)=2$$

$$\underset{x\to \pi }{\lim }\tan x=0$$

$${\log }_{1\mathrm{/}3}\frac{1}{81}={\log }_{1\mathrm{/}3}(1\mathrm{/}3{)}^4=4$$

$\bullet$ Assemble the matrix:

$$A=\left(\begin{array}{cc}1& 2\\ 0& 4\end{array}\right)$$

$\bullet$ Apply $\det (A^2)=(\det A{)}^2$:

$$\det A=4\Rightarrow \det (A^2)=16$$

$m^2-4=0\Rightarrow m\in \{-2,2\}$.

$\det A=m(m+1)-6=m^2+m-6=(m-2)(m+3)=0\Rightarrow m\in \{-3,2\}$.

After both moves, the value is $0.8\cdot 1.2=0.96$ — a $4\mathrm{\%}$ net loss.

${\displaystyle \frac{1}{n}}\sum (x_i+5)={\displaystyle \frac{1}{n}}\sum x_i+5=\bar{x}+5$.

Folosind schimbarea de bază, ${\log }_k(k+1)={\displaystyle \frac{\ln (k+1)}{\ln k}}$. Produsul telescopează: ${\displaystyle \frac{\ln 3}{\ln 2}}\cdot {\displaystyle \frac{\ln 4}{\ln 3}}\cdots {\displaystyle \frac{\ln 256}{\ln 255}}={\displaystyle \frac{\ln 256}{\ln 2}}={\log }_{2}256=8$.