Daily math problems for August 30, 2026 — Algebra, Trigonometry, Determinants & more

Factoring gives $(x-2)(x-3)=0$, so the roots are $2$ and $3$. Their sum is $5$. (Equivalently, $-b\mathrm{/}a=5\mathrm{/}1$.)

$\sin x=\frac{1}{2}$ at $x=\frac{\pi }{6}$ (Q1) and $x=\frac{5\pi }{6}$ (Q2). Two solutions on $[0,2\pi )$.

$\det =2\cdot 5\cdot (-3)=-30$.

Aplicând regula produsului: ${\log }_{6}4+{\log }_{6}9={\log }_{6}36={\log }_6{6}^2=2$.

$A^{-1}=\left(\begin{array}{cc}1& 0\\ 0& 1\mathrm{/}2\end{array}\right)$. So $X=B\cdot A^{-1}=\left(\begin{array}{cc}2\cdot 1& 4\cdot 1\mathrm{/}2\\ 6\cdot 1& 8\cdot 1\mathrm{/}2\end{array}\right)=\left(\begin{array}{cc}2& 2\\ 6& 4\end{array}\right)$.

Slope $=-3$. Line through $(1,4)$: $y-4=-3(x-1)\Rightarrow y=-3x+7$.

$\bullet$ Use the fundamental limit:

$$\underset{n\to \mathrm{\infty }}{\lim }{(1+{\displaystyle \frac{2}{n}})}^n=e^2$$

$\bullet$ Rewrite the expression as a cube:

$${(1+{\displaystyle \frac{2}{n}})}^{3n}={\left[{(1+{\displaystyle \frac{2}{n}})}^n\right]}^3$$

$\bullet$ Take the limit:

$$(e^2{)}^3=e^6$$

$a=5$, $b=3$, $c=\sqrt{25-9}=4$. So $e={\displaystyle \frac{4}{5}}$.

$!4=4!(1-1+{\displaystyle \frac{1}{2}}-{\displaystyle \frac{1}{6}}+{\displaystyle \frac{1}{24}})=24\cdot {\displaystyle \frac{9}{24}}=9$.

The units are the residues coprime to $8$: $\{1,3,5,7\}$, four elements.