Daily math problems for October 30, 2026 — Fundamental Algebra, Algebra, Euclidean Geometry & more

$y=2x-1\Rightarrow y+1=2x\Rightarrow x={\displaystyle \frac{y+1}{2}}$.

$f(-2)=(-2{)}^3=-8$.

$12+8=20{\text{ cm}}^2$.

$\bullet$ Add columns $2$ and $3$ into column $1$ to factor out $(x+4)$:

$$\det =(x+4)\mid \begin{array}{ccc}1& 2& 2\\ 1& x& 2\\ 1& 2& x\end{array}\mid$$

$\bullet$ Subtract row $1$ from rows $2$ and $3$, then expand:

$$\det =(x+4)(x-2{)}^2$$

$\bullet$ Solve $(x+4)(x-2{)}^2=0$:

$$x=-4\text{ or }x=2$$

$f^{\mathrm{\prime }}(x)=5(x^2+1{)}^4\cdot 2x=10x(x^2+1{)}^4$.

$d={\displaystyle \frac{\mathrm{\mid }3\cdot 2-4\cdot 3+5\mathrm{\mid }}{\sqrt{9+16}}}={\displaystyle \frac{\mathrm{\mid }-1\mathrm{\mid }}{5}}={\displaystyle \frac{1}{5}}$.

Sum $=5\cdot 7+8\cdot 8+7\cdot 9=35+64+63=162$. Mean $={\displaystyle \frac{162}{20}}=8.1$.

$\det =(b-a)(c-a)(c-b)$. (It's zero iff at least two of $a,b,c$ are equal.)

Domain constraint: $x^2-4>0\Rightarrow x\in (-\mathrm{\infty },-2)\cup (2,+\mathrm{\infty })$.

Inequality: ${\log }_{1\mathrm{/}3}(x^2-4)>0={\log }_{1\mathrm{/}3}1$. Since the base ${\displaystyle \frac{1}{3}}<1$, ${\log }_{1\mathrm{/}3}$ is decreasing, so the inequality flips: $x^2-4<1$, i.e. $x^2<5$, so $x\in (-\sqrt{5},\sqrt{5})$.

Intersection with the domain: $[(-\mathrm{\infty },-2)\cup (2,+\mathrm{\infty })]\cap (-\sqrt{5},\sqrt{5})=(-\sqrt{5},-2)\cup (2,\sqrt{5})$.

$y^2-6y+8=0\Rightarrow y\in \{2,4\}$. Then $x^2=2\Rightarrow x=\pm \sqrt{2}$ and $x^2=4\Rightarrow x=\pm 2$. Four distinct real solutions.