Daily math problems for July 26, 2026 — Combinatorics, Calculus, Logaritmi & more

$4!=24$.

$F(x)={\displaystyle \frac{x^2}{2}}$. ${\int }_0^{1}xdx=F(1)-F(0)={\displaystyle \frac{1}{2}}$.

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

By Viète's formulas, $r+s+t=6$ and $rs+rt+st=11$. Hence $r^2+s^2+t^2=36-22=14$.

$\sigma (c\cdot x)=\mathrm{\mid }c\mathrm{\mid }\sigma (x)$. With $c=2$: $\sigma (2x)=2\sigma$.

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

$\bullet$ Differentiate both sides implicitly (product rule on $xy$):

$$2x+y+x{y}^{\mathrm{\prime }}+2y{y}^{\mathrm{\prime }}=0$$

$\bullet$ Substitute $(x,y)=(1,1)$:

$$2+1+y^{\mathrm{\prime }}+2y^{\mathrm{\prime }}=0$$

$\bullet$ Solve for $y^{\mathrm{\prime }}$:

$$3y^{\mathrm{\prime }}=-3\Rightarrow y^{\mathrm{\prime }}=-1$$

$z^2=-4\Rightarrow z^2+4=0\Rightarrow (z-2i)(z+2i)=0$. Solutions: $z=2i$ or $z=-2i$.

$P(A)=P(B)={\displaystyle \frac{1}{2}}$, $P(R\mathrm{\mid }A)={\displaystyle \frac{3}{4}}$, $P(R\mathrm{\mid }B)={\displaystyle \frac{1}{4}}$.

$P(A\mathrm{\mid }R)={\displaystyle \frac{{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{1}{2}}}{{\displaystyle \frac{3}{4}}\cdot {\displaystyle \frac{1}{2}}+{\displaystyle \frac{1}{4}}\cdot {\displaystyle \frac{1}{2}}}}={\displaystyle \frac{3\mathrm{/}8}{4\mathrm{/}8}}={\displaystyle \frac{3}{4}}$.

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