Daily math problems for August 26, 2026 — Trigonometric identities, Algebra, Logaritmi & more

$\sin x\cos x={\displaystyle \frac{\sin x\cos x}{{\sin }^{2}x+{\cos }^{2}x}}={\displaystyle \frac{\tan x}{1+{\tan }^{2}x}}={\displaystyle \frac{2}{1+4}}={\displaystyle \frac{2}{5}}$.

$\bullet$ Balance each element. Carbon: $a=6d$. Hydrogen: $2b=12d\Rightarrow b=6d$. Oxygen: $2a+b=2c+6d$, giving $c=6d$.

$\bullet$ Choose the smallest positive integer solution $d=1$:

$$a=b=c=6,d=1$$

$\bullet$ The balanced equation is:

$$6{\text{CO}}_2+6{\text{H}}_2\text{O}\to 6{\text{O}}_2+{\text{C}}_6{\text{H}}_{12}{\text{O}}_6$$

$\bullet$ Add the coefficients:

$$a+b+c+d=6+6+6+1=19$$

Prin schimbarea de bază, ${\displaystyle \frac{\ln 27}{\ln 3}}={\log }_{3}27={\log }_3{3}^3=3$.

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

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

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

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

The "one-step subgroup test": $H$ is a subgroup iff non-empty and $a,b\in H\Rightarrow a{b}^{-1}\in H$.

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

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

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