Daily math problems for November 2, 2026 — Fundamental Algebra, Descriptive Statistics & Sampling, Algebra & more

$3+4=7$; $2\times 7=14$; $20-14=6$.

An upward (rising-left-to-right) trend means higher $x$ is associated with higher $y$ — a positive correlation.

${\displaystyle \frac{3}{4}}\times 100\mathrm{\%}=75\mathrm{\%}$.

A homomorphism is injective iff its kernel is trivial.

$\bullet$ Read off Vieta's relations for $x^3-2x^2+3x-4=0$:

$$s_1=x_1+x_2+x_3=2$$

$$s_2=x_1{x}_2+x_1{x}_3+x_2{x}_3=3$$

$\bullet$ Apply the identity:

$$x_1^2+x_2^2+x_3^2=s_1^2-2s_2$$

$\bullet$ Evaluate:

$$4-6=-2$$

${\log }_{2}6+{\log }_2{\displaystyle \frac{8}{3}}={\log }_2\text{\hspace{0.17em}⁣}(6\cdot {\displaystyle \frac{8}{3}})={\log }_{2}16=4$.

$\bullet$ Write $2^{1\mathrm{/}n}=e^{\ln 2\mathrm{/}n}$ and use $e^u-1\sim u$ for $u\to 0$:

$$2^{1\mathrm{/}n}-1\sim \frac{\ln 2}{n}$$

$\bullet$ Multiply by $n$:

$$n\cdot \frac{\ln 2}{n}=\ln 2$$

$\cos A={\displaystyle \frac{25+9-49}{2\cdot 5\cdot 3}}={\displaystyle \frac{-15}{30}}=-{\displaystyle \frac{1}{2}}$. Since $A\in (0,\pi )$, $A={\displaystyle \frac{2\pi }{3}}$.

By parts: $\int x\ln xdx={\displaystyle \frac{x^2}{2}}\ln x-\int {\displaystyle \frac{x}{2}}dx={\displaystyle \frac{x^2}{2}}\ln x-{\displaystyle \frac{x^2}{4}}$. Evaluating on $[0,1]$ (with $x^2\ln x\to 0$ as $x\to 0^{+}$) gives $0-\frac{1}{4}-0=-\frac{1}{4}$.

With $Y=X^2$: $Y^2-5Y+4=0\Rightarrow Y\in \{1,4\}$. Then $X^2=1\Rightarrow X=\pm 1$ and $X^2=4\Rightarrow X=\pm 2$. Four real solutions.