Daily math problems for November 5, 2026 — Fundamental Algebra, Euclidean Geometry, Determinants & more

$6=2\cdot 3$ and $8=2^3$.

Highest powers: $2^3$ and $3$.

$\text{LCM}=8\cdot 3=24$.

$C=2\pi \times 7=14\pi$ cm.

${180}^{\circ }÷3={60}^{\circ }$.

$\overrightarrow{AB}=(3,5)$, $\overrightarrow{AC}=(-1,3)$. $\det =3\cdot 3-5\cdot (-1)=14$. Area $={\displaystyle \frac{14}{2}}=7$.

$\bullet$ Both numerator and denominator $\to 0$. Use $\tan u\sim u$ and $\sin u\sim u$ as $u\to 0$:

$$\underset{x\to 1}{\lim }\frac{\tan (x^2-1)}{\sin (x^2-3x+2)}=\underset{x\to 1}{\lim }\frac{x^2-1}{x^2-3x+2}$$

$\bullet$ Factor and cancel the common $(x-1)$:

$$\frac{(x-1)(x+1)}{(x-1)(x-2)}=\frac{x+1}{x-2}$$

$\bullet$ Substitute $x=1$:

$$\frac{2}{-1}=-2$$

$\bullet$ Apply $\ln (1+u)\sim u$ when $u\to 0$:

$$\ln \text{\hspace{0.17em}⁣}(1+\frac{1}{2^n})\sim \frac{1}{2^n}$$

$\bullet$ The product reduces to a geometric ratio:

$$3^n\cdot \frac{1}{2^n}={\left(\frac{3}{2}\right)}^n$$

$\bullet$ Since $3\mathrm{/}2>1$, this diverges:

$$\underset{n\to \mathrm{\infty }}{\lim }{\left(\frac{3}{2}\right)}^n=+\mathrm{\infty }$$

$x^2+3=(x+1{)}^2=x^2+2x+1\Rightarrow 2=2x\Rightarrow x=1$. Check: $\sqrt{4}=2=1+1$ ✓.

$2x^2+x+1=2(x^2-x+2)\Rightarrow 2x^2+x+1=2x^2-2x+4\Rightarrow 3x=3\Rightarrow x=1$.

Verify the domain: at $x=1$, $2x^2+x+1=4>0$ and $x^2-x+2=2>0$, so both logs are defined. The solution set is $\{1\}$.

$\sin {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{\sqrt{3}}{2}}$ and $\cos {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\sqrt{3}}{2}}$. Sum: $\sqrt{3}$.

By Pascal: $\left(\genfrac{}{}{0px}{}{7}{3}\right)+\left(\genfrac{}{}{0px}{}{7}{4}\right)=\left(\genfrac{}{}{0px}{}{8}{4}\right)={\displaystyle \frac{8!}{4!4!}}=70$.