Daily math problems for October 29, 2026 — Algebra, Fundamental Algebra, Geometry & more

Require $x-4\ge 0$, i.e. $x\ge 4$.

Domain: $[4,\mathrm{\infty })$.

$3\times 4=12$, then $2+12=14$.

Centre $(h,k)=(2,-3)$ and $r=5$ (so $r^2=25$):

$(x-2{)}^2+(y-(-3){)}^2=25⟹(x-2{)}^2+(y+3{)}^2=25$.

$f(x)=x+{\displaystyle \frac{1}{x}}$. As $x\to \pm \mathrm{\infty }$, ${\displaystyle \frac{1}{x}}\to 0$, so the slant asymptote is $y=x$.

$\bullet$ Domain: both arguments positive, so $x>2$.

$\bullet$ Combine logs:

$${\log }_3((x-2)(x+6))=2$$

$\bullet$ Exponentiate and expand:

$$(x-2)(x+6)=9$$

$$x^2+4x-21=0$$

$\bullet$ Factor:

$$(x+7)(x-3)=0$$

$\bullet$ $x=-7$ fails the domain $x>2$. Only $x=3$ remains.

Take any $g\mathrm{\ne }e$. The subgroup $⟨g⟩$ has order dividing $p$. It's not $1$ (since $g\mathrm{\ne }e$), so it has order $p$, i.e. $⟨g⟩=G$. So $G$ is cyclic, hence abelian.

$u=x^2+1\Rightarrow du=2xdx$. When $x=0$, $u=1$; when $x=1$, $u=2$. ${\int }_1^2{u}^{3}du={\displaystyle \frac{16-1}{4}}={\displaystyle \frac{15}{4}}$.

${\displaystyle \frac{\sin 3x}{\tan 5x}}={\displaystyle \frac{\sin 3x}{3x}}\cdot {\displaystyle \frac{5x}{\tan 5x}}\cdot {\displaystyle \frac{3}{5}}\to 1\cdot 1\cdot {\displaystyle \frac{3}{5}}={\displaystyle \frac{3}{5}}$.

$n+1=11\Rightarrow n=10$.

$\bullet$ RHS must be non-negative, so $x\ge -2$.

$\bullet$ Case 1 ($\mathrm{\mid }x\mathrm{\mid }\ge 2$, so $x^2-4\ge 0$):

$$x^2-4=x+2\Rightarrow x^2-x-6=0$$

$\bullet$ Factor and solve:

$$(x-3)(x+2)=0\Rightarrow x=3,-2$$

$\bullet$ Case 2 ($\mathrm{\mid }x\mathrm{\mid }<2$, so $x^2-4<0$):

$$4-x^2=x+2\Rightarrow x^2+x-2=0$$

$\bullet$ Factor and solve:

$$(x+2)(x-1)=0\Rightarrow x=1$$

(only $x=1$ lies strictly inside $\mathrm{\mid }x\mathrm{\mid }<2$; $x=-2$ already appears in Case 1).

$\bullet$ Distinct valid solutions:

$$\{-2,1,3\}$$

$\bullet$ Three real solutions.