Daily math problems for November 8, 2026 — Euclidean Geometry, Fundamental Algebra, Algebra & more

${360}^{\circ }÷4={90}^{\circ }$.

Scale factor $=8÷2=4$; longer side $=3\times 4=12$ cm.

${\displaystyle \frac{3x}{x}}=3$ for $x\mathrm{\ne }0$.

Lagrange: in a finite group, the order of any subgroup (and hence the order of any element, which is the order of the cyclic subgroup it generates) divides $\mathrm{\mid }G\mathrm{\mid }$.

The inverse of $\widehat{3}$ in ${\mathbb{Z}}_7$ is $\widehat{5}$ (since $3\cdot 5=15\equiv 1$). So $\widehat{x}=\widehat{5}\cdot \widehat{2}=\widehat{10}=\widehat{3}$.

$\mathrm{\mid }x-3\mathrm{\mid }<2\iff -2<x-3<2\iff 1<x<5$, so the set is $(1,5)$.

${\lim }_{x\to 0^{-}}f=-1$ and ${\lim }_{x\to 0^{+}}f=1$ — finite but unequal limits, hence a jump discontinuity.

${\lim }_{x\to \mathrm{\infty }}{e}^{-x}=0$, so $y=0$ is a horizontal asymptote at $+\mathrm{\infty }$.

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

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

$\mathrm{\mid }z-1\mathrm{\mid }=\mathrm{\mid }z-(-1)\mathrm{\mid }$ describes the set of points equidistant from $1$ and $-1$. That's the perpendicular bisector of the segment between them, i.e. the imaginary axis $\mathrm{Re}(z)=0$.