Daily math problems for October 10, 2026 — Algebra, Euclidean Geometry, Parametric & Polar Calculus & more

$8^{2\mathrm{/}3}={\left(8^{1\mathrm{/}3}\right)}^2=2^2=4$.

(Or: $8^2=64$, then $\sqrt[3]{64}=4$.)

$15÷5=3$.

$$A={\displaystyle \frac{1}{2}}{\int }_0^{2\pi }{2}^{2}d\theta ={\displaystyle \frac{1}{2}}(4)(2\pi )=4\pi \mathrm{.}$$

Consistent with $\pi r^2=\pi (2{)}^2=4\pi$ — the curve $r=2$ is a circle of radius $2$.

Substitute $u=x^2+1$, giving $du=2xdx$, so $xdx=\frac{1}{2}du$. The limits become $x=0\Rightarrow u=1$ and $x=2\Rightarrow u=5$:

$${\int }_0^2\frac{x}{(x^2+1{)}^2}dx=\frac{1}{2}{\int }_1^5{u}^{-2}du$$

$$\frac{1}{2}{[-\frac{1}{u}]}_1^5=\frac{1}{2}(1-\frac{1}{5})=\frac{2}{5}$$

The terminal side of ${\displaystyle \frac{5\pi }{6}}$ is in Quadrant II, where cosine is negative. Its reference angle is

$$\pi -\frac{5\pi }{6}=\frac{\pi }{6}\mathrm{.}$$

The unit-circle value at the reference angle is

$$\cos \frac{\pi }{6}=\frac{\sqrt{3}}{2}\mathrm{.}$$

Applying the Quadrant II sign gives

$$\cos \frac{5\pi }{6}=-\frac{\sqrt{3}}{2}\mathrm{.}$$

A standard fundamental limit: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{\ln (1+x)}{x}}=1}$.

$u=x^2\Rightarrow du=2xdx$. So $\int \cos udu=\sin u+C=\sin (x^2)+C$.

By Stirling, $\sqrt[n]{n!}\sim {\displaystyle \frac{n}{e}}(2\pi n{)}^{1\mathrm{/}(2n)}$. The factor $(2\pi n{)}^{1\mathrm{/}(2n)}\to 1$, so ${\displaystyle \frac{\sqrt[n]{n!}}{n}}\to {\displaystyle \frac{1}{e}}$.

$A=\left(\begin{array}{cc}2& 3\\ 1& -1\end{array}\right)$ — coefficients of $x$ and $y$ in each equation.

Convert each base to modulus–argument form: $$1+i\sqrt{3}=2\mathrm{cis}{\displaystyle \frac{\pi }{3}},1-i=\sqrt{2}\mathrm{cis}\text{\hspace{0.17em}⁣}(-{\displaystyle \frac{\pi }{4}})$$ Apply De Moivre to each power: $$(1+i\sqrt{3}{)}^5=2^5\mathrm{cis}{\displaystyle \frac{5\pi }{3}}=32\mathrm{cis}{\displaystyle \frac{5\pi }{3}}$$ $$(1-i{)}^3=(\sqrt{2}{)}^3\mathrm{cis}\text{\hspace{0.17em}⁣}(-{\displaystyle \frac{3\pi }{4}})=2\sqrt{2}\mathrm{cis}\text{\hspace{0.17em}⁣}(-{\displaystyle \frac{3\pi }{4}})$$ Divide moduli and subtract arguments: $$\mathrm{\mid }z\mathrm{\mid }={\displaystyle \frac{32}{2\sqrt{2}}}=8\sqrt{2}$$ $$\arg z={\displaystyle \frac{5\pi }{3}}-(-{\displaystyle \frac{3\pi }{4}})={\displaystyle \frac{29\pi }{12}}\equiv {\displaystyle \frac{5\pi }{12}}(\text{mod }2\pi )$$ $$z=8\sqrt{2}\mathrm{cis}{\displaystyle \frac{5\pi }{12}}$$