Daily math problems for September 14, 2026 — Descriptive Statistics & Sampling, Algebra, Calculus & more

Sorted: $\{1,3,5,7,9\}$. The middle value (3rd of 5) is $5$.

A homomorphism preserves the operation: $\phi (a\cdot b)=\phi (a)\ast \phi (b)$.

$\bullet$ Simplify the limit by factoring numerator and denominator:

$$\underset{k\to x}{\lim }\frac{(k-4)(k+4)}{(k-4)(k-1)}=\frac{x+4}{x-1}$$

$\bullet$ The determinant has a row of zeros (with ${\log }_{3}1=0$), so it contributes $0$:

$$f(x)=\frac{x+4}{x-1}$$

$\bullet$ Evaluate at $x=4$ and $x=0$:

$$f(4)=\frac{8}{3},f(0)=-4$$

$\bullet$ Sum them:

$$f(4)+f(0)=\frac{8}{3}-4=-\frac{4}{3}$$

${\sin }^2\theta =1-{\displaystyle \frac{4}{9}}={\displaystyle \frac{5}{9}}$, so $\sin \theta ={\displaystyle \frac{\sqrt{5}}{3}}$. Hence $\tan \theta ={\displaystyle \frac{\sin \theta }{\cos \theta }}={\displaystyle \frac{\sqrt{5}\mathrm{/}3}{2\mathrm{/}3}}={\displaystyle \frac{\sqrt{5}}{2}}$.

Middle coefficient: $\left(\genfrac{}{}{0px}{}{8}{4}\right)=70$.

For any $x\in [-1,1]$, $\arcsin x+\arccos x={\displaystyle \frac{\pi }{2}}$ (a standard identity, since $\arccos x={\displaystyle \frac{\pi }{2}}-\arcsin x$).

The factor $(X-2)$ appears with exponent $3$, so the multiplicity is $3$.

$P=\left(\genfrac{}{}{0px}{}{4}{2}\right)\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{3}}\right)}^2\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{2}{3}}\right)}^2=6\cdot {\displaystyle \frac{1}{9}}\cdot {\displaystyle \frac{4}{9}}={\displaystyle \frac{24}{81}}={\displaystyle \frac{8}{27}}$.

Cauchy's equation with continuity gives $f(x)=cx$ for some constant $c$. Since $f(1)=3$, $c=3$, so $f(5)=15$.

$e^{a+b}=e^a{e}^b$ ✓ (homomorphism). $e^x$ is strictly increasing (injective) and surjective onto $(0,\mathrm{\infty })$. Hence an isomorphism. (Inverse: $\ln$.)