Daily math problems for August 9, 2026 — Trigonometry, Calculus, Matrices & more

$\sin {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{\sqrt{3}}{2}}$ — a notable value from the unit circle.

$\bullet$ 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:

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

$f(0)=-1<0$ and $f(1)=1>0$. Since $f$ is continuous, by the IVT there exists $c\in (0,1)$ with $f(c)=0$.

$\bullet$ Expand along the first row:

$$\det A=m(m^2-1)-1\cdot (m-0)+0$$

$\bullet$ Simplify:

$$\det A=m^3-2m=m(m^2-2)$$

$\bullet$ Set to zero:

$$m(m^2-2)=0$$

$\bullet$ Solutions:

$$m\in \{-\sqrt{2},0,\sqrt{2}\}$$

Using the closed forms $\cos {36}^{\circ }=\frac{1+\sqrt{5}}{4}$ and $\cos {72}^{\circ }=\frac{\sqrt{5}-1}{4}$, the difference is $\frac{(1+\sqrt{5})-(\sqrt{5}-1)}{4}=\frac{2}{4}=\frac{1}{2}$.

$2\sin x\cos x-\sin x=0\Rightarrow \sin x(2\cos x-1)=0$. $\sin x=0$: $x\in \{0,\pi \}$. $\cos x=1\mathrm{/}2$: $x\in \{\pi \mathrm{/}3,5\pi \mathrm{/}3\}$. Total: $4$.

$A=2000\cdot {1.05}^2=2000\cdot 1.1025=\mathrm{\$}\mathrm{2,205}$.

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.

$S_n={\displaystyle \frac{n(3n+7)}{2}}$. Compute: $S_{17}={\displaystyle \frac{17\cdot 58}{2}}=493<500$ and $S_{18}={\displaystyle \frac{18\cdot 61}{2}}=549\ge 500$. Hence the smallest $n$ is $18$.