Daily math problems for August 2, 2026 — Trigonometry, Financial Mathematics, Matrices & more

$\sin \text{\hspace{0.17em}⁣}(\pi +{\displaystyle \frac{\pi }{6}})=-\sin {\displaystyle \frac{\pi }{6}}=-{\displaystyle \frac{1}{2}}$.

${\displaystyle \frac{100-80}{80}}\cdot 100\mathrm{\%}={\displaystyle \frac{20}{80}}\cdot 100\mathrm{\%}=25\mathrm{\%}$.

Add componentwise: $\left(\begin{array}{cc}1& 1\\ 8& 6\end{array}\right)$.

$\bullet$ Rewrite as an exponential:

$$x^{1\mathrm{/}\sqrt{x}}=e^{\ln x\mathrm{/}\sqrt{x}}$$

$\bullet$ Apply L'Hôpital to the exponent:

$$\underset{x\to \mathrm{\infty }}{\lim }\frac{\ln x}{\sqrt{x}}=\underset{x\to \mathrm{\infty }}{\lim }\frac{2}{\sqrt{x}}=0$$

$\bullet$ Exponentiate:

$$L=e^0=1$$

$x_1+x_2=-{\displaystyle \frac{-7}{2}}={\displaystyle \frac{7}{2}}$. (And the product is ${\displaystyle \frac{c}{a}}={\displaystyle \frac{3}{2}}$.)

$\bullet$ Integration by parts with $u=x$, $dv=e^{x}dx$, so $du=dx$, $v=e^x$:

$$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx$$

$\bullet$ Evaluate the remaining integral:

$$x{e}^x-e^x+C$$

$\bullet$ Factor:

$$(x-1)e^x+C$$

$\gcd (2,6)=2\mathrm{\ne }1$, so $\widehat{2}$ has no multiplicative inverse in ${\mathbb{Z}}_6$. ($\widehat{1}$ and $\widehat{5}$ are coprime to $6$ and invertible.)

$u=x$, $du=dx$, $v=\sin x$. $\int x\cos xdx=x\sin x-\int \sin xdx=x\sin x+\cos x+C$.

$\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{0.12}{12}})}^{12}-1={1.01}^{12}-1\approx 1.1268-1=0.1268\approx 12.68\mathrm{\%}$.

$x^2>9\iff \mathrm{\mid }x\mathrm{\mid }>3\iff x<-3$ or $x>3$. Solution: $(-\mathrm{\infty },-3)\cup (3,+\mathrm{\infty })$.