Daily math problems for October 1, 2026 — Matrices, Trigonometry, Calculus & more

$\mid \begin{array}{cc}3& 1\\ 2& 2\end{array}\mid =3\cdot 2-1\cdot 2=6-2=4$.

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

${\displaystyle \frac{(x-2)(x+2)}{x-2}}=x+2\to 4$ as $x\to 2$.

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $=\frac{1}{2}\cdot 3\cdot 4=6$.

$\bullet$ Find vertex $A=AB\cap AC$ by solving $x-2y=-3$ and $2x-y=3$:

$$A=(3,3)$$

$\bullet$ Find vertex $B=AB\cap BC$ by solving $x-2y=-3$ and $3x+2y=-1$:

$$B=(-1,1)$$

$\bullet$ Find vertex $C=AC\cap BC$ by solving $2x-y=3$ and $3x+2y=-1$:

$$C=(\frac{5}{7},-\frac{11}{7})$$

$\bullet$ Apply the shoelace-style area formula:

$$\mathcal{A}=\frac{1}{2}\mid x_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\mid$$

$\bullet$ Plugging the coordinates gives $\frac{1}{2}\cdot \frac{96}{7}$:

$$\mathcal{A}=\frac{48}{7}$$

$\mathrm{\mid }A\mathrm{\mid }=4$, so $A$ has $2^4=16$ subsets (including $\mathrm{\varnothing }$ and $A$ itself).

The graph is a semicircle; rotating about the $x$-axis sweeps out a sphere of radius $R$. (Volume $={\displaystyle \frac{4}{3}}\pi R^3$.)

${\int }_{-\mathrm{\infty }}^{\mathrm{\infty }}{e}^{-x^2}dx=\sqrt{\pi }$. By even symmetry, ${\int }_0^{\mathrm{\infty }}{e}^{-x^2}dx={\displaystyle \frac{\sqrt{\pi }}{2}}$.

$f^{\mathrm{\prime }}(x)={\displaystyle \frac{(1\mathrm{/}x)\cdot x-\ln x\cdot 1}{x^2}}={\displaystyle \frac{1-\ln x}{x^2}}$. At $x=1$: ${\displaystyle \frac{1-0}{1}}=1$.

$A^2=\left(\begin{array}{cc}0\cdot 0+1\cdot 0& 0\cdot 1+1\cdot 0\\ 0\cdot 0+0\cdot 0& 0\cdot 1+0\cdot 0\end{array}\right)=\left(\begin{array}{cc}0& 0\\ 0& 0\end{array}\right)$. ($A$ is nilpotent.)