Daily math problems for September 29, 2026 — Calculus, Trigonometry, Algebra & more

Polynomial functions are continuous everywhere on $\mathbb{R}$.

$v(t)=3t^2-12t+9$. At $t=2$: $v=12-24+9=-3$ m/s (the particle is moving backward).

Area $={\displaystyle \frac{1}{2}}\cdot 3\cdot 4=6$.

${\displaystyle \frac{p}{q}}$ runs over $\text{\hspace{0.17em}⁣}\{\pm 1,\pm {\displaystyle \frac{1}{2}}\}$.

A standard antiderivative: $\int {\displaystyle \frac{dx}{1+x^2}}=\arctan x+C$.

$V=\pi {\int }_0^1{x}^{4}dx=\pi \cdot {\displaystyle \frac{1}{5}}={\displaystyle \frac{\pi }{5}}$.

$A^{-1}=\left(\begin{array}{cc}1\mathrm{/}2& 0\\ 0& 1\mathrm{/}3\end{array}\right)$. So $X=\left(\begin{array}{c}4\mathrm{/}2\\ 6\mathrm{/}3\end{array}\right)=\left(\begin{array}{c}2\\ 2\end{array}\right)$.

$\bullet$ Evaluate each of the nine limits to fill the matrix:

$$\left(\begin{array}{ccc}\sqrt{3}\mathrm{/}2& 2& 1\\ 3& 4& 4\\ 0& 3& 1\end{array}\right)$$

$\bullet$ Expand along row 1. The cofactor of the $(1,1)$ entry is $4\cdot 1-4\cdot 3=-8$:

$$\frac{\sqrt{3}}{2}\cdot (-8)=-4\sqrt{3}$$

$\bullet$ Cofactors of the $(1,2)$ and $(1,3)$ entries are $3\cdot 1-4\cdot 0=3$ and $3\cdot 3-4\cdot 0=9$:

$$-2\cdot 3+1\cdot 9=3$$

$\bullet$ Add the three contributions:

$$\det =3-4\sqrt{3}$$

Invertibility ⇔ non-zero determinant.

${\displaystyle \frac{2x^2+3x+1}{x-1}}=2x+5+{\displaystyle \frac{6}{x-1}}$. As $x\to \pm \mathrm{\infty }$, the remainder vanishes, leaving the slant asymptote $y=2x+5$.