Daily math problems for July 29, 2026 — Analytic Geometry, Calculus, Powers, Radicals, Logarithms & more

$r^2=49\Rightarrow r=7$.

$f$ is continuous on $[-1,1]$ and $f(-1)=f(1)=1$, but $f$ is not differentiable at $0$. Rolle's hypotheses fail.

${\displaystyle \frac{1}{\sqrt{3}-1}}\cdot {\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}+1}}={\displaystyle \frac{\sqrt{3}+1}{3-1}}={\displaystyle \frac{\sqrt{3}+1}{2}}$.

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

Invertibility ⇔ non-zero determinant.

$\bullet$ 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:

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

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

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.

$7^2=49$, $7^4={49}^2=2401\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Since $2024=4\cdot 506$, we have $7^{2024}=(7^4{)}^{506}\equiv 1^{506}=1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Last two digits: $01$.

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.