Daily math problems for July 11, 2026 — Calculus, Analytic Geometry, Functions & more

$V=\pi {\int }_0^3{x}^{2}dx=\pi \cdot {\displaystyle \frac{27}{3}}=9\pi$. (Equivalently, $V_{\text{cone}}={\displaystyle \frac{1}{3}}\pi r^{2}h={\displaystyle \frac{1}{3}}\pi \cdot 9\cdot 3=9\pi$.)

$a(t)=x^{\mathrm{\prime }\mathrm{\prime }}(t)=6t-12$. At $t=1$: $a=-6$ m/s².

Area $={\displaystyle \frac{1}{2}}\mathrm{\mid }1(6-10)+3(10-2)+5(2-6)\mathrm{\mid }={\displaystyle \frac{1}{2}}\mathrm{\mid }-4+24-20\mathrm{\mid }=0$. Hence collinear.

$f$ is strictly increasing on $\mathbb{R}$, hence injective. For every $y\in \mathbb{R}$, $x=\sqrt[3]{y}$ satisfies $f(x)=y$, hence surjective. So $f$ is bijective, with $f^{-1}(y)=\sqrt[3]{y}$.

$x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1{)}^2(x+2)$. So ${\displaystyle \frac{(x-1{)}^2(x+2)}{x-1}}=(x-1)(x+2)\to 0$ as $x\to 1$.

$\bullet$ Simplify each logarithm:

$${\log }_{\sqrt{2}}\text{\hspace{0.17em}⁣}\left(\frac{1}{x-3}\right)=-2{\log }_2(x-3)$$

$${\log }_{100}({10}^x)=\frac{x}{2}$$

$\bullet$ Combine the numerator:

$${\log }_2(x+25)-2{\log }_2(x-3)={\log }_2\frac{x+25}{(x-3{)}^2}$$

$\bullet$ The equation becomes:

$$\frac{{\log }_2\frac{x+25}{(x-3{)}^2}}{x\mathrm{/}2}=\frac{2}{7}$$

$${\log }_2\frac{x+25}{(x-3{)}^2}=\frac{x}{7}$$

$\bullet$ Test $x=7$:

$${\log }_2\frac{32}{16}={\log }_{2}2=1=\frac{7}{7}$$

$\bullet$ The astroid meets the axes at $(\pm a,0)$ and $(0,\pm a)$, so it fits in a square of side $2a$.

$\bullet$ Given side $=4$:

$$2a=4\Rightarrow a=2$$

$\bullet$ Apply the astroid area formula:

$$\mathcal{A}=\frac{3\pi a^2}{8}=\frac{3\pi \cdot 4}{8}=\frac{3\pi }{2}$$

Using $1-\cos x=2{\sin }^2(x\mathrm{/}2)$: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\underset{x\to 0}{\lim }{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2={\displaystyle \frac{1}{2}}\cdot 1={\displaystyle \frac{1}{2}}}$.

$\vec{u}\cdot \vec{v}=a+(-2)=a-2=0\Rightarrow a=2$.

${10}^3-10=990$, and $990\mathrm{/}6=165$. The induction proof factors $n^3-n=(n-1)n(n+1)$, which is the product of three consecutive integers and is therefore divisible by both $2$ and $3$, hence by $6$.