Daily math problems for August 11, 2026 — Linear Function, Complex Numbers, Functions & more

$3x=14-5=9\Rightarrow x=3$.

$(1+i{)}^8=\text{\hspace{0.17em}⁣}{\left[\sqrt{2}\right]}^8\cdot \text{\hspace{0.17em}⁣}(\cos 2\pi +i\sin 2\pi )=16\cdot (1+0)=16$.

$f$ is invertible iff its slope $a-2\mathrm{\ne }0$. So $f$ fails to be invertible exactly when $a=2$ (it then becomes the constant $5$, neither injective nor surjective).

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

$\bullet$ 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:

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

$4p=8\Rightarrow p=2$. Focus: $(2,0)$.

The order of $g$ is the smallest positive integer $n$ such that $g^n=e$ (or infinity if no such $n$ exists).

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

$\bullet$ 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 $x=7$:

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

${10}^3-10=990$, 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 $6$.