Daily math problems for September 17, 2026 — Analytic Geometry, Complex Numbers, Functions & more

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

$\mathrm{\mid }z\mathrm{\mid }=\sqrt{3^2+4^2}=\sqrt{25}=5$.

$g(2)=5$, then $f(5)=2\cdot 5-1=9$. Hence $(f\circ g)(2)=9$.

Matrices form a ring (with $0$ matrix and identity matrix), but multiplication is non-commutative for $n\ge 2$. It's a non-commutative ring with unity.

Midpoint: $(2,0)$. Segment $AB$ is horizontal, so the bisector is vertical through $(2,0)$: $x=2$.

$d=\sqrt{(-3{)}^2+4^2}=\sqrt{25}=5$.

$\bullet$ Integration by parts with $u=x$, $dv=e^{x}dx$, so $du=dx$, $v=e^x$:

$$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx$$

$\bullet$ Evaluate the remaining integral:

$$x{e}^x-e^x+C$$

$\bullet$ Factor:

$$(x-1)e^x+C$$

$2\sin x\cos x-\sqrt{3}\cos x=0\Rightarrow \cos x(2\sin x-\sqrt{3})=0$.

$\bullet$ $\cos x=0\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{2}},{\displaystyle \frac{3\pi }{2}}\}$

$\bullet$ $\sin x={\displaystyle \frac{\sqrt{3}}{2}}\Rightarrow x\in \text{\hspace{0.17em}⁣}\{{\displaystyle \frac{\pi }{3}},{\displaystyle \frac{2\pi }{3}}\}$

Total: $4$ solutions on $[0,2\pi )$.

$\bullet$ Square the first matrix using $i^2=-1$:

$${\left(\begin{array}{cc}3& 2\\ 0& -1\end{array}\right)}^2=\left(\begin{array}{cc}9& 4\\ 0& 1\end{array}\right)$$

$\bullet$ Evaluate the entries of the second matrix: $\lim x^2\mathrm{/}x!=0$, ${\log }_{3}81=4$, $i^{2022}=-1$, $\tan 225\mathrm{°}=1$:

$$\left(\begin{array}{cc}0& 4\\ -1& 1\end{array}\right)$$

$\bullet$ Add and take the determinant:

$$\left(\begin{array}{cc}9& 8\\ -1& 2\end{array}\right),\det =18-(-8)=26$$

$(1+i{)}^2=2i$, so $(1+i{)}^{10}=(2i{)}^5=32\cdot i^5=32i$.