Daily math problems for September 3, 2026 — Complex Numbers, Calculus, Trigonometry & more

$\mathrm{\mid }3+4i\mathrm{\mid }=\sqrt{3^2+4^2}=\sqrt{9+16}=\sqrt{25}=5$.

${\lim }_{x\to \mathrm{\infty }}{x}^2=\mathrm{\infty }$, so there is no horizontal asymptote.

$\sin {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{1}{2}}$, and ${\displaystyle \frac{\pi }{6}}$ is in the principal range, so $\arcsin {\displaystyle \frac{1}{2}}={\displaystyle \frac{\pi }{6}}$.

$\mathrm{\mid }x-2\mathrm{\mid }<3\iff -3<x-2<3\iff -1<x<5$. The solution set is $(-1,5)$.

$\bullet$ Write $2^{1\mathrm{/}n}=e^{\ln 2\mathrm{/}n}$ and use $e^u-1\sim u$ for $u\to 0$:

$$2^{1\mathrm{/}n}-1\sim \frac{\ln 2}{n}$$

$\bullet$ Multiply by $n$:

$$n\cdot \frac{\ln 2}{n}=\ln 2$$

The three distinct numbers are $225,252,522$. Sum $=225+252+522=999$.

$f^{\mathrm{\prime }}(x)=(x{)}^{\mathrm{\prime }}{e}^x+x(e^x{)}^{\mathrm{\prime }}=e^x+x{e}^x=e^x(x+1)$.

Subtract: $5y=15\Rightarrow y=3$. Then $3x=y=3\Rightarrow x=1$. So $A=1+9=10$.

$P(-1)>0$ and $P(1)<0$: $2+m>0$ and $-2+m<0$, giving $-2<m<2$.

The positive divisors of $12$ are $\{1,2,3,4,6,12\}$, so $\mathrm{\mid }A\mathrm{\mid }=6$.