Daily math problems for October 18, 2026 — Euclidean Geometry, Algebra, Infinite Series & more

${\displaystyle \frac{90}{360}}={\displaystyle \frac{1}{4}}$.

$-2x+5\le 11$

$-2x\le 6$

Divide both sides by $-2$ (flip the sign): $x\ge -3$.

$$\underset{n\to \mathrm{\infty }}{\lim }\mid {\displaystyle \frac{a_{n+1}}{a_n}}\mid =\underset{n\to \mathrm{\infty }}{\lim }{\displaystyle \frac{\mathrm{\mid }x\mathrm{\mid }}{n+1}}=0<1\text{for every }x\mathrm{.}$$

The series converges for all $x\in \mathbb{R}$, so $R=\mathrm{\infty }$ (this is the Maclaurin series of $e^x$).

Apply the power law to the first term. $$2{\log }_{5}10={\log }_5{10}^2={\log }_{5}100$$

Now use the subtraction (quotient) law. $${\log }_{5}100-{\log }_{5}4={\log }_5\frac{100}{4}={\log }_{5}25$$

Finally write $25$ as a power of $5$. $${\log }_{5}25={\log }_5{5}^2=2$$

$a^2=16$, so $a=4$. The vertices are $(\pm a,0)=(\pm 4,0)$.

$\bullet$ Compute the determinant directly:

$${\cos }^{2}x-{\sin }^{2}x=\cos 2x$$

$\bullet$ From the hypothesis:

$$\cos 2x=\frac{3}{5}$$

$\bullet$ Apply the Pythagorean identity:

$${\sin }^{2}2x=1-\frac{9}{25}=\frac{16}{25}$$

$\bullet$ Taking the negative root:

$$\sin 2x=-\frac{4}{5}$$

$I_n=x^n{e}^x{\mid }_0^1-n{\int }_0^1{x}^{n-1}{e}^{x}dx=e-n{I}_{n-1}$.

$3\cdot (-5)=(m-1)(-(m+1))$, i.e. $-15=-(m-1)(m+1)=1-m^2$, so $m^2=16$ and $m\in \{-4,4\}$.

$P=\left(\genfrac{}{}{0px}{}{4}{2}\right)\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{1}{3}}\right)}^2\text{\hspace{0.17em}⁣}{\left({\displaystyle \frac{2}{3}}\right)}^2=6\cdot {\displaystyle \frac{1}{9}}\cdot {\displaystyle \frac{4}{9}}={\displaystyle \frac{24}{81}}={\displaystyle \frac{8}{27}}$.

$V=\pi {\int }_0^1(x^2-x^4)dx=\pi \text{\hspace{0.17em}⁣}{[{\displaystyle \frac{x^3}{3}}-{\displaystyle \frac{x^5}{5}}]}_0^1=\pi \text{\hspace{0.17em}⁣}({\displaystyle \frac{1}{3}}-{\displaystyle \frac{1}{5}})={\displaystyle \frac{2\pi }{15}}$.