Daily math problems for October 16, 2026 — Euclidean Geometry, Infinite Series, Algebra & more

$C=2\pi r$ (equivalently $\pi d$).

Here $a=3$ and $r=\frac{1}{2}$, so $\mathrm{\mid }r\mathrm{\mid }<1$.

$$\sum _{n=0}^{\mathrm{\infty }}3{\left(\frac{1}{2}\right)}^n={\displaystyle \frac{a}{1-r}}={\displaystyle \frac{3}{1-\frac{1}{2}}}={\displaystyle \frac{3}{\frac{1}{2}}}=6.$$

Here $b=-7$, $c=12$.

By Vieta, sum of roots $=-b=-(-7)=7$.

(Check by factoring: $(x-3)(x-4)=0$, so roots are $3$ and $4$, summing to $7$.)

The general term in the expansion is $$\left(\genfrac{}{}{0px}{}{5}{r}\right)(2{)}^{5-r}(3x{)}^r\mathrm{.}$$

The term in $x^3$ comes from $r=3$: $$\left(\genfrac{}{}{0px}{}{5}{3}\right)(2{)}^{5-3}(3{)}^3{x}^3\mathrm{.}$$

Evaluate each factor: $$\left(\genfrac{}{}{0px}{}{5}{3}\right)=10,2^2=4,3^3=27.$$

Multiply: $$10\times 4\times 27=1080.$$

So the coefficient of $x^3$ is $1080$.

For all $n\ge 1$, $0<{\displaystyle \frac{1}{n^2+1}}\le {\displaystyle \frac{1}{n^2}}$.

$\sum {\displaystyle \frac{1}{n^2}}$ is a convergent $p$-series ($p=2>1$).

By direct comparison, $\sum {\displaystyle \frac{1}{n^2+1}}$ also converges.

$f(3)=3^{-2}={\displaystyle \frac{1}{3^2}}={\displaystyle \frac{1}{9}}$.

This is the standard definition of $e$: $\text{\hspace{0.17em}⁣}{(1+{\displaystyle \frac{1}{n}})}^n\to e$.

The three edges from origin are $(3,0,0),(0,4,0),(0,0,5)$. Determinant: $3\cdot 4\cdot 5=60$. Volume: $60\mathrm{/}6=10$.

$f^{\mathrm{\prime }}(x)=3x^2-3$ vanishes at $x=\pm 1$. $f(-1)=-1+3+1=3>0$ and $f(1)=1-3+1=-1<0$. The local max is positive and the local min is negative, so the cubic crosses the $x$-axis three times.

$\bullet$ 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 $x=7$:

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