Daily math problems for November 6, 2026 — Fundamental Algebra, Euclidean Geometry, Financial Mathematics & more

$2(x+3)+4=2x+6+4=2x+10$.

$2(3x-5)=6x-10$.

Angles around a point add up to ${360}^{\circ }$.

${\displaystyle \frac{72}{7}}\approx 10.3$ years.

Sum $=40$, $n=8$, so $\bar{x}=5$. Squared deviations: $9,1,1,1,0,0,4,16$ — sum $=32$. Variance $={\displaystyle \frac{32}{8}}=4$.

${\displaystyle \frac{1+i}{1-i}}\cdot {\displaystyle \frac{1+i}{1+i}}={\displaystyle \frac{(1+i{)}^2}{1-i^2}}={\displaystyle \frac{2i}{2}}=i$.

Let $E$ be the expected number of flips to reach two heads in a row from the start. Condition on outcomes:

- With probability $\frac{1}{2}$, the first flip is T. We've used one flip and are back at the start: contributes $\frac{1}{2}(1+E)$. - With probability $\frac{1}{4}$, the first two flips are HT. We've used two flips and are back at the start: contributes $\frac{1}{4}(2+E)$. - With probability $\frac{1}{4}$, the first two flips are HH — done in two flips: contributes $\frac{1}{4}\cdot 2$.

So:

$$E=\frac{1}{2}(1+E)+\frac{1}{4}(2+E)+\frac{1}{4}\cdot 2$$

Expanding: $E=\frac{1}{2}+\frac{1}{2}E+\frac{1}{2}+\frac{1}{4}E+\frac{1}{2}=\frac{3}{2}+\frac{3}{4}E$. Therefore $\frac{1}{4}E=\frac{3}{2}$, i.e. $E=6$.

Sanity check: "flip until first head" has expected value $2$; requiring a *consecutive* second head triples it.

Substitute $x=\tan \theta$. Then $dx={\sec }^2\theta d\theta$ and $1+x^2={\sec }^2\theta$, so the ${\sec }^2\theta$ on top and bottom cancel cleanly. The bounds $0\to 1$ map to $\theta :0\to \pi \mathrm{/}4$:

$$I={\int }_0^{\pi \mathrm{/}4}\ln (1+\tan \theta )d\theta \mathrm{.}$$

Reflect by $\varphi =\pi \mathrm{/}4-\theta$. The angle-difference formula gives $\tan (\pi \mathrm{/}4-\varphi )={\displaystyle \frac{1-\tan \varphi }{1+\tan \varphi }}$, so

$$1+\tan \theta =1+\frac{1-\tan \varphi }{1+\tan \varphi }=\frac{2}{1+\tan \varphi }\mathrm{.}$$

Taking logs: $\ln (1+\tan \theta )=\ln 2-\ln (1+\tan \varphi )$.

Pair the integral with its reflected twin. Renaming the dummy variable $\varphi \to \theta$ (the bounds are unchanged because the substitution is a reflection of $[0,\pi \mathrm{/}4]$ onto itself):

$$I={\int }_0^{\pi \mathrm{/}4}[\ln 2-\ln (1+\tan \theta )]d\theta =\frac{\pi }{4}\ln 2-I\mathrm{.}$$

Solve. $2I={\displaystyle \frac{\pi }{4}}\ln 2$, so

$$\boxed{{\displaystyle I=\frac{\pi }{8}\ln 2\mathrm{.}}}$$

The reflection trick — pairing $f(\theta )$ with $f(\pi \mathrm{/}4-\theta )$ — works whenever the integrand simplifies under that reflection. Worth keeping in your toolbox alongside the half-angle and Weierstrass substitutions.

Euler's classical result: ${\displaystyle \sum _{n=1}^{\mathrm{\infty }}{\displaystyle \frac{1}{n^2}}={\displaystyle \frac{{\pi }^2}{6}}}$.

Left branch ($x<0$): $f(x)=x+1\in (-\mathrm{\infty },1)$, strictly increasing. Right branch ($x\ge 0$): $f(x)=2x+1\in [1,+\mathrm{\infty })$, strictly increasing.

The two images $(-\mathrm{\infty },1)$ and $[1,+\mathrm{\infty })$ are disjoint and together cover all of $\mathbb{R}$, so $f$ is surjective. Each branch is strictly increasing, and the ranges don't overlap, so no two distinct $x$-values give the same $f(x)$ — $f$ is injective. Hence $f$ is bijective.

(Note: $f(-1)=0$ and $f(0)=1$, so distractor A is factually false.)