Daily math problems for November 7, 2026 — Euclidean Geometry, Calculus, Trigonometry & more

Similar shapes are the same shape with all lengths in the same ratio (a scale factor).

A semicircle is half a circle, ${\displaystyle \frac{1}{2}}$.

$10\times 6-2\times 2=60-4=56{\text{ m}}^2$.

${\displaystyle \frac{1}{3}}{\int }_0^3{x}^{2}dx={\displaystyle \frac{1}{3}}\cdot 9=3$.

${\displaystyle \frac{7\pi }{4}}\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{3\pi }{2}},2\pi )$, which is the fourth quadrant.

$\overrightarrow{AB}=(4,0)$ and $\overrightarrow{DC}=(4,0)$ — same vector, so $ABCD$ is a parallelogram. Base $=4$, height $=3$. Area $=12$.

With $a=2x$ and $b=3$: $T_{k+1}=\left(\genfrac{}{}{0px}{}{n}{k}\right)(2x{)}^{n-k}{3}^k$.

$n\ln \text{\hspace{0.17em}⁣}(1+{\displaystyle \frac{1}{n^2}})\sim n\cdot {\displaystyle \frac{1}{n^2}}={\displaystyle \frac{1}{n}}\to 0$, so the limit equals $e^0=1$.

$7^2=49$, $7^4={49}^2=2401\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Since $2024=4\cdot 506$, we have $7^{2024}=(7^4{)}^{506}\equiv 1^{506}=1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. Last two digits: $01$.

Base along the $x$-axis has length $3$; the third vertex sits at height $4$. Area $={\displaystyle \frac{1}{2}}\cdot 3\cdot 4=6$.