Daily math problems for October 25, 2026 — Euclidean Geometry, Infinite Series, Descriptive Statistics & Sampling & more

$30-4=26{\text{ cm}}^2$.

For (c): ${\lim }_{n\to \mathrm{\infty }}{\displaystyle \frac{n}{2n+1}}={\displaystyle \frac{1}{2}}\mathrm{\ne }0$, so the series diverges by the $n$-th term test.

For (a), (b), (d) the term limits are all $0$, so the test is inconclusive (and (d), the harmonic series, actually diverges by other means).

Boys $=20-12=8$.

Fraction of boys $={\displaystyle \frac{8}{20}}={\displaystyle \frac{2}{5}}$.

${\displaystyle \frac{dx}{dt}}=2t$ and ${\displaystyle \frac{dy}{dt}}=3t^2$, so ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{3t^2}{2t}}={\displaystyle \frac{3t}{2}}$.

At $t=2$: ${\displaystyle \frac{dy}{dx}}={\displaystyle \frac{3(2)}{2}}=3$.

Factor the quadratic: $$x^2-5x+6=(x-2)(x-3)$$ The roots are at $x=2$ and $x=3$. The leading coefficient is positive, so the parabola opens upward and the expression is negative between the roots: $$2<x<3$$

$I_3$ has $1$s on the main diagonal and $0$s elsewhere.

$\vec{v}=2\vec{u}$, so the two vectors are collinear (parallel).

$\bullet$ The astroid meets the axes at $(\pm a,0)$ and $(0,\pm a)$, so it fits in a square of side $2a$.

$\bullet$ Given side $=4$:

$$2a=4\Rightarrow a=2$$

$\bullet$ Apply the astroid area formula:

$$\mathcal{A}=\frac{3\pi a^2}{8}=\frac{3\pi \cdot 4}{8}=\frac{3\pi }{2}$$

$(x-3)(x+3)\le 0\iff x\in [-3,3]$. The integers in this interval are $\{-3,-2,-1,0,1,2,3\}$ — seven values.

The cancellation law states: $a\ast b=a\ast c\Rightarrow b=c$. (Even without inverses, this property may or may not hold; in a group it always does.)