Daily math problems for October 27, 2026 — Euclidean Geometry, Descriptive Statistics & Sampling, Parametric & Polar Calculus & more

$A=\pi r^2$.

A random selection makes every member equally likely, so the probability is the favorable count over the total count.

$$P(\text{plays})=\frac{20}{50}=\frac{2}{5}$$

$r^{\mathrm{\prime }}(t)=⟨2t,3t^2⟩$, so $r^{\mathrm{\prime }}(1)=⟨2,3⟩$.

Speed $=\sqrt{2^2+3^2}=\sqrt{4+9}=\sqrt{13}$.

Growing by $10\mathrm{\%}$ each year multiplies the balance by $1.10$, so after $2$ years: $$2000\times (1.10{)}^2$$ $$2000\times 1.21=2420$$ The balance after $2$ years is $\mathrm{\$}\mathrm{2,420}$.

(Adding a flat $10\mathrm{\%}$ of the original $\mathrm{\$}200$ each year is *linear* growth and gives only $\mathrm{\$}\mathrm{2,400}$; applying the increase just once gives $\mathrm{\$}\mathrm{2,200}$.)

$x={\displaystyle \frac{\det (A_1)}{\det (A)}}={\displaystyle \frac{-10}{-5}}=2$.

$\phi$ is surjective (every $\widehat{k}$ has $k$ as preimage) and a homomorphism. But $\phi (0)=\phi (n)=\widehat{0}$, so it's not injective.

For $x\in (0,1)$, multiplying by $x$ shrinks the value: $x^3=x\cdot x^2<x^2$ since $0<x<1$.

$\bullet$ Compute the determinant of the coefficient matrix:

$$\det \left(\begin{array}{ccc}1& 1& 1\\ 1& m& 1\\ 1& 1& m\end{array}\right)=1(m^2-1)-1(m-1)+1(1-m)$$

$\bullet$ Simplify:

$$\det =m^2-2m+1=(m-1{)}^2$$

$\bullet$ Determinant vanishes only at $m=1$. At $m=1$ all three rows of the coefficient matrix become $(1,1,1)$ but the right-hand side is $(1,2,3{)}^T$:

$$x+y+z=1,2,3$$

$\bullet$ Three identical equations with different constants are inconsistent, so the system has no solution iff $m=1$.

Associative ✓, identity $1$ ✓. But $2$ has no integer multiplicative inverse. Hence a monoid but not a group.

$P^{\mathrm{\prime }}(x)>0$ everywhere, so $P$ is strictly increasing on $\mathbb{R}$. Combined with ${\lim }_{\pm \mathrm{\infty }}P=\pm \mathrm{\infty }$, $P$ has exactly one real root by IVT.