Daily math problems for October 21, 2026 — Fundamental Algebra, Geometry, Parametric & Polar Calculus & more

$y=3x\Rightarrow x={\displaystyle \frac{y}{3}}$.

Here $a^2=25$ and $b^2=9$, so $a=5$ and $b=3$.

The semi-major axis is $\max (a,b)=5$.

${\displaystyle \frac{dx}{dt}}=3$, ${\displaystyle \frac{dy}{dt}}=4$.

Integrand: $\sqrt{3^2+4^2}=\sqrt{25}=5$ (constant).

$$L={\int }_0^{2}5dt=10.$$

Set the divisor to zero: $$2x+1=0⟹x=-\frac{1}{2}\mathrm{.}$$

By the remainder theorem the remainder equals $f\text{\hspace{0.17em}⁣}(-\frac{1}{2})$: $$f\text{\hspace{0.17em}⁣}(-\frac{1}{2})=4(-\frac{1}{8})+2\left(\frac{1}{4}\right)+6(-\frac{1}{2})+5.$$

Evaluate term by term: $$-\frac{1}{2}+\frac{1}{2}-3+5=2.$$

So the remainder is $2$.

Both lines have slope $2$ but different $y$-intercepts ($3$ and $-1$).

Same slope, different intercept ⇒ parallel and distinct (never meet).

$\frac{d}{dx}\sin (3x)=\cos (3x)\cdot 3=3\cos (3x)$.

Avem $125=5^3$, deci $5^{x+1}=5^3\Rightarrow x+1=3\Rightarrow x=2$.

Matrices form a ring (with $0$ matrix and identity matrix), but multiplication is non-commutative for $n\ge 2$. It's a non-commutative ring with unity.

$V=\pi {\int }_0^2(4x^2+4x+1)dx=\pi \text{\hspace{0.17em}⁣}{[{\displaystyle \frac{4x^3}{3}}+2x^2+x]}_0^2=\pi \text{\hspace{0.17em}⁣}({\displaystyle \frac{32}{3}}+8+2)=\pi \cdot {\displaystyle \frac{32+30}{3}}={\displaystyle \frac{62\pi }{3}}$.

The tetrahedron with these vertices has volume $1\mathrm{/}6\mathrm{\ne }0$, so the four points are NOT coplanar — they form a non-degenerate tetrahedron.