Daily math problems for July 22, 2026 — Analytic Geometry, Polynomials, Quadratic Function & more

$\bullet$ Compute coordinates of $A$. Note ${\log }_{4}8=\frac{3}{2}$ and treat ${\log }_{1337}\mathrm{\mid }i\mathrm{\mid }=0$:

$$A=(2\cdot \frac{3}{2},0)=(3,0)$$

$\bullet$ Coordinates of $B$ from ${\log }_{2}32=5$ and $P_2=2$:

$$B=(5,2)$$

$\bullet$ Coordinates of $C$ from $0!=1$ and ${\log }_{2}512=9$:

$$C=(1,9)$$

$\bullet$ Apply the area formula:

$$\mathcal{A}=\frac{1}{2}\mathrm{\mid }{x}_A(y_B-y_C)+x_B(y_C-y_A)+x_C(y_A-y_B)\mathrm{\mid }$$

$$=\frac{1}{2}\mathrm{\mid }3(-7)+5(9)+1(-2)\mathrm{\mid }$$

$$=\frac{1}{2}\cdot 22=11$$

$\mathrm{deg}(Q)=3-1=2$.

Base $=6$, height $=4$. Area $={\displaystyle \frac{1}{2}}\cdot 6\cdot 4=12$.

By Viète: $x_1+x_2=5$, $x_1{x}_2=4$. Hence $x_1^2+x_2^2=25-8=17$.

${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{(1+x{)}^r-1}{x}}=r}$. (This generalizes the bonomial-derivative-at-zero.)

$0\le I_n\le {\displaystyle \frac{1}{n+1}}\to 0$, so $I_n\to 0$ by the squeeze theorem.

Vertical: $x=\pm 1$ (two). Horizontal: $\lim f(x)=1$ (one). Total: $3$. (No slant — the rational function has equal-degree numerator and denominator.)

A ring homomorphism preserves both operations: addition AND multiplication.

$\mathrm{\mid }{A}_n\mathrm{\mid }={\displaystyle \frac{n!}{2}}$ (one of the standard properties of the alternating group).

$2{\cos }^{2}x+\cos x-1=0\Rightarrow (2\cos x-1)(\cos x+1)=0$. $\cos x=1\mathrm{/}2$: 2 solutions ($\pi \mathrm{/}3,5\pi \mathrm{/}3$). $\cos x=-1$: 1 solution ($\pi$). Total: $3$.