Daily math problems for September 6, 2026 — Polynomials, Logic, Powers, Radicals, Logarithms & more

$P(2)=8-4+5=9$.

${\left[{\displaystyle \frac{4\cdot 5}{2}}\right]}^2={10}^2=100$. (Verify directly: $1+8+27+64=100$.)

$\sqrt[3]{8}=2$, so $8^{2\mathrm{/}3}=2^2=4$.

$R(\theta )R(-\theta )=I_2$, so $R(\theta {)}^{-1}=R(-\theta )$.

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

Pair each subset $S$ with $S\mathrm{△}\{1\}$ (i.e., toggle the element $1$). This pairing has no fixed points and flips the parity of the sum, so even-sum and odd-sum subsets are equinumerous: $2^{10}\mathrm{/}2=512$.

$5\cdot 4=20$, which is $A_5^2$.

$7^2=49$, $7^4={49}^2=2401\equiv 1(\mathrm{m}\mathrm{o}\mathrm{d}100)$. 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: $01$.

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

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