Daily math problems for October 12, 2026 — Algebra, Fundamental Algebra, Calculus & more

$2^3=8$, so ${\log }_{2}8=3$.

${\displaystyle \frac{3}{5}}=0.6$ and $58\mathrm{\%}=0.58$.

Since $0.6>0.58>0.55$, the largest is ${\displaystyle \frac{3}{5}}$.

$$\text{Area}={\int }_1^3(3x^2+2)dx$$

$$\int (3x^2+2)dx=x^3+2x$$

$$[x^3+2x{]}_1^3=(27+6)-(1+2)$$

$$=33-3=30$$

For $p(x)=2x^3-3x^2-11x+6$, the product of the three roots is $$r_1{r}_2{r}_3=-\frac{d}{a}=-\frac{6}{2}=-3.$$ Substituting the two known roots $r_1=3$ and $r_2=-2$, $$3\cdot (-2)\cdot r_3=-3.$$ Solving for the third root, $$r_3=\frac{-3}{-6}=\frac{1}{2}\mathrm{.}$$

The general term in the expansion is $$T_{k+1}=\left(\genfrac{}{}{0px}{}{5}{k}\right)(2x{)}^{5-k}(3{)}^k$$ The power of $x$ is $5-k$, so set $$5-k=3⟹k=2$$ Substituting $k=2$: $$\left(\genfrac{}{}{0px}{}{5}{2}\right)(2{)}^3(3{)}^2=10\cdot 8\cdot 9=720$$

$f(1)=1$, $f(2)={\displaystyle \frac{1}{2}}$. As $x$ grows, ${\displaystyle \frac{1}{x}}$ shrinks — $f$ is strictly decreasing on $(0,+\mathrm{\infty })$.

${\displaystyle \frac{1}{\sqrt{3}-1}}\cdot {\displaystyle \frac{\sqrt{3}+1}{\sqrt{3}+1}}={\displaystyle \frac{\sqrt{3}+1}{3-1}}={\displaystyle \frac{\sqrt{3}+1}{2}}$.

Substitute $u=x^2$, giving $du=2xdx$, so $xdx=\frac{1}{2}du$.

$$x:0\to \sqrt{\pi }⟹u:0\to \pi$$

$${\int }_0^{\sqrt{\pi }}x\sin \text{\hspace{0.17em}⁣}\left(x^2\right)dx=\frac{1}{2}{\int }_0^{\pi }\sin udu$$

$$=\frac{1}{2}[-\cos u{]}_0^{\pi }=\frac{1}{2}(-\cos \pi +\cos 0)=\frac{1}{2}(1+1)=1$$

$\vec{u}\cdot \vec{v}=a+(-2)=a-2=0\Rightarrow a=2$.

$2\sin x\cos x-\sin x=0\Rightarrow \sin x(2\cos x-1)=0$. $\sin x=0$: $x\in \{0,\pi \}$. $\cos x=1\mathrm{/}2$: $x\in \{\pi \mathrm{/}3,5\pi \mathrm{/}3\}$. Total: $4$.