Daily math problems for July 23, 2026 — Calculus, Vectors, Sequences & more

$v(t)=3t^2-12t+9$. At $t=2$: $v=12-24+9=-3$ m/s (the particle is moving backward).

Multiply each component by $2$: $2\vec{u}=6\vec{i}-4\vec{j}$.

${\int }_0^1{e}^{x}dx=e^1-e^0=e-1$.

Since $b^2=ac$, the product $abc=b\cdot ac=b\cdot b^2=b^3=64$, so $b=4$.

$y^2-6y+8=0\Rightarrow y\in \{2,4\}$. Then $x^2=2\Rightarrow x=\pm \sqrt{2}$ and $x^2=4\Rightarrow x=\pm 2$. Four distinct real solutions.

$\sin {\displaystyle \frac{\pi }{3}}={\displaystyle \frac{\sqrt{3}}{2}}$ and $\cos {\displaystyle \frac{\pi }{6}}={\displaystyle \frac{\sqrt{3}}{2}}$. Sum: $\sqrt{3}$.

$3x+4y=25$.

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

$\cos A={\displaystyle \frac{25+9-49}{2\cdot 5\cdot 3}}={\displaystyle \frac{-15}{30}}=-{\displaystyle \frac{1}{2}}$. Since $A\in (0,\pi )$, $A={\displaystyle \frac{2\pi }{3}}$.

$2x^2+x+1=2(x^2-x+2)\Rightarrow 2x^2+x+1=2x^2-2x+4\Rightarrow 3x=3\Rightarrow x=1$.

Verify the domain: at $x=1$, $2x^2+x+1=4>0$ and $x^2-x+2=2>0$, so both logs are defined. The solution set is $\{1\}$.