Daily math problems for October 28, 2026 — Fundamental Algebra, Algebra, Euclidean Geometry & more

$12÷4=3$ and $2\times 3=6$, then $3+6=9$.

$\sqrt{50}=\sqrt{25\cdot 2}=\sqrt{25}\cdot \sqrt{2}=5\sqrt{2}$.

Vertically opposite angles are equal.

$f$ is invertible iff its slope $a-2\mathrm{\ne }0$. So $f$ fails to be invertible exactly when $a=2$ (it then becomes the constant $5$, neither injective nor surjective).

$(x\ast y)\ast z=(x+y-xy)+z-(x+y-xy)z=x+y+z-xy-xz-yz+xyz$. By symmetry, $x\ast (y\ast z)$ gives the same expression. Hence associative.

$\int x{e}^{x}dx=x{e}^x-\int e^{x}dx=x{e}^x-e^x+C=(x-1)e^x+C$.

$A=1000\cdot {1.1}^3=1000\cdot 1.331=\mathrm{\$}\mathrm{1,331}$.

$\int \ln xdx=x\ln x-x+C$. Evaluate: $(e\cdot 1-e)-(1\cdot 0-1)=0-(-1)=1$.

Using $1-\cos x=2{\sin }^2(x\mathrm{/}2)$: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{2{\sin }^2(x\mathrm{/}2)}{x^2}}={\displaystyle \frac{1}{2}}\underset{x\to 0}{\lim }{\left({\displaystyle \frac{\sin (x\mathrm{/}2)}{x\mathrm{/}2}}\right)}^2={\displaystyle \frac{1}{2}}\cdot 1={\displaystyle \frac{1}{2}}}$.

$b_2=2$, $b_4=8$, $b_6=32$. The sum is $2+8+32=42$. (These three form a GP with first term $2$ and ratio $4$.)