Daily math problems for October 15, 2026 — Fundamental Algebra, Algebra, Trigonometry & more

$(5+1{)}^2=6^2=36$, then $36÷4=9$.

Factor the quadratic: $$x^2-7x+12=(x-3)(x-4)=0$$

So the solutions are: $$x=3\text{or}x=4$$

Their sum is: $$3+4=7$$

This matches Vieta's formula, $-b\mathrm{/}a=-(-7)\mathrm{/}1=7$.

We need $\theta \in [0,\pi ]$ with $\cos \theta =\frac{1}{2}$.

From the unit circle, $\cos \text{\hspace{0.17em}⁣}\left(\frac{\pi }{3}\right)=\frac{1}{2}$.

$\frac{\pi }{3}$ lies in $[0,\pi ]$, so $\arccos \text{\hspace{0.17em}⁣}\left(\frac{1}{2}\right)={\displaystyle \frac{\pi }{3}}$.

Take $u=x,dv=e^{x}dx$, giving $du=dx,v=e^x$: $${\int }_0^{1}x{e}^{x}dx=[x{e}^x{]}_0^1-{\int }_0^1{e}^{x}dx$$ $$=(1\cdot e-0)-[e^x{]}_0^1=e-(e-1)=1$$

Apply the quotient rule to merge the difference into a single logarithm: $${\log }_{2}80-{\log }_{2}5={\log }_2\frac{80}{5}$$ Simplify the argument and evaluate, since $2^4=16$: $${\log }_{2}16=4$$

Composition is associative; ${\mathrm{id}}_E$ is the identity. Only bijective functions have inverses; non-bijective ones don't. Hence a monoid, but not a group. (The group is $S_E=$ bijections.)

First term $a=3$, ratio $r=\frac{1}{2}$. $S_{\mathrm{\infty }}={\displaystyle \frac{3}{1-\frac{1}{2}}}=6$.

Cauchy's equation with continuity gives $f(x)=cx$ for some constant $c$. Since $f(1)=3$, $c=3$, so $f(5)=15$.

For any $x\in [-1,1]$, $\arcsin x+\arccos x={\displaystyle \frac{\pi }{2}}$ (a standard identity, since $\arccos x={\displaystyle \frac{\pi }{2}}-\arcsin x$).

$S_n={\displaystyle \frac{n(3n+7)}{2}}$. Compute: $S_{17}={\displaystyle \frac{17\cdot 58}{2}}=493<500$ and $S_{18}={\displaystyle \frac{18\cdot 61}{2}}=549\ge 500$. Hence the smallest $n$ is $18$.