Daily math problems for October 4, 2026 — Trigonometry, Fundamental Algebra, Systems of Linear Equations & more

$\mathrm{\mid }\sin x\mathrm{\mid }\le 1$ for any real $x$, so $\sin x=2$ has no solutions.

Let $b$ be the cost of the ball in cents. The bat costs $b+100$. Their total is $110$, so:

$$b+(b+100)=110⟹2b=10⟹b=5$$

The ball costs 5 cents. Most people answer 10 — the trick is that the bat–ball *difference* is $100$, not the bat's price. Linear-equation discipline beats gut every time.

Cramer: $x_i={\displaystyle \frac{\det (A_i)}{\det (A)}}$.

The positive divisors of $12$ are $\{1,2,3,4,6,12\}$, so $\mathrm{\mid }A\mathrm{\mid }=6$.

$\bullet$ Use the fundamental limit:

$$\underset{n\to \mathrm{\infty }}{\lim }{(1+{\displaystyle \frac{2}{n}})}^n=e^2$$

$\bullet$ Rewrite the expression as a cube:

$${(1+{\displaystyle \frac{2}{n}})}^{3n}={\left[{(1+{\displaystyle \frac{2}{n}})}^n\right]}^3$$

$\bullet$ Take the limit:

$$(e^2{)}^3=e^6$$

${\sum }_{k=0}^6(-1{)}^k\left(\genfrac{}{}{0px}{}{6}{k}\right)=(1-1{)}^6=0$.

By the Fundamental Theorem of Algebra (or Moivre), $z^n=1$ has exactly $n$ distinct complex roots, the $n$th roots of unity ${\epsilon }_k=e^{2\pi ik\mathrm{/}n}$, $k=0,\dots ,n-1$.

Subtract: $5y=15\Rightarrow y=3$. Then $3x=y=3\Rightarrow x=1$. So $A=1+9=10$.

$P(-1)>0$ and $P(1)<0$: $2+m>0$ and $-2+m<0$, giving $-2<m<2$.

$\mathrm{\mid }x-2\mathrm{\mid }<3\iff -3<x-2<3\iff -1<x<5$. The solution set is $(-1,5)$.