Daily math problems for May 8, 2026 — Geometry, Compound Interest, Algebra

For any triangle, the inradius is $r=A\mathrm{/}s$, where $A$ is the area and $s$ is the semi-perimeter.

Hypotenuse: $\sqrt{3^2+4^2}=5$. Area: ${\displaystyle \frac{1}{2}}\cdot 3\cdot 4=6$. Semi-perimeter: $(3+4+5)\mathrm{/}2=6$.

$r=6\mathrm{/}6=1$.

For right triangles specifically there's also a shortcut: $r=(a+b-c)\mathrm{/}2=(3+4-5)\mathrm{/}2=1$.

Let $B_n$ be the balance after the $n$-th payment. $B_0=10000$, and $B_n=1.06\cdot B_{n-1}-2000$.

$B_1=10600-2000=8600$ $B_2=9116-2000=7116$ $B_3=7542.96-2000=5542.96$ $B_4=5875.54-2000=3875.54$ $B_5=4108.07-2000=2108.07$ $B_6=2234.55-2000=234.55$

After year 6 there's only \234.55 left, so a 6th payment of \234.55 finishes it. Total: 6 years (the last payment is small).

Factor numerator and denominator: $$\frac{x^2-9}{x^2-6x+9}=\frac{(x-3)(x+3)}{(x-3{)}^2}$$

Cancel one factor of $(x-3)$ (valid since $x\mathrm{\ne }3$): $$=\frac{x+3}{x-3}$$