Daily math problems for August 19, 2026 — Elementary Functions, Sequences, Calculus & more

$x+2=9\Rightarrow x=7$. Check: $\sqrt{9}=3$ ✓.

$f(3)=3^{-2}={\displaystyle \frac{1}{3^2}}={\displaystyle \frac{1}{9}}$.

$5\cdot 2^{n-1}=320\Rightarrow 2^{n-1}=64=2^6\Rightarrow n-1=6\Rightarrow n=7$.

$V=\pi {\int }_0^2(4x^2+4x+1)dx=\pi \text{\hspace{0.17em}⁣}{[{\displaystyle \frac{4x^3}{3}}+2x^2+x]}_0^2=\pi \text{\hspace{0.17em}⁣}({\displaystyle \frac{32}{3}}+8+2)=\pi \cdot {\displaystyle \frac{32+30}{3}}={\displaystyle \frac{62\pi }{3}}$.

In $V_4$, every non-identity element has order $2$. (This distinguishes it from ${\mathbb{Z}}_4$, which has an element of order $4$.)

$\sin x+\cos x=\sqrt{2}\text{\hspace{0.17em}⁣}(\sin x\cdot {\displaystyle \frac{\sqrt{2}}{2}}+\cos x\cdot {\displaystyle \frac{\sqrt{2}}{2}})=\sqrt{2}\sin \text{\hspace{0.17em}⁣}(x+{\displaystyle \frac{\pi }{4}})$.

Since $\sin$ has maximum $1$, the maximum of $f$ is $\sqrt{2}$, attained at $x={\displaystyle \frac{\pi }{4}}$.

$n+1=11\Rightarrow n=10$.

We need $a_{12}=a_{21}$, i.e. $a=3$.

A square homogeneous system has non-trivial (non-zero) solutions iff $\det (A)=0$.

The sum telescopes: ${\sum }_{k=1}^{99}({\displaystyle \frac{1}{k}}-{\displaystyle \frac{1}{k+1}})=1-{\displaystyle \frac{1}{100}}={\displaystyle \frac{99}{100}}$.