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

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

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

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

$V = \pi \int_0^2 (4x^2 + 4x + 1) \, dx = \pi\!\left[\dfrac{4x^3}{3} + 2x^2 + x\right]_0^2 = \pi\!\left(\dfrac{32}{3} + 8 + 2\right) = \pi \cdot \dfrac{32 + 30}{3} = \dfrac{62\pi}{3}$.

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

$\sin x + \cos x = \sqrt{2}\!\left(\sin x \cdot \dfrac{\sqrt{2}}{2} + \cos x \cdot \dfrac{\sqrt{2}}{2}\right) = \sqrt{2}\sin\!\left(x + \dfrac{\pi}{4}\right)$.

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

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

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: $\sum_{k=1}^{99} \left(\dfrac{1}{k} - \dfrac{1}{k+1}\right) = 1 - \dfrac{1}{100} = \dfrac{99}{100}$.