Daily math problems for August 27, 2026 — Analytic Geometry, Logs, Calculus & more

In slope-intercept form $y=mx+b$, the slope is $m=3$.

$\bullet$ Rewrite $50$ using $100$:

$$50={\displaystyle \frac{100}{2}}$$

$\bullet$ Apply $\lg (M\mathrm{/}N)=\lg M-\lg N$:

$$\lg 50=\lg 100-\lg 2$$

$\bullet$ Since $\lg 100=2$ and $\lg 2=a$:

$$\lg 50=2-a$$

As $n\to \mathrm{\infty }$, ${\displaystyle \frac{1}{n}}\to 0$.

$\bullet$ Compute each entry:

$$f(1)=2\ln 1+3=3$$

$$f^{\mathrm{\prime }}(x)=\frac{2}{x}+6x\Rightarrow f^{\mathrm{\prime }}(2)=13$$

$${\log }_2(13+3)={\log }_{2}16=4$$

$\bullet$ Also $\sin 7\pi =0$ and $\mathrm{\mid }3-4i\mathrm{\mid }=5$, giving:

$$A=\left(\begin{array}{cc}3& 4\\ 0& 5\end{array}\right)$$

$\bullet$ Use $\det (A^2)=(\det A{)}^2$:

$$\det A=15\Rightarrow \det (A^2)=225$$

$y=3x-6\Rightarrow x={\displaystyle \frac{y+6}{3}}$, so $f^{-1}(y)={\displaystyle \frac{y+6}{3}}$.

${\lim }_{x\to 0^{-}}f=-1$ and ${\lim }_{x\to 0^{+}}f=1$ — finite but unequal limits, hence a jump discontinuity.

${\displaystyle \frac{7\pi }{4}}\in \text{\hspace{0.17em}⁣}({\displaystyle \frac{3\pi }{2}},2\pi )$, which is the fourth quadrant.

${\mathbb{Z}}_n$ is a field iff every non-zero element is invertible, iff every element from $1$ to $n-1$ is coprime to $n$, iff $n$ is prime.

By parts: $\int x\ln xdx={\displaystyle \frac{x^2}{2}}\ln x-\int {\displaystyle \frac{x}{2}}dx={\displaystyle \frac{x^2}{2}}\ln x-{\displaystyle \frac{x^2}{4}}$. Evaluating on $[0,1]$ (with $x^2\ln x\to 0$ as $x\to 0^{+}$) gives $0-\frac{1}{4}-0=-\frac{1}{4}$.

The exponent ${\displaystyle \frac{6-3k}{2}}$ is an integer when $k$ is even: $k\in \{0,2,4,6\}$ — four values.