Daily math problems for August 8, 2026 — Logs, Quadratic Function, Probability & more

${\log }_{2}8=3$ deoarece $2^3=8$.

$(x-2)(x+2)<0\iff x\in (-2,2)$.

$P={\displaystyle \frac{3}{8}}\cdot {\displaystyle \frac{2}{7}}={\displaystyle \frac{6}{56}}={\displaystyle \frac{3}{28}}$.

$\sigma$: $1\to 2\to 3\to 4\to 1$. So ${\sigma }^2$: $1\to 3$, $3\to 1$, $2\to 4$, $4\to 2$. That's $(13)(24)$.

$A={\displaystyle \frac{36\sqrt{3}}{4}}=9\sqrt{3}$.

$x^3-3x+2=(x-1)(x^2+x-2)=(x-1)(x-1)(x+2)=(x-1{)}^2(x+2)$. So ${\displaystyle \frac{(x-1{)}^2(x+2)}{x-1}}=(x-1)(x+2)\to 0$ as $x\to 1$.

$\bullet$ Take reciprocals of each equation, using $\frac{1}{x}+\frac{1}{y}=\frac{x+y}{xy}$:

$$\frac{1}{x}+\frac{1}{y}=\frac{3}{5},\frac{1}{x}+\frac{1}{z}=\frac{2}{3},\frac{1}{y}+\frac{1}{z}=\frac{1}{4}$$

$\bullet$ Let $a=1\mathrm{/}x$, $b=1\mathrm{/}y$, $c=1\mathrm{/}z$. Sum all three:

$$2(a+b+c)=\frac{3}{5}+\frac{2}{3}+\frac{1}{4}=\frac{91}{60}$$

$$a+b+c=\frac{91}{120}$$

$\bullet$ Subtract each pair-sum from $a+b+c$:

$$a=\frac{61}{120},b=\frac{11}{120},c=\frac{19}{120}$$

$\bullet$ Recover $x,y,z$ and add:

$$x+y+z=\frac{120}{61}+\frac{120}{11}+\frac{120}{19}$$

Direct substitution gives $\frac{0}{0}$. Differentiate top and bottom: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x-1}{2x}}}$ — still $\frac{0}{0}$. Once more: ${\displaystyle \underset{x\to 0}{\lim }{\displaystyle \frac{e^x}{2}}={\displaystyle \frac{1}{2}}}$.

The slope is $8$ score points per additional hour. For a $0.5$-hour increase, the predicted change is $0.5\times 8=4$.

$f^{\mathrm{\prime }}(x)=3x^2-3a$, so $f^{\mathrm{\prime }}(2)=12-3a=0\Rightarrow a=4$. Check: $f^{\mathrm{\prime }\mathrm{\prime }}(x)=6x$, so $f^{\mathrm{\prime }\mathrm{\prime }}(2)=12>0$, confirming a local minimum.