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

$\log_2 8 = 3$ deoarece $2^3 = 8$.

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

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

$\sigma$: $1 \to 2 \to 3 \to 4 \to 1$. So $\sigma^2$: $1 \to 3$, $3 \to 1$, $2 \to 4$, $4 \to 2$. That's $(1\;3)(2\;4)$.

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

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

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

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

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

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

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

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

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

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

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

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

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

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