Solutions and detailed explanations to the math questions on this site are presented below.
Rewrite the logarithmic equation using exponential form:
\[ x^{-\frac{3}{2}} = \frac{1}{8} \] \[ x^{-\frac{3}{2}} = \frac{1}{2^3} = 2^{-3} \] \[ (\sqrt{x})^{-3} = 2^{-3} \] \[ \sqrt{x} = 2 \quad \Rightarrow \quad x = 4 \]
\[ 20\% \text{ of } 2 = \frac{20}{100} \cdot 2 = 0.4 \]
If \(\log_4(x) = 12\), then \[ x = 4^{12} \] \[ \frac{x}{4} = 4^{11} \] \[ \log_2\left(\frac{x}{4}\right) = \log_2\left(4^{11}\right) = \log_2\left((2^2)^{11}\right) = \log_2(2^{22}) = 22 \]
A population \(P\) increasing at a rate of 2% per year is modeled by: \[ P = P_0 e^{0.02 t} \] where \(t\) is years after \(t = 0\), and \(P_0\) is the initial population. If \(t=0\) corresponds to January 1, 2000, and \(P = 2{,}000{,}000\) at \(t=4\): \[ 2{,}000{,}000 = P_0 e^{0.02 \cdot 4} \] \[ P_0 = \frac{2{,}000{,}000}{e^{0.08}} \approx 1{,}846{,}000 \] Hence, \(P_0\) is the population on January 1, 2000.
If the graph of \(f\) is a parabola with vertex on the \(x\)-axis and opening upwards, its range is: \[ [0, +\infty) \] The graph of \(f(x-5)\) is shifted 5 units right; range remains unchanged. The graph of \(-f(x-5)\) is reflected on the \(x\)-axis: \[ (-\infty, 0] \] For \(g(x) = 2 - f(x-5)\), we shift \(-f(x-5)\) up by 2 units: \[ (-\infty, 2] \]
Since \(f(x) < 0\), \[ g(x) = |f(x)| = -f(x) \] The graph of \(g(x) = -f(x)\) is the reflection of \(f(x)\) over the \(x\)-axis.
Solve \(2y - 6 = -4 f(x-3)\) for \(y\): \[ y = -2 f(x-3) + 3 \] The graph of \(y = -2 f(x-3) + 3\) is obtained by:
Rewrite \[ y - 2x^2 = 8x + 5 \quad \Rightarrow \quad y = 2x^2 + 8x + 5 \] Complete the square: \[ y = 2(x^2 + 4x + 4) - 8 + 5 = 2(x+2)^2 - 3 \] Vertex at \((-2, -3)\). Translating 3 units left and 2 units up: \[ \text{New vertex: } (-2-3, -3+2) = (-5, -1) \]
Slopes of the lines \(ax + by = c\) and \(bx - ay = c\): \[ m_1 = -\frac{a}{b}, \quad m_2 = \frac{b}{a} \] \[ m_1 m_2 = -1 \quad \Rightarrow \text{lines are perpendicular} \]
If \(a\) and \(A\) have different signs, the parabolas open in opposite directions. Since \(b^2 - 4ac < 0\) and \(B^2 - 4AC < 0\), neither intersects the \(x\)-axis. Hence, they do not intersect.
\(\sin(0) = 0\), \(\cos(0) = 1\). From \(x=0\) to \(x=\pi/2\), \(\sin(x)\) increases to 1, \(\cos(x)\) decreases to 0. From \(x=\pi/2\) to \(x=\pi\), \(\sin(x)\) decreases to 0, \(\cos(x)\) decreases to -1. Hence both decrease on \((\pi/2, \pi)\).
If \(f(x)=0\) at \(x=-2,0,3\), then \(f(x-2)=0\) when \[ x-2 = -2, 0, 3 \quad \Rightarrow \quad x = 0, 2, 5 \]
If \(f(x)=0\) at \(x=-4, 8, 11\), then \(f(2x)=0\) when \[ 2x = -4, 8, 11 \quad \Rightarrow \quad x = -2, 4, \frac{11}{2} \]
Number of ways to select 2 teachers from 5 and 4 students from 10: \[ C(5,2) \cdot C(10,4) = 10 \cdot 210 = 2100 \] where \(C(n,r) = \frac{n!}{r!(n-r)!}\).
With books C and D fixed, arrange remaining 3 books: \[ 3! = 6 \]
Adding 5 to all data values increases the mean by 5: \[ \text{new mean} = 15 \] Standard deviation remains unchanged.
Z-score: \[ z = \frac{x - m}{s} = \frac{0.8s + m - m}{s} = 0.8 \] Percentage above Jane: \[ 1 - 0.7881 = 0.2119 = 21.19\% \] Number of students above Jane: \[ 0.2119 \cdot 500 \approx 106 \]
Since \(f(x)\) is odd: \[ f(-x) = -f(x) \] Let \(g(x) = |f(x)|\): \[ g(-x) = |f(-x)| = |-f(x)| = |f(x)| = g(x) \] Hence, \(|f(x)|\) is even.
Period of \(\sin(3x)\): \[ \frac{2\pi}{3} \] Reflecting negative parts for \(|\sin(3x)|\) halves the period: \[ \frac{2\pi/3}{2} = \frac{\pi}{3} \]
Volume of water displaced by a ball: \[ \pi \cdot 2^2 \cdot 0.6 = 2.4 \pi \text{ cm}^3 \] Volume of ball: \[ \frac{4}{3} \pi r^3 = 2.4 \pi \quad \Rightarrow \quad r = 1.2 \text{ cm} \]
Using trigonometric identity \(2\sin x \cos x = \sin(2x)\), period: \[ \frac{2\pi}{2} = \pi \]
Binomial probability: \[ p = 0.9, \quad q = 0.1 \] \[ P = C(10,7) \cdot 0.9^7 \cdot 0.1^3, \quad C(10,7) = \frac{10!}{7!3!} = 120 \] \[ P \approx 0.057 \]