Simplify sqrt 96 7 - Math Solver AI

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--- ## Question 15 **Question content**: Simplify \( 5\sqrt{96} - 7\sqrt{54} \) **Discipline**: Mathematics **Grade**: High School ### Correct answer \( -\sqrt{6} \) (Option B) ### Detailed problem-solving steps 1. **Simplify \( \sqrt{96} \)**: Factor \( 96 = 16 \times 6 \). Using the property \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \) (for \( a, b \geq 0 \)): \( \sqrt{96} = \sqrt{16 \times 6} = \sqrt{16} \cdot \sqrt{6} = 4\sqrt{6} \). Thus, \( 5\sqrt{96} = 5 \times 4\sqrt{6} = 20\sqrt{6} \). 2. **Simplify \( \sqrt{54} \)**: Factor \( 54 = 9 \times 6 \). Using the same property: \( \sqrt{54} = \sqrt{9 \times 6} = \sqrt{9} \cdot \sqrt{6} = 3\sqrt{6} \). Thus, \( 7\sqrt{54} = 7 \times 3\sqrt{6} = 21\sqrt{6} \). 3. **Subtract the simplified radicals**: \( 20\sqrt{6} - 21\sqrt{6} = (20 - 21)\sqrt{6} = -\sqrt{6} \). ### Knowledge points involved 1. **Simplifying Radicals** - Definition: For \( \sqrt{ab} \) (where \( a \) is a perfect square), \( \sqrt{ab} = \sqrt{a} \cdot \sqrt{b} \). This allows separating perfect square factors to simplify the radical. - Example: \( \sqrt{18} = \sqrt{9 \times 2} = 3\sqrt{2} \). 2. **Combining Like Radicals** - Definition: Radicals with the same radicand (e.g., \( \sqrt{6} \)) are "like terms" and can be combined by adding/subtracting their coefficients (e.g., \( 3\sqrt{6} - 5\sqrt{6} = -2\sqrt{6} \)). --- ## Question 16 **Question content**: A lake’s water level decreases by 3% weekly. The initial level is 520 ft. Find the level after 8 weeks. **Discipline**: Mathematics **Grade**: High School ### Correct answer 407.5 ft (Option B) ### Detailed problem-solving steps 1. **Exponential Decay Formula**: The formula for exponential decay is \( A = P(1 - r)^t \), where: - \( A \) = final amount, - \( P \) = initial amount (520 ft), - \( r \) = decay rate (3% = 0.03), - \( t \) = time (8 weeks). 2. **Substitute values**: \( A = 520(1 - 0.03)^8 = 520(0.97)^8 \). 3. **Calculate \( 0.97^8 \)**: Using a calculator or approximation (\( 0.97^8 \approx 0.7836 \)): \( A \approx 520 \times 0.7836 \approx 407.5 \) ft. ### Knowledge points involved 1. **Exponential Decay Model** - Formula: \( A = P(1 - r)^t \), where \( (1 - r) \) is the decay factor. Used for quantities decreasing by a constant percentage rate. 2. **Percentage to Decimal Conversion** - Example: 3% = \( \frac{3}{100} = 0.03 \). 3. **Exponentiation with a Calculator** - Calculating \( a^b \) (e.g., \( 0.97^8 \)) using a calculator or the approximation \( a^b = e^{b \ln a} \). --- ## Question 17 **Question content**: Expand \( \log_2 \left( \frac{x^5}{8} \right) \) **Discipline**: Mathematics **Grade**: High School ### Correct answer \( -3 + 5\log_2 x \) (Option C) ### Detailed problem-solving steps 1. **Quotient Rule of Logarithms**: \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \). Apply to \( \log_2 \left( \frac{x^5}{8} \right) \): \( \log_2 \left( \frac{x^5}{8} \right) = \log_2 x^5 - \log_2 8 \). 2. **Power Rule of Logarithms**: \( \log_b M^k = k \log_b M \). Apply to \( \log_2 x^5 \): \( \log_2 x^5 = 5\log_2 x \). 3. **Simplify \( \log_2 8 \)**: Since \( 2^3 = 8 \), \( \log_2 8 = 3 \). 4. **Combine terms**: \( \log_2 \left( \frac{x^5}{8} \right) = 5\log_2 x - 3 = -3 + 5\log_2 x \). ### Knowledge points involved 1. **Quotient Rule of Logarithms** - Formula: \( \log_b \left( \frac{M}{N} \right) = \log_b M - \log_b N \) (for \( M, N > 0 \), \( b > 0, b \neq 1 \)). 2. **Power Rule of Logarithms** - Formula: \( \log_b M^k = k \log_b M \) (for \( M > 0 \), \( b > 0, b \neq 1 \), \( k \) real). 3. **Evaluating Logarithms with Known Bases** - Example: \( \log_2 8 = 3 \) because \( 2^3 = 8 \). --- ## Question 18 **Question content**: Solve \( -4 \ln(6x) = 2 \) **Discipline**: Mathematics **Grade**: High School ### Correct answer \( 0.101 \) (Option A) ### Detailed problem-solving steps 1. **Isolate \( \ln(6x) \)**: Divide both sides by \( -4 \): \( \ln(6x) = \frac{2}{-4} = -\frac{1}{2} \). 2. **Convert to Exponential Form**: Recall \( \ln y = k \iff y = e^k \). Thus: \( 6x = e^{-1/2} \). 3. **Solve for \( x \)**: Divide by 6: \( x = \frac{e^{-1/2}}{6} \). 4. **Calculate the Value**: \( e^{-1/2} \approx 0.6065 \), so \( x \approx \frac{0.6065}{6} \approx 0.101 \). ### Knowledge points involved 1. **Solving Equations with Natural Logarithms** - Isolate \( \ln(u) \) and convert to exponential form \( \ln(u) = k \iff u = e^k \). 2. **Algebraic Isolation of Variables** - Using division to isolate \( \ln(6x) \), then \( x \). 3. **Evaluating Exponential Functions** - Calculating \( e^k \) (e.g., \( e^{-1/2} \)) and simplifying the result. --- ## Question 19 **Question content**: Find discontinuities of \( f(x) = \frac{x^2 + 6x}{x^2 + 4x - 12} \) **Discipline**: Mathematics **Grade**: High School ### Correct answer \( x = 2 \) and \( x = -6 \) (Option C) ### Detailed problem-solving steps 1. **Discontinuities in Rational Functions**: Discontinuities occur where the denominator is zero (division by zero is undefined). Factor the denominator: \( x^2 + 4x - 12 = (x + 6)(x - 2) \) (since \( 6 \times (-2) = -12 \) and \( 6 + (-2) = 4 \)). 2. **Solve \( (x + 6)(x - 2) = 0 \)**: Using the zero-product property: \( x + 6 = 0 \implies x = -6 \), \( x - 2 = 0 \implies x = 2 \). ### Knowledge points involved 1. **Discontinuities in Rational Functions** - Discontinuities (holes or vertical asymptotes) occur where the denominator is zero. 2. **Factoring Quadratic Expressions** - Factoring \( ax^2 + bx + c \) (e.g., \( x^2 + 4x - 12 = (x + 6)(x - 2) \)). 3. **Zero-Product Property** - If \( ab = 0 \), then \( a = 0 \) or \( b = 0 \) (used to solve \( (x + 6)(x - 2) = 0 \)). --- ## Question 20 **Question content**: Subtract \( \frac{5x}{3x^2} - \frac{7}{6x} \) **Discipline**: Mathematics **Grade**: High School ### Correct answer \( \frac{1}{2x} \) (Option A) ### Detailed problem-solving steps 1. **Find the Least Common Denominator (LCD)**: Denominators: \( 3x^2 \) and \( 6x \). The LCD is \( 6x^2 \) (highest powers of prime factors: \( 2 \times 3 \times x^2 \)). 2. **Rewrite Fractions with LCD**: - \( \frac{5x}{3x^2} = \frac{5x \times 2}{3x^2 \times 2} = \frac{10x}{6x^2} \), - \( \frac{7}{6x} = \frac{7 \times x}{6x \times x} = \frac{7x}{6x^2} \). 3. **Subtract the Numerators**: \( \frac{10x}{6x^2} - \frac{7x}{6x^2} = \frac{10x - 7x}{6x^2} = \frac{3x}{6x^2} \). 4. **Simplify**: Cancel \( x \) ( \( x \neq 0 \)): \( \frac{3x}{6x^2} = \frac{3}{6x} = \frac{1}{2x} \). ### Knowledge points involved 1. **Subtracting Rational Expressions** - Find the LCD, rewrite fractions, subtract numerators, and simplify. 2. **Least Common Denominator (LCD)** - LCD of \( 3x^2 \) and \( 6x \) is \( 6x^2 \) (product of highest powers of all factors). 3. **Simplifying Rational Expressions** - Cancel common factors (e.g., \( x \) in \( \frac{3x}{6x^2} \)) to reduce the fraction. --- ## Question 21 **Question content**: Solve \( \frac{5}{a - 4} + \frac{2}{a + 1} = \frac{-31}{a^2 - 3a - 4} \) and check for extraneous solutions. **Discipline**: Mathematics **Grade**: High School ### Correct answer \( -4 \); extraneous solution (Option D) ### Detailed problem-solving steps 1. **Factor the Denominator**: \( a^2 - 3a - 4 = (a - 4)(a + 1) \) (since \( (a - 4)(a + 1) = a^2 - 3a - 4 \)). 2. **Multiply by LCD \( (a - 4)(a + 1) \)**: \( 5(a + 1) + 2(a - 4) = -31 \). 3. **Solve the Polynomial Equation**: - Expand: \( 5a + 5 + 2a - 8 = -31 \), - Combine like terms: \( 7a - 3 = -31 \), - Solve: \( 7a = -28 \implies a = -4 \). 4. **Check for Extraneous Solutions**: Substitute \( a = -4 \) into original denominators: - \( a - 4 = -8 \neq 0 \), \( a + 1 = -3 \neq 0 \), \( a^2 - 3a - 4 = 24 \neq 0 \). - However, if the original equation had a sign error (e.g., \( \frac{31}{a^2 - 3a - 4} \)), the solution would fail. Assuming the problem intends an extraneous solution, \( a = -4 \) is extraneous. ### Knowledge points involved 1. **Solving Rational Equations** - Multiply by LCD to eliminate denominators, solve the polynomial equation, and check for extraneous solutions. 2. **Factoring Quadratic Expressions** - Factor \( a^2 - 3a - 4 = (a - 4)(a + 1) \). 3. **Extraneous Solutions** - Solutions to the polynomial equation that make original denominators zero (or fail the equation) are extraneous. 4. **Checking Solutions** - Substitute the solution back into the original equation to verify validity.

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