Let's tackle each question step-by-step.
Question 1: Logarithms
To evaluate \(0.5972 \times 0.8467\) and \(0.7835 \times 0.646\) using logarithms:
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Compute \( \log(0.5972) + \log(0.8467) \): \[ \log(0.5972) \approx -0.2193 \quad (\text{from log table}) \] \[ \log(0.8467) \approx -0.0709 \quad (\text{from log table}) \] \[ \log(0.5972 \times 0.8467) = -0.2193 + -0.0709 = -0.2902 \] Thus, \[ 0.5972 \times 0.8467 \approx 10^{-0.2902} \approx 0.5154 \]
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Compute \( \log(0.7835) + \log(0.646) \): \[ \log(0.7835) \approx -0.104 \quad (\text{from log table}) \] \[ \log(0.646) \approx -0.1875 \quad (\text{from log table}) \] \[ \log(0.7835 \times 0.646) = -0.104 + -0.1875 = -0.2915 \] Thus, \[ 0.7835 \times 0.646 \approx 10^{-0.2915} \approx 0.5115 \]
Question 2: Simplifying the indices
To simplify \( \frac{729^{4/3}}{(1/243)^{-2/5}} \div 27^{1/3} \):
- \( 729 = 3^6 \), thus \( 729^{4/3} = (3^6)^{4/3} = 3^{8} \).
- \( 243 = 3^5 \), thus \( (1/243)^{-2/5} = (3^{-5})^{-2/5} = 3^{2} \).
- \( 27 = 3^3 \), thus \( 27^{1/3} = (3^3)^{1/3} = 3^1 \).
Putting it all together: \[ \frac{3^8}{3^2} \div 3^1 = 3^{8 - 2 - 1} = 3^5. \] The final answer is \( 3^5 \).
Question 3: Finding \( t \) and the equation of the line
Given the gradient \( m = \frac{3}{2} \): Using the points \((3, t)\) and \((t, 8)\): \[ \frac{8 - t}{t - 3} = \frac{3}{2} \] Cross-multiply: \[ 2(8 - t) = 3(t - 3) \] Expanding: \[ 16 - 2t = 3t - 9 \] Rearranging gives: \[ 16 + 9 = 5t \implies 25 = 5t \implies t = 5. \]
Now we can express the equation of the line using the point-slope form. We can use point \((3, 5)\): Using \( y - y_1 = m(x - x_1) \): \[ y - 5 = \frac{3}{2}(x - 3). \] Multiplying and rearranging: \[ 2y - 10 = 3x - 9 \implies 3x - 2y + 1 = 0. \] Thus, the equation is \( 3x - 2y + 1 = 0 \).
Question 4: Equation of a line perpendicular to \( 2y + x - 3 = 0 \)
First, convert it to slope-intercept: \[ 2y = -x + 3 \implies y = -\frac{1}{2}x + \frac{3}{2}, \] The slope is \(-\frac{1}{2}\), thus the perpendicular slope is \(2\).
Using point \((-3, 5)\): \[ y - 5 = 2(x + 3) \] This expands to: \[ y - 5 = 2x + 6 \implies y - 2x - 11 = 0 \implies 2x - y + 11 = 0. \]
Question 5: Evaluate the expression
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Question 6: Solve the equation \( 9^{(x+1)} + 3^{(2x+1)} = 36 \)
Convert \( 9 \) to powers of \( 3 \): \[ (3^2)^{(x+1)} + 3^{(2x+1)} = 36 \implies 3^{(2(x+1))} + 3^{(2x + 1)} = 6^2. \] Let \( 3^{(x+1)} = y \), then the equation becomes: \[ y^2 + 3y = 36. \] This is: \[ y^2 + 3y - 36 = 0. \] Using the quadratic formula: \[ y = \frac{-3 \pm \sqrt{3^2 + 4 \cdot 36}}{2} = \frac{-3 \pm 15}{2} \implies y = 6 \text{ or } y = -9. \] Thus: \[ 3^{(x+1)} = 6 \implies x + 1 = 1 \implies x = 0. \]
Question 7: Volume of the cylindrical tank
The formula for the volume of a cylinder is \( V = \pi r^2 h \). The radius \( r = \frac{1.4 m}{2} = 0.7 m \), and height \( h = 0.8 m \): \[ V = \pi (0.7^2)(0.8) = \pi (0.49)(0.8) \approx 1.226 \text{ m}^3. \] Since \( 1 m^3 = 1000 L \): \[ V \approx 1226 L. \] Thus, fraction of tank filled is: \[ \frac{492.8}{1226} \approx 0.402 , \text{or} , 40.2%. \]
Question 8: Percentage discount
\[ \text{Discount} = M - SP = 450 - 393.75 = 56.25. \] Percentage discount: \[ \frac{56.25}{450} \times 100 \approx 12.5%. \]
Question 9: Syengo's expenditure
His total expenditure on food, rent, and transport: Let salary = \( S \). Food = \( \frac{1}{3}S \), rent = \( \frac{1}{4}S \), remainder = \( S - \left( \frac{1}{3}S + \frac{1}{4}S \right) = S \left( 1 - \frac{7}{12} \right) = \frac{5}{12}S \). Transport = \( \frac{3}{5} \times \frac{5}{12}S = \frac{1}{4}S \), giving \[ \frac{1}{4}S = 1800, \implies S = 7200. \]
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