Boat Travels 3 Miles Upstream In The Same Time
Boat Travels 3 Miles Upstream In Thesame Timeit Takes To Travel 4
The assignment involves multiple mathematical problems, including solving equations, understanding proportional relationships, simplifying expressions, and analyzing functions. The core tasks include finding the speed of a current given certain travel conditions, determining the domain of a rational function, dividing and simplifying algebraic expressions, solving equations algebraically, expressing roots in simplest radical form, rationalizing denominators, simplifying expressions with positive exponents, analyzing inverse proportionality, solving quadratic equations by completing the square, and applying word problem strategies to real-world scenarios. This comprehensive assignment aims to assess proficiency in algebraic manipulation, functions, and problem-solving skills.
Paper For Above instruction
The various mathematical problems presented span fundamental concepts in algebra, functions, and problem-solving strategies. Each problem requires a clear understanding of concepts such as distance, speed, and time relationships; properties of rational functions; methods of solving equations; properties of radicals; and the principles of inverse proportionality. Addressing these topics not only supports the development of algebraic proficiency but also enhances critical thinking skills needed to interpret real-world situations mathematically.
1. Finding the Speed of the Current
The first problem involves a boat traveling upstream and downstream, with the boat's speed in still water known. The goal is to find the speed of the current. When a boat travels upstream, its effective speed relative to the land is reduced by the speed of the current; downstream, it increases by that speed. The problem states that the time taken to travel 3 miles upstream equals the time to travel 4 miles downstream.
Let c be the speed of the current in miles per hour. The boat's speed in still water is given as 7 mph. Then, the upstream speed is (7 - c) mph, and the downstream speed is (7 + c) mph.
The time for upstream travel is t1 = 3 / (7 - c), and downstream travel is t2 = 4 / (7 + c). Since these times are equal, set up the equation:
3 / (7 - c) = 4 / (7 + c)
Cross-multiplied, it yields:
3(7 + c) = 4(7 - c)
21 + 3c = 28 - 4c
Combine like terms:
3c + 4c = 28 - 21
7c = 7
c = 1 mph
Therefore, the speed of the current is 1 mile per hour.
2. Domain of the Rational Function
The function provided is f(x) = 3 / (x - 2). To find its domain, identify values of x that make the denominator zero since division by zero is undefined. Setting the denominator to zero:
x - 2 = 0
x = 2
Thus, the domain of f(x) excludes x = 2, and in interval notation, it is:
(-∞, 2) ∪ (2, ∞)
3. Divide and Simplify
Dividing polynomials, for example, (4x^3 + 2x^2) ÷ 2x:
(4x^3 + 2x^2) / 2x = (4x^3 / 2x) + (2x^2 / 2x) = 2x^2 + x
Hence, the simplified result is 2x^2 + x.
4. Solve an Equation and Verify
Suppose the equation is x^2 + 4x = 0. Factoring gives:
x(x + 4) = 0
Solutions are x = 0 or x = -4.
Verify by substitution:
For x=0: 0^2 + 4(0) = 0, which is true.
For x = -4: (-4)^2 + 4(-4) = 16 - 16 = 0, which is also true.
5. Express in Simplest Radical Form
Evaluate √50. Simplify radicals:
√50 = √(25×2) = √25 × √2 = 5√2
6. Rationalize the Denominator
Given an expression, for example, 3 / √2, rationalize by multiplying numerator and denominator by √2:
(3 / √2) × (√2 / √2) = (3√2) / 2
7. Simplify and Express with Positive Exponents Only
For example, simplify x^-3 y^2 / x^2 y^-4:
((x^-3) / (x^2)) × (y^2 / y^-4) = x^{-3-2} y^{2-(-4)} = x^{-5} y^{6} = (1 / x^5) y^6
8. Inverse Proportionality and Electric Bill
The electric bill B is inversely proportional to the thermostat setting T. The relation is B = k / T. Given when T=70°, B=$120, find k:
120 = k / 70 → k = 120 × 70 = 8400
When T=75°, the bill is:
B = 8400 / 75 = 112
Hence, the electric bill at 75° is $112.
9. Solve Quadratic by Completing the Square
Solve x^2 – 6x = 3
Rewrite as:
x^2 – 6x + ? = 3 + ?
Complete the square by adding (6/2)^2 = 9 to both sides:
x^2 – 6x + 9 = 3 + 9 = 12
(x – 3)^2 = 12
Take square roots:
x – 3 = ±√12 = ±2√3
Solutions:
x = 3 ± 2√3
10. House Cleaning Time Problem
Let the father's time to clean alone be t hours, then the son's time is t + 1 hours. The rates are 1/t and 1/(t+1).
Combined rate: 1/t + 1/(t+1) = 1/3
Find common denominator and solve:
(t+1 + t) / [t(t+1)] = 1/3
(2t + 1) / [t(t+1)] = 1/3
Cross-multiplied:
3(2t + 1) = t(t+1)
6t + 3 = t^2 + t
Rearranged to standard quadratic form:
t^2 + t – 6t – 3 = 0 → t^2 – 5t – 3 = 0
Apply quadratic formula:
t = [5 ± √(25 + 12)] / 2 = [5 ± √37] / 2
Since time can't be negative, take the positive root:
t = (5 + √37) / 2 ≈ (5 + 6.08)/2 ≈ 11.08/2 ≈ 5.54 hours
Thus, the son takes approximately 5.5 hours to clean alone.
Conclusion
These problems demonstrate key algebraic concepts, including solving equations, understanding functions, radical simplification, proportional relationships, and applying these skills to real-world problems. Mastery of these areas provides a solid foundation for advanced mathematics and practical problem-solving.
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