Problem 35
Question
Find a formula for \(g\) by scaling the output of \(f\). Let \(f(n)\) give the average time in seconds required for a computer to process \(n\) megabytes (MB) of data, and \(g(n)\) the time in microseconds. Use the fact that \(1 \mathrm{~s}\) equals \(1,000,000 \mu \mathrm{s}\).
Step-by-Step Solution
Verified Answer
Answer: The formula for \(g(n)\) is \(g(n) = 1,000,000 \cdot f(n)\).
1Step 1: Recall the relationship between seconds and microseconds
In this problem, we are given the relationship between seconds and microseconds: \(1 \mathrm{~s}\) equals \(1,000,000 \mu \mathrm{s}\). It means that to convert a value from seconds to microseconds, we need to multiply it by \(1,000,000\).
2Step 2: Find the formula for \(g(n)\) using the relationship from Step 1
Since \(f(n)\) gives the average time to process \(n\) MB of data in seconds and we want to find the time in microseconds, we will multiply the output of \(f(n)\) by the conversion factor of \(1,000,000\). Therefore, the formula for \(g(n)\) will be:
\[g(n) = 1,000,000 \cdot f(n)\]
This formula will give us the average time to process \(n\) MB of data in microseconds.
Key Concepts
Scaling FunctionsTime ConversionAlgebraic Formulas
Scaling Functions
Scaling functions is a vital concept in various mathematical applications, including algebra. Essentially, scaling involves multiplying or dividing a function's output by a constant to alter its magnitude. This method is used to transform units, resize graphs, and even adjust functions for optimization purposes.
When it comes to unit conversion, like in our exercise, scaling is integral. Here, the function \(f(n)\) signifies the time taken to process data in seconds. To discover the equivalent time in microseconds, you must scale \(f(n)\) by multiplying it by the conversion factor of 1,000,000. This changes the scale of measurement without altering the underlying relationship of the function itself.
Understanding scaling is crucial because it enables us to interpret function outputs in various unit expressions, ensuring flexibility and adaptability in problem-solving.
When it comes to unit conversion, like in our exercise, scaling is integral. Here, the function \(f(n)\) signifies the time taken to process data in seconds. To discover the equivalent time in microseconds, you must scale \(f(n)\) by multiplying it by the conversion factor of 1,000,000. This changes the scale of measurement without altering the underlying relationship of the function itself.
Understanding scaling is crucial because it enables us to interpret function outputs in various unit expressions, ensuring flexibility and adaptability in problem-solving.
Time Conversion
Time conversion is a frequently encountered task in mathematics and everyday life. It involves changing a time value from one unit to another, such as from seconds to microseconds. Converting time is straightforward: you multiply or divide the original time value by a specific conversion factor.
In the case of our exercise, you begin with time in seconds, as provided by \(f(n)\). Since 1 second equals 1,000,000 microseconds, multiply \(f(n)\) by the factor of 1,000,000 to find the equivalent time in microseconds. So, the relationship can be expressed as:
In the case of our exercise, you begin with time in seconds, as provided by \(f(n)\). Since 1 second equals 1,000,000 microseconds, multiply \(f(n)\) by the factor of 1,000,000 to find the equivalent time in microseconds. So, the relationship can be expressed as:
- microseconds = seconds x 1,000,000
Algebraic Formulas
Algebraic formulas are fundamental tools in mathematics used for solving equations and describing relationships between quantities. They consist of variables, numbers, and operations brought together to express a particular mathematical idea or law.
In our exercise, the formula \(g(n) = 1,000,000 \cdot f(n)\) is an example of using algebraic expressions to achieve unit conversion. Here, the algebraic operation of multiplication is employed to transition from one time unit (seconds) to another (microseconds).
Mastering algebraic formulas allows you to manipulate these mathematical relationships efficiently. This involves interpreting, transforming, and sometimes even deriving new formulas to suit specific problem contexts. By understanding the underpinning algebraic logic, converting between different mathematical forms, such as units or scales, becomes straightforward.
In our exercise, the formula \(g(n) = 1,000,000 \cdot f(n)\) is an example of using algebraic expressions to achieve unit conversion. Here, the algebraic operation of multiplication is employed to transition from one time unit (seconds) to another (microseconds).
Mastering algebraic formulas allows you to manipulate these mathematical relationships efficiently. This involves interpreting, transforming, and sometimes even deriving new formulas to suit specific problem contexts. By understanding the underpinning algebraic logic, converting between different mathematical forms, such as units or scales, becomes straightforward.
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