Mary works at a jewelry store. She receives a base salary every week plus a commission based on how much she sells. Mary’s income function can be modeled byP(x) is her income and x is the amount of her sales in dollars.How much will Mary earn in a week in which she sells $4000 worth of jewelry? Show your work. The image displays a piecewise function for $ P(x) $, described as follows:\[ P(x) = \begin{cases} 400 + 0.06x & \text{if } 0 \leq x \leq 5000 \\700 + 0.09(x - 5000) & \text{if } x > 5000 \end{cases} \]This function can be interpreted in an educational context by explaining that $ P(x) $ represents a value that changes based on the variable $ x $. 1. **For $ x $ values between 0 and 5000 (inclusive):** - The function $ P(x) = 400 + 0.06x $ applies. - This indicates a linear relationship with a starting value of 400 and a slope of 0.06, meaning as $ x $ increases, $ P(x) $ increases at a rate of 0.06 times $ x $.2. **For $ x $ values greater than 5000:** - The function changes to $ P(x) = 700 + 0.09(x - 5000) $. - Here, the baseline value starts at 700, and the slope increases to 0.09. This segment accounts for values greater than 5000, starting the calculation of added value from 5000.This piecewise function demonstrates how functions can be defined with different expressions over different intervals of the input variable, a common approach in mathematical modeling.

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Mary works at a jewelry store. She receives a base salary every week plus a commission based on how much she sells. Mary’s income function can be modeled by P(x) is her income and x is the amount of her sales in dollars. How much will Mary earn in a week in which she sells $4000 worth of jewel

Mary works at a jewelry store. She receives a base salary every week plus a commission based on how much she sells. Mary’s income function can be modeled by P(x) is her income and x is the amount of her sales in dollars. How much will Mary earn in a week in which she sells $4000 worth of jewel

Advanced Engineering Mathematics

10th Edition

ISBN:9780470458365

Author:Erwin Kreyszig

Publisher:Erwin Kreyszig

Chapter2: Second-order Linear Odes

Section: Chapter Questions

Problem 1RQ

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Question

Mary works at a jewelry store. She receives a base salary every week plus a commission based on how much she sells. Mary’s income function can be modeled by
P(x) is her income and x is the amount of her sales in dollars.
How much will Mary earn in a week in which she sells $4000 worth of jewelry? Show your work.

The image displays a piecewise function for $ P(x) $, described as follows:

\[
P(x) =
\begin{cases}
400 + 0.06x & \text{if } 0 \leq x \leq 5000 \\
700 + 0.09(x - 5000) & \text{if } x > 5000
\end{cases}
\]

This function can be interpreted in an educational context by explaining that $ P(x) $ represents a value that changes based on the variable $ x $.

1. **For $ x $ values between 0 and 5000 (inclusive):**
- The function $ P(x) = 400 + 0.06x $ applies.
- This indicates a linear relationship with a starting value of 400 and a slope of 0.06, meaning as $ x $ increases, $ P(x) $ increases at a rate of 0.06 times $ x $.

2. **For $ x $ values greater than 5000:**
- The function changes to $ P(x) = 700 + 0.09(x - 5000) $.
- Here, the baseline value starts at 700, and the slope increases to 0.09. This segment accounts for values greater than 5000, starting the calculation of added value from 5000.

This piecewise function demonstrates how functions can be defined with different expressions over different intervals of the input variable, a common approach in mathematical modeling.

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