Math Problem Statement

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Find a real-life example of (1) direct and (1) inverse variation and write a short explanation of how it works. a. identify the variables, b. write the direct and inverse variation equation, c. create and solve a problem related to your scenario.

Solution

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Real-life examples of direct and inverse variation:

1. Direct Variation Example:

Scenario: Water flow rate and time to fill a tank.

  • Explanation: In this case, the time it takes to fill a tank is directly proportional to the amount of water required to fill it. If the flow rate increases, the tank fills faster, and if it decreases, it takes longer. Mathematically, this means that time tt varies directly with the amount of water VV.

a. Identify the variables:

  • Flow rate rr (liters per minute)
  • Time tt (minutes)

b. Direct variation equation: V=rtV = r \cdot t Where VV is the total volume of water, rr is the flow rate, and tt is the time.

c. Problem: If the flow rate is 5 liters per minute, how long will it take to fill a 200-liter tank?

Solution: Using the equation V=rtV = r \cdot t: 200=5t200 = 5 \cdot t Solving for tt: t=2005=40 minutest = \frac{200}{5} = 40 \text{ minutes} It will take 40 minutes to fill the tank.

2. Inverse Variation Example:

Scenario: Pressure and volume of a gas (Boyle's Law).

  • Explanation: In a closed container, the pressure PP of a gas is inversely proportional to its volume VV, assuming constant temperature. If the volume decreases, the pressure increases, and vice versa.

a. Identify the variables:

  • Pressure PP (in atmospheres)
  • Volume VV (in liters)

b. Inverse variation equation: PV=kP \cdot V = k Where kk is a constant.

c. Problem: If the pressure of a gas is 2 atmospheres when the volume is 6 liters, what will the pressure be if the volume decreases to 3 liters?

Solution: Using the equation PV=kP \cdot V = k, we first find kk when P=2P = 2 and V=6V = 6: 26=122 \cdot 6 = 12 Now, find the new pressure when the volume is 3 liters: P3=12P \cdot 3 = 12 Solving for PP: P=123=4 atmospheresP = \frac{12}{3} = 4 \text{ atmospheres} The pressure will increase to 4 atmospheres when the volume is reduced to 3 liters.


Would you like any further details on these examples or help with something else?

Related Questions:

  1. How does the concept of direct variation apply to speed and distance?
  2. Can you provide another real-world example of inverse variation?
  3. How does the constant kk change in direct and inverse variation problems?
  4. What happens when both variables in an inverse variation change simultaneously?
  5. How does temperature affect the relationship between pressure and volume of a gas?

Tip:

Always verify whether the relationship between variables is linear or nonlinear to properly categorize it as direct or inverse variation.

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Math Problem Analysis

Mathematical Concepts

Direct Variation
Inverse Variation

Formulas

Direct variation: y = kx
Inverse variation: xy = k

Theorems

Boyle's Law
Direct Proportionality Theorem

Suitable Grade Level

Grades 7-9