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Answer the given questions by setting up and solving the appropriate proportions. Given that \(10^{4} \mathrm{cm}^{2}=10^{6} \mathrm{mm}^{2},\) what area in square centimeters is \(2.50 \times 10^{5} \mathrm{mm}^{2} ?\)

Short Answer

Expert verified
The area is \(2.5 \times 10^{3} \text{ cm}^2\).

Step by step solution

01

Understand the given conversion

We are provided with a conversion: \(10^{4} \text{ cm}^2 = 10^{6} \text{ mm}^2\). This implies that for every \(1 \text{ cm}^2\), there are \(100 \text{ mm}^2\) as \(10^{6} \text{ mm}^2 / 10^{4} \text{ cm}^2 = 100\text{ mm}^2\).
02

Set up the proportion

To find how many square centimeters correspond to \(2.50 \times 10^{5} \text{ mm}^2\), we set up a proportion: \(\frac{10^{4} \text{ cm}^2}{10^{6} \text{ mm}^2} = \frac{x \text{ cm}^2}{2.50 \times 10^{5} \text{ mm}^2}\), where \(x\) is the area in square centimeters.
03

Solve the proportion

Cross-multiplying the proportion gives us: \[ 10^{4} \times (2.50 \times 10^{5}) = x \times 10^{6} \]Simplifying the left side, we have \[ 2.5 \times 10^{9} = x \times 10^{6}\]Divide both sides by \(10^{6}\):\[ x = \frac{2.5 \times 10^{9}}{10^{6}} = 2.5 \times 10^{3} \text{ cm}^2\].

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Key Concepts

These are the key concepts you need to understand to accurately answer the question.

Unit Conversion
Unit conversion is a fundamental part of mathematics that allows you to change a measurement from one unit to another. This seems complex, but it's simply about understanding that different units can represent the same quantity, much like speaking two different languages to communicate the same idea.
In this exercise, we are converting from square millimeters to square centimeters. From the problem, we know that \(10^4 \text{ cm}^2\) is equivalent to \(10^6 \text{ mm}^2\).
  • First, notice that you divide \(10^6 \text{ mm}^2\) by \(10^4\) to get \(100 \text{ mm}^2\) per \(1 \text{ cm}^2\).
  • This gives us a conversion factor: \(1\) square centimeter is equivalent to \(100\) square millimeters.
This knowledge helps us translate areas measured in square millimeters to square centimeters by dividing by \(100\). Understanding these conversions is crucial in fields like science and engineering, where measurements often need to switch units.
Problem Solving
Problem solving in mathematics is about identifying what you know and how it connects to what you're trying to find. In this exercise, the essential problem is finding how many square centimeters is \(2.50 \times 10^5 \text{ mm}^2\).
Following these steps can aid in problem solving:
  • Recognize the known values and conversion factors from the problem, which here includes the conversion between square centimeters and square millimeters.
  • Establish a relationship or equation based on these known values, guiding us to understand how one measurement converts to another.
  • Create a proportion, which is a simple equation where two ratios are equal, allowing us to find the unknowns by comparing known quantities.
Mathematical problem solving combines logic, understanding of concepts, and sometimes a bit of creativity to find solutions.
Cross Multiplication
Cross multiplication is a technique used to solve equations where two ratios are set equal to each other, also known as proportions. It is a powerful tool that simplifies the process of finding unknown values, especially in conversion problems.
Given our proportion \[\frac{10^4 \text{ cm}^2}{10^6 \text{ mm}^2} = \frac{x \text{ cm}^2}{2.50 \times 10^5 \text{ mm}^2}\]we cross-multiply:
- Multiply the top of one fraction by the bottom of the other and set them equal:\[10^4 \times (2.50 \times 10^5) = x \times 10^6\]
- After cross-multiplying, simplify the equation to solve for \(x\). This involves dividing both sides by \(10^6\) to isolate \(x\), finding that \(x = 2.5 \times 10^3 \text{ cm}^2\).
Using cross multiplication helps you tackle proportion problems efficiently, as it systematically eliminates the fractions, making computations straightforward.

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Most popular questions from this chapter

Solve the given applied problems involving variation. The amount of heat \(H\) required to melt ice is proportional to the mass \(m\) of ice that is melted. If it takes \(2.93 \times 10^{5} \mathrm{J}\) to melt \(875 \mathrm{g}\) of ice, how much heat is required to melt \(625 \mathrm{g} ?\)

Solve the given applied problems involving variation. The power \(P\) in an electric circuit varies jointly as the resistance \(R\) and the square of the current \(I\). If the power is \(10.0 \mathrm{W}\) when the current is \(0.500 \mathrm{A}\) and the resistance is \(40.0 \Omega\), find the power if the current is \(2.00 \mathrm{A}\) and the resistance is \(20.0 \Omega\)

Answer the given questions by setting up and solving the appropriate proportions. When a bullet is fired from a loosely held rifle, the ratio of the mass of the bullet to that of the rifle equals the negative of the reciprocal of the ratio of the velocity of the bullet to that of the rifle. If a \(3.0 \mathrm{kg}\) rifle fires a \(5.0 \mathrm{g}\) bullet and the velocity of the bullet is \(300 \mathrm{m} / \mathrm{s}\) what is the recoil velocity of the rifle?

Find the required value by setting up the general equation and then evaluating. Find \(y\) for \(x=6\) and \(z=0.5\) if \(y\) varies directly as \(x\) and inversely as \(z,\) and \(y=60\) when \(x=4\) and \(z=10\)

Solve the given applied problems involving variation. The time \(t\) required to test a computer memory unit varies directly as the square of the number \(n\) of memory cells in the unit. If a unit with 4800 memory cells can be tested in 15.0 s, how long does it take to test a unit with 8400 memory cells?

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