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Chapter 1: Problem 4

Change the numbers from scientific notation to ordinary notation. $$6.8 \times 10^{7}$$

Short Answer

Expert verified
The ordinary notation of \(6.8 \times 10^{7}\) is 68,000,000.

Step by step solution

01

Understanding Scientific Notation

Scientific notation is a way to express very large or very small numbers. In scientific notation, a number is written as the product of a coefficient (a number between 1 and 10) and a power of ten. For example, \(6.8 \times 10^{7}\) means \(6.8\) multiplied by \(10\) raised to the power of \(7\).
02

Applying the Power of Ten

The expression \(10^{7}\) represents 10 million (10 followed by 7 zeros). This means we need to multiply \(6.8\) by 10 million.
03

Multiplying the Numbers

Now, multiply \(6.8\) by \(10,000,000\). This can be done by moving the decimal point in \(6.8\) seven places to the right. This turns \(6.8\) into \(68,000,000\).So, \(6.8 \times 10^{7} = 68,000,000\).

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Key Concepts

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

Expressing Large Numbers
When dealing with large numbers, scientific notation becomes an essential tool. It simplifies the way we write and work with exceedingly large or small figures. Rather than jotting down all the zeros manually, we use a straightforward approach combining a coefficient and a ten to an exponent. For instance, a number like 68,000,000 can be expressively written as \(6.8 \times 10^7\) in scientific notation. This method not only keeps numbers compact but also aids in quick calculations and comparisons.

To express a large number in scientific notation, follow these steps:
  • Identify the coefficient by moving the decimal point to create a number between 1 and 10.
  • Count the number of places the decimal point has moved鈥攖his becomes the exponent of 10.
  • Combine them in the format: coefficient \(\times 10^{\text{exponent}}\).
This approach helps prevent errors when transcribing or calculating large numbers.
Multiplying by Powers of Ten
Multiplying by powers of ten is a fundamental math concept that simplifies the process of scaling numbers up. When you multiply a number by ten raised to a certain power, you're essentially moving the decimal point to the right for each step corresponding to the power. For example, if you multiply a number by \(10^3\), you're moving the decimal three steps to the right, effectively multiplying the number by 1,000.

Here's how it works:
  • For every power of ten, the decimal point moves one place to the right.
  • \(10^1 = 10\); move the decimal one place.
  • \(10^2 = 100\); move it two places.
  • \(10^3 = 1,000\); move it three places, and so on.


Using powers of ten not only simplifies multiplication but also keeps numbers organized and easy to handle.
Converting to Ordinary Notation
Converting scientific notation back to ordinary notation involves reversing the process of expressing large numbers. This practical skill is vital as it allows us to visualize how big or small a number really is in its full form. Suppose we have a number in scientific notation like \(6.8 \times 10^7\). To convert it to its ordinary form, follow these steps:

  • Understand the power of ten in the expression. Here, \(10^7\) signifies moving the decimal point 7 places to the right.
  • Shift the decimal point in the original coefficient, 6.8, seven places to the right.
  • Add zeroes if necessary to fill in for the shifted spaces.
  • Write the final number. For \(6.8 \times 10^7\), the converted form is 68,000,000.


By mastering this conversion, you ensure that scientific notation and ordinary numbers can be easily interchanged, assisting in a variety of mathematical tasks.

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Most popular questions from this chapter

Find the indicated values. For a car's cooling system, the equation \(p(C-n)+n=A\) is used. If \(p=0.25, C=15.0 L,\) and \(A=13.0 L,\) solve for \(n\) \((\text { in } L)\)

Find the value of each square root by use of a calculator. Each number is approximate. (a) \(\sqrt{0.0429^{2}-0.0183^{2}}\) (b) \(\sqrt{0.0429^{2}}-\sqrt{0.0183^{2}}\)

Find the value of each square root by use of a calculator. Each number is approximate. (a) \(\sqrt{1296+2304}\) (b) \(\sqrt{1296}+\sqrt{2304}\)

Solve the given problems. The speed (in \(\mathrm{m} / \mathrm{s}\) ) of sound in seawater is found by evaluating \(\sqrt{B / d}\) for \(B=2.18 \times 10^{9} \mathrm{Pa}\) and \(d=1.03 \times 10^{3} \mathrm{kg} / \mathrm{m}^{3} .\) Find this speed, which is important in locating underwater objects using sonar.

Make the indicated coversions. $$15.7 \mathrm{qt} \text { to } \mathrm{L}$$
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