Convert the following into correct scientific notation: (a) \(14.3 \times 10^{1} ; (b) 851 \times 10^{-2}\) (c) \(7500 \times 10^{-3}\)

In scientific notation, a coefficient is a number between 1 and 10. It's the part of the notation that you multiply by a power of 10. For example, in \(3.5 \times 10^4\), 3.5 is the coefficient.
Co-efficients in scientific notation help make larger or smaller numbers easier to read. This format helps us express very large or small values efficiently. The first step in converting to scientific notation is ensuring the coefficient is in the correct form. This often means moving the decimal point to create a number between 1 and 10.
Let's see a practical example with the number 851. If we move the decimal two places to the left, we convert 851 into 8.51. Now, 8.51 is our coefficient because it falls between 1 and 10.

The power of 10 in scientific notation shows how many times you would multiply or divide by 10. It's expressed as an exponent, like in \(1.2 \times 10^3\). Here, the power of 10 is 3.
If the exponent is positive, it indicates the number of places the decimal point moves to the right. For instance, moving the decimal point 3 places to the right in 1.2 would give you 1200.
If the exponent is negative, it indicates the number of places the decimal point moves to the left. For example, in 3.4 \( \times 10^{-3} \), moving the decimal point 3 places to the left turns 3.4 into 0.0034.
In the given exercise, when converting 7500 \( \times 10^{-3} \), you move the decimal point three places to the left, so 7500 becomes 7.5, and the exponent is adjusted accordingly.

Adjusting the exponent is key to converting numbers into correct scientific notation. As we move the decimal point, we need to adjust the exponent to reflect that movement correctly.
For example, in the solution of 14.3 \( \times 10^{1} \), we move the decimal point one place to the left to get 1.43. To keep the value the same, we add 1 to the original exponent of 1. So, \(14.3 \times 10^{1}\) becomes \(1.43 \times 10^{2}\).
Consider another example with 851 \( \times 10^{-2} \). Moving the decimal two places to the left changes 851 to 8.51. Hence, we add 2 to the original exponent of \(-2\), resulting in \(8.51 \times 10^{0}\). The same principle applies when converting 7500 \( \times 10^{-3} \), adjusting the exponent accordingly to maintain the value in scientific notation.