## Thinking Mathematically (6th Edition)

$1.2\times {{10}^{1}}$.
If a number is expressed as $a\times {{10}^{n}}$, where a is a number greater than or equal to 1 and less than 10, and n is an integer, then the number is said to be expressed in scientific notation. Hence all the four numbers, $1.2\times {{10}^{6}}$, $8.7\times {{10}^{-2}}$, $2.9\times {{10}^{6}}$ and $3\times {{10}^{-3}}$, involved in the given expression are expressed in scientific notation. Now, to compute the given expression, break the computation into three parts. First, compute the multiplications $\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)$ and $\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)$ separately, and then divide the result of the first multiplication $\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)$ by the result of the second multiplication $\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)$. Then, perform the multiplication $\left( a\times b \right)$ as the usual multiplication of any two numbers is performed and the multiplication $\left( {{10}^{x}}\times {{10}^{y}} \right)$ using the product rule for exponents, explained below. And then write the two results obtained after performing the above two multiplications, together, with the symbol $\times$ between them. Now, apply the procedure mentioned above and the product rule for exponents to compute the required multiplications: \begin{align} & \left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)=\left( 1.2\times 8.7 \right)\left( {{10}^{6}}\times {{10}^{-2}} \right) \\ & =\left( 10.44 \right)\left( {{10}^{6-2}} \right) \\ & =10.44\times {{10}^{4}} \end{align} Hence, $\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)=10.44\times {{10}^{4}}$. Similarly, \begin{align} & \left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)=\left( 2.9\times 3 \right)\left( {{10}^{6}}\times {{10}^{-3}} \right) \\ & =\left( 8.7 \right)\left( {{10}^{6-3}} \right) \\ & =8.7\times {{10}^{3}} \end{align} Hence, $\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)=8.7\times {{10}^{3}}$. Now, divide the result of the first multiplication $\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)$ by the result of the second multiplication $\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)$: $\frac{\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)}{\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)}=\frac{10.44\times {{10}^{4}}}{8.7\times {{10}^{3}}}$ Then, perform the division $\left( \frac{a}{b} \right)$ as the usual division of any two numbers is performed and the division $\left( \frac{{{10}^{x}}}{{{10}^{y}}} \right)$ using the quotient rule for exponents, explained below. And then write the two results obtained after performing the above two divisions, together, with the symbol $\times$ between them. Now, apply the procedure mentioned above and the quotient rule for exponents to compute the required division: \begin{align} & \frac{10.44\times {{10}^{4}}}{8.7\times {{10}^{3}}}=\left( \frac{10.44}{8.7} \right)\times \left( \frac{{{10}^{4}}}{{{10}^{3}}} \right) \\ & =\left( \frac{\left( 1.2 \right)\left( \right)}{} \right)\times \left( {{10}^{4-3}} \right) \\ & =1.2\times {{10}^{1}} \end{align} Hence, \begin{align} & \frac{\left( 1.2\times {{10}^{6}} \right)\left( 8.7\times {{10}^{-2}} \right)}{\left( 2.9\times {{10}^{6}} \right)\left( 3\times {{10}^{-3}} \right)}=\frac{10.44\times {{10}^{4}}}{8.7\times {{10}^{3}}} \\ & =1.2\times {{10}^{1}} \end{align}