Chapter MH, Problem 3PP

Solution

(a)

Interpretation: The following operations need to be performed.

  $\left(6.0\times {10}^{-4}\right)\times \left(4.0\times {10}^{-6}\right)$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $2.4\times {10}^{-9}$

The given operation is as follows:

  $\left(6.0\times {10}^{-4}\right)\times \left(4.0\times {10}^{-6}\right)$

The following method is used to solve it:

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

Thus,

  $\begin{array}{c}\left(6.0\times 4.0\right)\times \left({10}^{-4+\left(-6\right)}\right)=24.0\times {10}^{-4-6}\\ =24.0\times {10}^{-10}\\ =2.4\times {10}^{-9}\end{array}$

(b)

Interpretation: The following operations need to be performed.

  $\left(4.5\times {10}^9\right)\times \left(6.0\times {10}^{-10}\right)$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $2.7\times {10}^0$

The given operation is as follows:

  $\left(4.5\times {10}^9\right)\times \left(6.0\times {10}^{-10}\right)$

The following method is used to solve it:

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

Thus,

  $\begin{array}{c}\left(4.5\times 6.0\right)\times \left({10}^{9+\left(-10\right)}\right)=27.0\times {10}^{9-10}\\ =27.0\times {10}^{-1}\\ =2.7\times {10}^0\end{array}$

(c)

Interpretation: The following operations need to be performed.

  $\frac{4.5\times {10}^{-8}}{1.5\times {10}^{-4}}$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $3.0\times {10}^{-4}$

The given operation is as follows:

  $\frac{4.5\times {10}^{-8}}{1.5\times {10}^{-4}}$

The following method is used to solve it:

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

Thus,

  $\begin{array}{c}\frac{4.5\times {10}^{-8}}{1.5\times {10}^{-4}}=\frac{4.5}{1.5}\times \left({10}^{-8-\left(-4\right)}\right)\\ =3\times {10}^{-8+4}\\ =3.0\times {10}^{-4}\end{array}$

(d)

Interpretation: The following operations need to be performed.

  $\frac{9.6\times {10}^8}{1.6\times {10}^{-6}}$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $6.0\times {10}^{14}$

The given operation is as follows:

  $\frac{9.6\times {10}^8}{1.6\times {10}^{-6}}$

The following method is used to solve it:

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

Thus,

  $\begin{array}{c}\frac{9.6\times {10}^8}{1.6\times {10}^{-6}}=\frac{9.6}{1.6}\times \left({10}^{8-\left(-6\right)}\right)\\ =6.0\times {10}^{8+6}\\ =6.0\times {10}^{14}\end{array}$

(e)

Interpretation: The following operations need to be performed.

  $\frac{\left(2.5\times {10}^6\right)\left(7.2\times {10}^4\right)}{1.8\times {10}^{-5}}$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $1.0\times {10}^{16}$

The given operation is as follows:

  $\frac{\left(2.5\times {10}^6\right)\left(7.2\times {10}^4\right)}{1.8\times {10}^{-5}}$

The following methods are used to perform the operations:

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

Thus,

  $\begin{array}{c}\frac{\left(2.5\times {10}^6\right)\left(7.2\times {10}^4\right)}{1.8\times {10}^{-5}}=\left(\frac{2.5\times 7.2}{1.8}\right)\times \left({10}^{6+4-\left(-5\right)}\right)\\ =10\times {10}^{10+5}\\ =10\times {10}^{15}\\ =1.0\times {10}^{16}\end{array}$

(f)

Interpretation: The following operations need to be performed.

  $\frac{\left(6.2\times {10}^{12}\right)\left(6.0\times {10}^{-7}\right)}{1.2\times {10}^6}$

Concept Introduction: Scientific notation is used to express too-large or too-small numbers in decimal form. The general form is represented as follows:

  $m\times {10}^n$

Here, m is the decimal form of the number and ${10}^n$ is the exponent form. Here, for too-large numbers, n is positive and for too-small numbers, n is negative.

The following method is used to perform operations on numbers in scientific forms.

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

  $3.1\times {10}^0$

The given operation is as follows:

  $\frac{\left(6.2\times {10}^{12}\right)\left(6.0\times {10}^{-7}\right)}{1.2\times {10}^6}$

The following methods are used to perform the operations:

  $\left(m\times {10}^n\right)\times \left(p\times {10}^q\right)=\left(m\times p\right)\times \left({10}^{n+q}\right)$

And,

  $\frac{\left(m\times {10}^n\right)}{\left(p\times {10}^q\right)}=\left(\frac{m}{p}\right)\times \left({10}^{n-q}\right)$

Thus,

  $\begin{array}{c}\frac{\left(6.2\times {10}^{12}\right)\left(6.0\times {10}^{-7}\right)}{1.2\times {10}^6}=\left(\frac{6.2\times 6.0}{1.2}\right)\times \left({10}^{12+\left(-7\right)+\left(-6\right)}\right)\\ =31\times {10}^{12-7-6}\\ =31\times {10}^{-1}\\ =3.1\times {10}^0\end{array}$