Chapter MH, Problem 6PP

Solution

(a)

Interpretation: The following operations need to be performed.

  $\left(2.475\text{ m}\right)+\left(3.5\text{ m}\right)+\left(\text{4}\text{.65 m}\right)$

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,

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

  $\text{8}\text{.958 m}$

The given operation is as follows:

  $\left(2.475\text{ m}\right)+\left(3.5\text{ m}\right)+\left(\text{4}\text{.65 m}\right)$

The given numbers can be simply added as follows:

  $\left(2.475\text{ m}\right)+\left(3.5\text{ m}\right)+\left(\text{4}\text{.65 m}\right)=\left(2.475+3.5+4.65\right)\text{ m}$

Thus,

  $\left(2.475+3.5+4.65\right)\text{ m}=\text{8}\text{.958 m}$

(b)

Interpretation: The following operations need to be performed.

  $\left(3.45\text{ m}\right)+\left(3.658\text{ m}\right)+\left(\text{47 m}\right)$

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,

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

  $\text{54}\text{.108 m}$

The given operation is as follows:

  $\left(3.45\text{ m}\right)+\left(3.658\text{ m}\right)+\left(\text{47 m}\right)$

The given numbers can be simply added as follows:

  $\left(3.45\text{ m}\right)+\left(3.658\text{ m}\right)+\left(\text{47 m}\right)=\left(3.45+3.658+47\right)\text{ m}$

Thus,

  $\left(3.45+3.658+47\right)\text{ m}=\text{54}\text{.108 m}$

(c)

Interpretation: The following operations need to be performed.

  $\left(5.36\times {10}^{-4}\text{ g}\right)-\left(6.381\times {10}^{-5}\text{ g}\right)$

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,

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

  $-5.845\times {10}^{-3}$

The given operation is as follows:

  $\left(5.36\times {10}^{-4}\text{ g}\right)-\left(6.381\times {10}^{-5}\text{ g}\right)$

The above expression can be rewritten as follows:

  $\left(5.36\times {10}^{-4}\text{ g}\right)-\left(63.81\times {10}^{-4}\text{ g}\right)$

Taking ${10}^{-4}$ common,

  $\left(5.36-63.81\right)\left({10}^{-4}\right)$

Or,

  $-58.45\times {10}^{-4}$

Or,

  $-5.845\times {10}^{-3}$

(d)

Interpretation: The following operations need to be performed.

  $\left(6.46\times {10}^{12}\text{ m}\right)-\left(6.32\times {10}^{11}\text{ m}\right)$

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,

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

  $5.828\times {10}^{12}\text{ m}$

The given operation is as follows:

  $\left(6.46\times {10}^{12}\text{ m}\right)-\left(6.32\times {10}^{11}\text{ m}\right)$

The above expression can be rewritten as follows:

  $\left(6.46\times {10}^{12}\text{ m}\right)-\left(0.632\times {10}^{12}\text{ m}\right)$

Taking ${10}^{12}$ common,

  $\left(6.46-0.632\right)\times {10}^{12}\text{ m}$

Or,

  $5.828\times {10}^{12}\text{ m}$

(e)

Interpretation: The following operations need to be performed.

  $\left(6.6\times {10}^{12}\text{ m}\right)\times \left(5.34\times {10}^{18}\text{ m}\right)$

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,

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

  $\text{3}\text{.52}\times {\text{10}}^{31}m$

The given operation is as follows:

  $\left(6.6\times {10}^{12}\text{ m}\right)\times \left(5.34\times {10}^{18}\text{ m}\right)$

The following method is used to solve the above expression.

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

Thus,

  $\left(6.6\times 5.34\right){10}^{12+18}\text{ m}$

Or,

  $\begin{array}{c}\left(6.6\times 5.34\right)\times {10}^{12+18}\text{ m}=35.244\times {10}^{30}\text{ m}\\ =\text{3}\text{.52}\times {\text{10}}^{31}m\end{array}$

(f)

Interpretation: The following operations need to be performed.

  $\frac{5.634\times {10}^{11}\text{ m}}{3.0\times {10}^{12}\text{ m}}$

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,

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

  $1.878\times {10}^{-1}\text{ m}$

The given operation is as follows:

  $\frac{5.634\times {10}^{11}\text{ m}}{3.0\times {10}^{12}\text{ m}}$

The following method is used to solve the above expression.

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

Thus,

  $\left(\frac{5.634}{3.0}\right)\times {10}^{11-12}\text{ m}=1.878\times {10}^{-1}\text{ m}$

(g)

Interpretation: The following operations need to be performed.

  $\frac{\left(4.765\times {10}^{11}\text{ m}\right)\left(5.3\times {10}^{-4}\text{ m}\right)}{7.0\times {10}^{-5}\text{ m}}$

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,

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

  $3.6\times {10}^{12}\text{ m}$

The given operation is as follows:

  $\frac{\left(4.765\times {10}^{11}\text{ m}\right)\left(5.3\times {10}^{-4}\text{ m}\right)}{7.0\times {10}^{-5}\text{ m}}$

The following method is used to solve the above expression.

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

Or,

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

Thus,

  $\begin{array}{c}\frac{\left(4.765\times {10}^{11}\text{ m}\right)\left(5.3\times {10}^{-4}\text{ m}\right)}{7.0\times {10}^{-5}\text{ m}}=\left(\frac{4.765\times 5.3}{7.0}\right)\times {10}^{11+\left(-4\right)-\left(-5\right)}\text{ m}\\ =3.6\times {10}^{11-4+5}\text{ m}\\ =3.6\times {10}^{12}\text{ m}\end{array}$