Chapter 3.3, Problem 7GP

Solution

The total number of heartbeats over the lifetime of a white rhino.

  $7.919\times {10}^8$

Given:

The total number of heartbeats over a lifetime for a white rhino is given by the function $\left(1.446\times {10}^9\right)m^{-0.05}$ . The mass of the body is $m=1.7\times {10}^5$ kilograms.

Concept Used:

To write a number in scientific notation, move the decimal point after the first nonzero integer and raise the exponent of 10 by the number of digits we moved the decimal point.

If the decimal point was moved to the left, then raise the positive exponent and if the decimal point moved to the right, then raise the negative exponent.

Calculation:

In order to find the white rhino’s number of heartbeats over its lifetime, use the given function and substitute $m=1.7\times {10}^5$ into it and simplify further as shown below

  $\begin{array}{c}\left(1.446\times {10}^9\right)m^{-0.05}=\left(1.446\times {10}^9\right){\left(1.7\times {10}^5\right)}^{-0.05}\\ =\frac{1.446\times {10}^9}{{\left(1.7\times {10}^5\right)}^{0.05}}\\ =\frac{1.446\times {10}^9}{{\left(170000\right)}^{0.05}}\\ =\frac{1.446\times {10}^9}{1.82609}\\ =\frac{1.446}{1.82609}\times {10}^9\\ =0.7919\times {10}^9\\ =7.919\times {10}^8\end{array}$

Thus, the total number of heartbeats over the lifetime of a white rhino is $7.919\times {10}^8$ .