Pearson Algebra 1 Common Core, 2011
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Pearson Algebra 1 Common Core, 2011 View details
4. Division Properties of Exponents
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Exercise 93 Page 445

Pay attention to units. There are 10^9 cubic meters in one cubic kilometer.

Approximately 1.6 * 10^6 grams per cubic meter

Practice makes perfect
The density of an object is defined as its mass divided by its volume. We are given Neptune's mass in kilograms and its radius in kilometers. We can start by finding the volume of Neptune. Since Neptune is spherical, its volume can be expressed as V= 43Ď€ r^3, where r is the radius.
V=4/3Ď€ r^3
V=4/3Ď€ ( 2.48 * 10^4)^3
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Simplify
V=4/3Ď€ (2.48)^3 * (10^4)^3
V=4/3Ď€ (2.48)^3 * 10^(12)
V=4/3Ď€ (15.25)* 10^(12)
V=63.88 * 10^(12)
We can write this in scientific notation by shifting the decimal one point over. 63.88 * 10^(12) ⇒ 6.388 * 10^(13)Thus, we have found that the volume of Neptune is 6.388 * 10^(13) cubic kilometers. We are almost ready to compute the density. Before we do that, however, we should adjust our units to fit the problem. The problem asks for density in grams per cubic meter. D=grams/cubic meters Currently our mass is in kilograms, so this must be changed to grams. There are 10^3 grams in one kilogram. 1.02 * 10^(26) kg * 10^3 g/1kg = 1.02 * 10^(29) g Our mass is 1.02 * 10^(29) grams. Now, let's convert volume from cubic kilometers to cubic meters. There are 10^9 cubic meters in one cubic kilometer. 6.388 * 10^(13) km^3 * 10^9 m^3/1km^3 = 6.388 * 10^(22) m^3 Our volume is 6.388 * 10^(22) cubic meters. Now that we have the proper units, we can finally compute the density by dividing mass by volume.
D=Mass/Volume
D=1.02 * 10^(29)/6.388 * 10^(22)
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Simplify right-hand side
D=1.02/6.388 * 10^(29)/10^(22)
D=1.02/6.388 * 10^7
D=0.16 * 10^7
This can be written in scientific notation by shifting the decimal one point to the right. 0.16 * 10^7 ⇒ 1.6 * 10^6 We have found that the density of Neptune is approximately 1.6 * 10^6 grams per cubic meter.