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# Units — Harder example

Watch Sal work through a harder Units problem.

## Want to join the conversation?

• Can you dumb this down for me •   The problem is essentially asking us "if 2.1 computers can fit in 1 cubic foot, then how many computers can fit in 1 cubic meter?" [Because it asks us what is equal to the maximum density of computers in cubic meters.] So, it's just asking us to convert 2.1 computer/ft^3, one density, to the equivalence of that in cubic meters, another density. Or think of it this way: It's asking us to convert from 2.1 computers/ft^3 --> 2.1 computers/x amount of m^3 --> x amount of computers in 1 m^3.

If 2.1 computers can fit in 1 cubic foot, then that same amount of computers would be able to fit into the equivalent of 1 cubic foot in cubic meters. So, this means that we can just find out the equivalent of 1 cubic foot (ft^3) in cubic meters (m^3) and then just replace the ft^3 in 2.1 computers/ft^3 [because this is what we're trying to convert] with what ft^3 is in m^3.

From there, we can just simplify the fraction and this will tell us how many computers would fit in just 1 cubic meter. For the first step, the problem already tells us that 1 meter = 3.28 feet. So we can just manipulate the equation to turn ft to ft^3 so that we can substitute ft^3 in 2.1 computers/ft^3.

1 meter = 3.28 feet
1 meter/3.28 = feet
(1 meter/3.28)^3 = (1 feet)^3
meter^3/3.28^3 = feet^3
m^3/3.28^3 = ft^3

So, 1 cubic foot = (m^3)/(3.28^3).

Substitute ft^3 in the measurement we're trying to convert.
2.1 computers / ft^3
2.1 computers / (m^3/3.28^3) Substitute.
(2.1 computers) x (3.28^3/m^3) Multiply
(2.1 computers x 3.28^3)/(m^3) Condense
(74.1038592 computers)/(m^3) Simplify

We know this is correct because it has our conversion in the desired units of computers per cubic meter (or computers/m^3). So the density 74.10318592 is the maximum number of computers per cubic meter.
• Me the whole time . . so confused
o • 1meter-3.28feet
Cube both sides, (since we are asked to leave our answer in computers per cubicmeter)

.:1cubicmeter-35.29cubicfeet

and we were told that the room must not exceed 2.1 computers per cubicfoot, i.e,

2.1computers-1cubicfoot.

now, the question is... how many computers will contain 1cubicmeters and 1cubicmeters is equal to 35.29cubicfeet

so, translating the above we have, how may computers will contain 35.29cubicfeet,

and 2.1coputers-1cubicfoot
xcomputer-35.29cubicfeet (i.e 1cubicmeter)
so, when we cross-multiply, we have x to be equal to 35.29 X 2.1 = 74.10computers per 35.29cubicfeet
= 74.10computers per 1cubicmeter.
• I get where he got 35.28ft^3 but why did he multiply it by 2.1ft^3 ? • im confused as to why he multiplied the 3.28^3 with 2.1, can anyone explain why ? • I don't understand why he multiplied by density? • Good Question. The density in this problem gives a ratio for the number of computers to cubic feet.One of the reasons he multiplied by the density was to get the units (ft3) to cancel, which leaves the cubic meters by themselves. Also, this method, known as the "factor label method" is a common means on unit conversion, which is what this problem involves (converting computers per cubic foot into computers per cubic meter.
(1 vote)
• I do not understand diagrams like this too well...how do you manually convert everything (If that makes sense) • I don't understand how he multiplied 2.1 x the volume. Could we have have divided the volume by how many computers we have to fit in?? • One way to look at this:

Given:
(1ft)^3 = 2.1 ----1
1m = 3.28ft ---- 2

Working:
1ft = (1/3.28)m ---- 2'
[(1/3.28)m]^3 = 2.1 ---- 2' into 1

[1/(3.28)^3]m^3 = 2.1
m^3 = 2.1(3.28)^3   