[latex]\Large\text{imperial }\rightarrow\text{ imperial}[/latex]
Or we can convert units of measurements such as the following...[latex]\Large1\text{ foot}=0.304\text{ metres}[/latex]
[latex]\Large\text{imperial }\rightarrow\text{ metric}[/latex]
In the trades we may get used to a certain way of doing things but understanding the terminology behind systems of measurement can help us when we work at jobs where the language is different. Once again, what we are doing is expanding our math vocabulary to help us understand the language of math. I’ll give you an example. When gas fitters learn about energy (or heat) calculations, they are given two numbers to work with. The imperial number is the British Thermal Units (BTU’s) while the metric version is kilowatts (kW). Students often have a tough time in the beginning getting the relationship right between the two. Their frustration builds and eventually they begin to wonder why they have to learn two different ways to say the same thing in the first place. The answer lies in the fact that in the gas fitting trades, students will run across appliances that are designated in both BTU’s and kW’s. If they are unable to work between the two, then there could be serious safety consequences if those appliances receive too little or, more importantly, too much gas. For reasons such as the one mentioned above, it's important to know what the different units of measurement mean as well as how to work between them. I know from my experience as a trades instructor that once students begin to understand the language of measurement and are able to both visualize and apply this language, the concepts of math and measurement become a lot easier. I think the first thing we should do at this point is go through the history of measurement and how the metric and imperial systems came to be.Unit | Multiplier |
---|---|
kilometre | 1,000 |
hectometre | 100 |
decametre | 10 |
metre (base unit) | 1 |
decimetre | 0.1 |
centimetre | 0.01 |
millimetre | 0.001 |
Common metric prefix | Multiplier |
---|---|
yotta | 1,000,000,000,000,000,000,000,000 |
zetta | 1,000,000,000,000,000,000,000 |
exa | 1,000,000,000,000,000,000 |
peta | 1,000,000,000,000,000 |
tera | 1,000,000,000,000 |
giga | 1,000,000,000 |
mega | 1,000,000 |
kilo | 1,000 |
hecto | 100 |
deca | 10 |
metre (base unit) | 1 |
deci | 0.1 |
centi | 0.01 |
milli | 0.001 |
micro | 0.000001 |
nano | 0.000000001 |
pico | 0.000000000001 |
femto | 0.000000000000001 |
atto | 0.000000000000000001 |
zepto | 0.000000000000000000001 |
yocto | 0.000000000000000000000001 |
Example
[latex]\Large2.3 \text{ metres}= \text{X centimetres}[/latex]
Similar to how we did things in the first four chapters we will go about this in steps. Step 1: Find the multiplier What we see is that going from centimetres to metres the multiplier is 0.01. What this is saying is a centimetre is 1/100th of a metre or that there are 100 centimetres in a metre. It’s important here to note that a centimetre is smaller than a metre and as this is the case then we would expect our answer to decrease. Step 2: Build a ratio[latex]\Large \dfrac{1\text{ m}}{2.3\text{ m}} = \dfrac{100\text{ cm}}{\text{X cm}}[/latex]
What this ratio states is that if 1 metre is equal to 100 centimetres then 2.3 metres is equal to X centimetres. Step 3: Cross multiply.[latex]\Large \begin{array}{c} \dfrac{1\text{ m}}{2.3\text{ m}} = \dfrac{100\text{ cm}}{\text{X cm}} \\ 1 \times \text{X} = 2.3 \times 100 \\ \text{X}=230 \\ \text{Answer}= 230\text{ centimetres}\end{array}[/latex]
We'll try another example.Example
[latex]\Large\text{multiplier} = 1000[/latex]
[latex]\Large1 \text{ kilometre} = 1000 \text{ metres}[/latex]
Step 2: Build a ratio[latex]\Large\dfrac{1 \text{ km}}{\text{X km}} = \dfrac{1000 \text{ m}}{1057 \text{ m}}[/latex]
Step 3: Cross multiply.[latex]\Large\begin{array}{c} \dfrac{1 \text{ km}}{\text{X km}}= \dfrac{1000 \text{ m}}{1057 \text{ m}} \\ 1\times 1057 = \text{X} \times 1000 \\ \text{X} = \dfrac{1057}{1000}=1.057 \\ \text{Answer} = 1.057 \text{ metres}\end{array}[/latex]
Question 1
Unit Name | Equivalent Values |
---|---|
Inch | 0.083 feet. 0.028 yards |
foot | 12 inches, 0.333 yards |
yard | 3 feet, 36 inches |
fathom | 6 feet, 72 inches |
rod | 5.50 yards, 16.5 feet |
furlong | 660 feet, 220 yards, 1/8 mile |
mile | 5280 feet, 1760 yards, 320 rods |
Nautical mile | 6,076 feet, 1.151 miles |
Example
[latex]\Large1\text{ foot}= 12 \text{ inches}[/latex]
Step 2: Build a ratio[latex]\Large\dfrac{1 \text{ foot}}{2 \text{ feet}}=\dfrac{12 \text{ inches}}{\text{X inches}}[/latex]
Step 3: Cross multiply[latex]\Large\begin{array}{c}\dfrac{1 \text{ foot}}{2 \text{ feet}}=\dfrac{12 \text{ inches}}{\text{X inches}} \\ 1\times \text{X} = 2 \times 12 \\ \text{X} = 24 \\ \text{Answer} = 24 \text{ inches}\end{array} [/latex]
Although you might have been able to do that in your head, it’s important to follow the steps involved and think about the answer you expect to get. This will help when the numbers are more involved and not as easy to figure out.Example
[latex]\Large 1 \text{ yard} = 36 \text{ inches}[/latex]
[latex] \Large \text{OR} [/latex]
[latex]\Large 1 \text{ inch}= 0.028 \text{ yards}[/latex]
In this question we are going from inches to yards so working with the number 0.028 will be easier for us. Step 2: Build a ratio[latex]\Large\dfrac{ 1 \text{ inch}}{247 \text{ inches}}= \dfrac{0.028 \text{ yards}}{\text{X yards}}[/latex]
Step 3: Cross multiply[latex]\Large\begin{array}{c}\dfrac{ 1 \text{ inch}}{247 \text{ inches}}= \dfrac{0.028 \text{ yards}}{\text{X yards}} \\ 1 \times \text{X} = 247 \times 0.028 \\ \text{X}= 6.916 \\ \text{Answer} = 6.916 \text{ yards}\end{array}[/latex]
Question 1
Metric | Imperial Equivalent |
---|---|
1 metre | 3.28 feet |
1 kilometre | 0.62 miles |
1 centimetre | 0.393 inches |
1 millimetre | 0.0394 inches |
Imperial | Metric Equivalent |
---|---|
1 foot | 0.305 metres |
1 mile | 1.61 kilometres |
1 inch | 2.54 centimetres |
1 inch | 25.4 millimetres |
[latex]\Large 1 \text{ mile} = 1.61 \text{ kilometres} [/latex]
Now what we need to figure out is the reverse. In this case how many miles there are in one kilometre. Once again ask yourself whether you think the answer should be bigger or smaller than 1. So to figure out our answer we need to do the following.[latex]\Large\begin{array}{c} \# \text{ kilometres} = \# \text{ miles} \times 1.61 \\ \downarrow \\ \# \text{ miles} = \dfrac{\# \text{ kilometres}}{1.61} \\ \downarrow \\ 1 \text{ mile} = 0.62 \text{ kilometres}\end{array}[/latex]
So we end up with 1 kilometre equaling 0.62 miles. We’ve just taken one constant to derive the other constant. You can do this with any of the numbers used to translate back and forth between metric and imperial. Let’s move on. Now what we will do is start to work between the imperial and metric systems and the easiest way to do this is by going through some example questions.Example
[latex]\Large\begin{array}{c} 1\text{ metre} = 3.28 \text{ feet} \\ 1 \text{ foot} = 0.305 \text{ metres}\end{array}[/latex]
As we are going from feet to metres we’ll go with 1 foot = 0.305 metres. Step 2: Build a ratio[latex]\Large \dfrac{1 \text{ foot}}{42 \text{ feet}}= \dfrac{0.305 \text{ metres}}{\text{X metres}} [/latex]
[latex]\Large\begin{array}{c} \dfrac{1 \text{ foot}}{42 \text{ feet}}= \dfrac{0.305 \text{ metres}}{\text{X metres}} \\ 1\times \text{X} = 42 \times 0.305 \\ \text{X} = 12.81 \\ \text{Answer} = 12.81 \text{metres}\end{array} [/latex]
Example
[latex]\Large\begin{array}{c} 1 \text{ centimetre}= 0.393 \text{ inches} \\ 1 \text{ inch} = 2.54 \text{ centimetres}\end{array} [/latex]
As we are going from centimetres to inches we’ll go with 1 centimetre = 0.393 inches Step 2: Build a ratio[latex]\Large \dfrac{1 \text{ cm}}{100 \text{ cm}}=\dfrac{0.393 \text{ in}}{\text{X in}}[/latex]
Step 3: Cross multiply[latex]\Large\begin{array}{c} \dfrac{1 \text{ cm}}{100 \text{ cm}}=\dfrac{0.393 \text{ in}}{\text{X in}} \\ 1 \times \text{X} = 100 \times 0.393 \\ \text{X} = 39.3 \\ \text{Answer} = 39.3 \text{cm}\end{array} [/latex]
Question 1
Question 2
Unit | Multiplier |
---|---|
kilogram (kg) | 1,000 |
gram (base unit) (g) | 1 |
centigram (cg) | 0.01 |
milligram (mg) | 0.001 |
[latex]\Large 1 \text{ metric tonne}= 1000 \text{ kilograms}[/latex]
Just for fun we’ll write down the metric weight of a few common (or maybe uncommon) items.Example | Weight |
---|---|
Usain Bolt (sprinter from Jamaica) | 94 kilograms |
Hummingbird | 4 grams (average) |
Grizzly bear | 300 kilograms (adult male average) |
An electron | 9.109×10−31 kilograms (very light) |
The sun | 1.989×10^{30} kilograms (very heavy) |
Example
[latex]\Large \dfrac{1 \text{ gram}}{48 \text{ grams}} = \dfrac{1000 \text{ mg}}{\text{X mg}}[/latex]
Step 3: Cross multiply.[latex]\Large \dfrac{1 \text{ gram}}{48 \text{ grams}} = \dfrac{1000 \text{ mg}}{\text{X mg}}[/latex]
[latex]\Large 1 \times \text{X} = 48 \times 1000[/latex]
[latex]\Large \text{x}= 48,000 [/latex]
[latex]\Large \text{Answer} = 48,000 \text{ grams}[/latex]
Examples
[latex]\Large \dfrac{1 \text{ tonne}}{2.4 \text{ tonnes}} = \dfrac{1000 \text{ kg}}{\text{X kg}}[/latex]
Step 3: Cross multiply[latex]\Large \begin{array}{c}\dfrac{1 \text{ tonne}}{2.4 \text{ tonnes}} = \dfrac{1000 \text{ kg}}{\text{X kg}} \\ 1 \times \text{X} = 2.4 \times 1000 \\ \text{X} = 2,400 \\ \text{Answer} = 2,400 \text{ kilograms}\end{array}[/latex]
Question 1
Unit Name | Equivalent Values |
---|---|
ounce (oz) | 1/16 or 0.0625 pounds |
pound (lb) | 16 ounces |
ton | 2000 pounds |
Example | Weight |
---|---|
Usain Bolt (sprinter from Jamaica) | 206.8 pounds |
Hummingbird | 0.14 ounces (average) |
Grizzly bear | 660 pounds (adult male average) |
An electron | 2 ×10^{−30 }pounds (still light) |
The sun | 4.385 ×10^{3}^{0 }pounds (still heavy) |
Example
[latex]\Large \dfrac{1 \text{ pound}}{61 \text{ pounds}}= \dfrac{16 \text{ ounces}}{\text{X ounces}}[/latex]
Step 3: Cross multiply[latex]\Large \begin{array}{c} \dfrac{1 \text{ pound}}{61 \text{ pound}}= \dfrac{16 \text{ ounces}}{\text{X ounces}} \\ 1 \times \text{X} = 61 \times 16 \\ \text{X} = 976 \\ \text{Answer} = 976 \text{ ounces}\end{array}[/latex]
Examples
[latex]\Large \begin{array}{c} \text{Feet}= \text{pieces} \times 8 \text{ feet/piece} \\ \text{Feet}= 103 \times 8 \\ \text{Feet}= 824 \end{array} [/latex]
Step 2: Find the total weight of the wood in ounces.[latex]\Large\begin{array}{c} \text{ounces}= \text{feet} \times 25.6 \text{ ounces/foot} \\ \text{ounces}= 824 \times 25.6 \\ \text{ounces}= 21,094\end{array} [/latex]
Step 3: Find the number that works between ounces and pounds remembering that we are going from ounces to pounds. In this case 1 ounce = 1/16 or 0.0625 pounds Step 4: Build a ratio[latex]\Large \dfrac{1 \text{ oz}}{21,094 \text{ oz}}= \dfrac{0.0625 \text{ lbs}}{\text{X lbs}} [/latex]
Step 5: Cross multiply[latex]\Large\begin{array}{c} \dfrac{1 \text{ oz}}{21,094 \text{ oz}}= \dfrac{0.0625 \text{ lbs}}{ \text{X lbs}} \\ 1 \times \text{X} = 21,094 \times 0.0625 \\ \text{X} = 1318.38 \\ \text{Answer} = 1318.38 \text{ lbs}\end{array}[/latex]
Question 1
Metric | Imperial Equivalent |
---|---|
kilogram (kg) | 2.2 pounds |
gram (g) | 0.035 ounces |
tonne | 2200 pounds |
Imperial | Metric Equivalent |
---|---|
pound (lb) | 0.454 kilograms |
ounce (oz) | 28.35 grams |
ton | 909 kilograms |
Example
[latex]\Large \dfrac{1 \text{ lb}}{410 \text{ lbs}}= \dfrac{0.454 \text{ kg}}{\text{X kg}}[/latex]
Step 3: Cross multiply[latex]\Large\begin{array}{c} \dfrac{1 \text{ lb}}{410 \text{ lbs}}= \dfrac{0.454 \text{ kg}}{\text{X kg}} \\ 1 \times \text{X}= 410 \times 0.454 \\ \text{X}= 186.14 \\ \text{Answer}= 186.14 \text{ kg}\end{array}[/latex]
Example
[latex]\Large \dfrac{1 \text{ oz}}{10.6 \text{ oz}}= \dfrac{28.35 \text{ g}}{\text{X g}}[/latex]
Step 3: Cross multiply[latex]\Large\begin{array}{c} \dfrac{1 \text{ oz}}{10.6 \text{ oz}}= \dfrac{28.35 \text{ g}}{\text{X g}} \\ 1 \times \text{X}= 10.6 \times 28.35 \\ \text{X}= 300.51 \\ \text{Answer}= 300.51 \text{ g}\end{array}[/latex]
Example
Looking back at our table we find that we don’t have a number that translates from pounds into grams. We do have a number that translates from pounds into kilograms though. We’ll start with that.
We see that 1 pound = 0.454 kilograms. We also see that 1 kilogram = 1000 grams. We can use both these numbers to calculate the number of grams. Step 2: Build a ratio from pounds into kilograms and cross multiply.[latex]\Large \begin{array}{c} \dfrac{1 \text{ lb}}{10 \text{ lbs}}= \dfrac{0.454 \text{ kg}}{\text{X kg}} \\ 1 \times \text{X}= 10 \times 0.454 \\ \text{X}= 4.54 \\ \text{Answer}= 4.54 \text{ kg}\end{array}[/latex]
Step 3: Translate kilograms into grams.[latex]\Large\begin{array}{c}\dfrac{1 \text{ kg}}{4.54 \text{ kg}}= \dfrac{1000 \text{ g}}{\text{X g}} \\ 1 \times \text{X}= 4.54 \times 1000 \\ \text{X}= 4540 \\ \text{Answer}= 4540 \text{ g}\end{array}[/latex]
Almost there!!!! Step 4: Take the number of grams in a 10 pound bag of flour and divide it by the number of grams of flour it takes to make one loaf of European crusty bread.[latex] \Large \begin{array}{rcl} \text{loaves of bread}& = & \dfrac{\text{grams in a 10 pound bag of flour}}{\text{grams in loaf of bread}} \\ & = & \dfrac{4540}{300.5} \\ & = & \text{15.10 loaves of bread}\end{array}[/latex]
Using our rounding skills developed in the earlier chapters we would round that down to 15 loaves. In fact, due to loss and waste during the bread making process we could assume that we would make even less loaves that that.Question 1
If you would like to know more about the different forms of energy check out the following link: Types of Energy (Solar Schools)
Metric | Imperial |
---|---|
kilowatts (kW) | British thermal unit (btu) |
calories (cal) | joules (J) |
Unit | Equivalent |
---|---|
kilowatt | 3412 btu's 860,421 cal |
British thermal unit | 0.293 kW 1055 J |
calorie | 4.186 J |
joules | 0.239 cal |
[latex] \begin{array}{rcl} \text{1 kilocalorie (1000 calories)} & \longrightarrow & \text{raises 1 kilogram (1000 grams) of water 1°C} \\ \text{1 calorie}& \longrightarrow & \text{raises 1 gram of water 1°C} \end{array}[/latex]
This same logic follows for the joule. One kilojoule equals 1000 joules. If we relate it to increasing the temperature of water what we would see is that is takes 1 joule to raise 1 gram of water 0.24°C. Following the same logic as the calorie and kilocalorie it would take 1 kilojoule to raise 1 kilogram of water 0.24°C.[latex] \begin{array}{rcl} \text{1 kilojoule (1000 joules)} & \longrightarrow & \text{raises 1 kilogram (1000 grams) of water 0.24°C} \\ \text{1 joule}& \longrightarrow & \text{raises 1 gram of water 0.24°C} \end{array}[/latex]
We can take this a bit further and talk about the heat energy in a British thermal unit. The heat energy in one BTU is enough to raise one pound of water 1 degree Fahrenheit. It’s generally considered to be the amount of heat energy in a match.Example
[latex]\Large \dfrac{\text {1 BTU}}{\text{14 BTU's}} = \dfrac{\text{1055 joules}}{\text{X joules}}[/latex]
Step 3: Cross multiply.[latex]\Large\begin{array}{c} \dfrac{\text {1 BTU}}{\text{14 BTU's}} = \dfrac{\text{1055 joules}}{\text{X joules}} \\ 1 \times \text{X} = 14 \times 1055 \\ \text{X} = 14,470 \\ \text{Answer} = 14,470 \text{ joules}\end{array}[/latex]
Example
[latex]\Large \dfrac{\text {1 kW}}{\text{X kW}} = \dfrac{\text{860,421 cal}}{\text{1,495,276 cal}}[/latex]
What you'll note here is that when you get to the stage where you cross multiply the equation doesn't end up as nice and easy to work with as we've had so far. We don't end up with the "1 x X." What we have to do here is manipulate the equation to solve for X. This is a little out of the scope of what we have gone through so far but you'll find a full explanation in the next chapter. Having said all that let's go through the motions to solve the equation. Step 3: Cross multiply[latex]\Large\begin{array}{c}\dfrac{\text {1 kW}}{\text{X kW}} = \dfrac{\text{860,421 cal}}{\text{1,495,276 cal}} \\ 1 \times 1,495,276 = \text{X} \times 860,421 \\ \text{X} = \dfrac{1,495,276}{860,421}= 1.738 \\ \text{Answer} = 1.738 \text{ kW}\end{array} [/latex]
Question 1
If you want to know more about absolute zero check out the link: Absolute Zero (Wikipedia)
So what we start with are the Celsius scale which is the metric version of temperature and the Fahrenheit scale which is the imperial version. You might notice that when you are watching T.V. in Canada you will see much smaller temperature numbers than when you watch T.V. from a U.S. channel. This is due to the fact that if we took a temperature reading on the Fahrenheit scale and found a similar temperature on the Celsius scale the Celsius temperature reading would work out to be much smaller. For example we could look at room temperature.[latex]\Large \text{Metric} = {20}^{\circ} \text{ Celsius} \qquad \text{Imperial} = {68}^{\circ} \text{ Fahrenheit}[/latex]
On another note does anyone watch T.V. anymore or is it just the internet and Netflix? Anyway, back to Celsius and Fahrenheit. What we want to do here is relate the metric and imperial scale with each other and then add the absolute scale equivalents. We’ll start with the boiling point of water. Water boils at 100°C and 212°F. Note the letter "C" represents Celsius while the letter F represent Fahrenheit. Next we find the freezing point of water. Water freezes at 0°C and at 32°F. We can put all those together into a small drawing. We’ll go through how to go from Celsius to Fahrenheit and back in just a bit but before we get to that we should add the absolute temperature scales.Metric | Kelvin |
Imperial | Rankine |
[latex]\Large \begin{array}{c} \text{°Kelvin} = \text{°Celsius} + 273\\ \text{°Rankine} = \text{°Fahrenheit} + 460 \end{array}[/latex]
Let’s go through a couple examples going from one to the other. These are going to be fairly straight forward so we won’t go through all the steps like we usually do.Example
[latex]\Large \begin{array}{c} \text{°Kelvin} = \text{°Celsius} + 273 \\ \text{°K} = 10 \text{°C} + 273 \\ \text{°K} = 283 \end{array} [/latex]
Example
[latex]\Large \begin{array}{c} \text{°Kelvin}= \text{°Celsius} + 273 \\ \text{°Celsius} = \text{°Kelvin} - 273 \\ \text{°C} = 25 - 273 \\ \text{°C} = -258 \end{array}[/latex]
Example
[latex]\Large \text{°Rankine} = \text{°Fahrenheit} + 460 [/latex]
[latex]\Large \text{°R} = 350 \text{°F} + 460[/latex]
[latex]\Large \text{°R}= 810 [/latex]
Example
[latex]\Large \text{°Rankine} = \text{°Fahrenheit} + 460 [/latex]
[latex]\Large \text{°Fahrenheit} = \text{°Rankine} - 460 [/latex]
[latex]\Large \text{°F} = 544 - 460 [/latex]
[latex]\Large \text{°F} = 84 [/latex]
[latex]\Large \text{°Fahrenheit} = \text{°Celsius} \times \dfrac{9}{5} + 32[/latex]
There are a couple of things to note here. One is the order the calculation needs to be done. The first thing to be done here is to multiply the degree Celsius by 9/5. After that add the 32. Doing it this way follows the rules of math and in the next chapter we go through a thorough explanation of those rules. For now just follow the path laid out. The second thing to note is that the fraction can also be stated as a number. Writing out the formula using a number instead of a fraction would look like this:[latex]\Large \begin{array}{rl} \text{First:} \quad & \dfrac{9}{5} = 1.8 \\ \text{Then:} \quad & \text{°Fahrenheit} = \text{°Celsius} \times 1.8 + 32 \end{array}[/latex]
You can use either of the two formulas as they both end up with the same answer. It just depends on which one you are more comfortable using. Now let’s go through a couple examples.Example
[latex]\Large \text{°Fahrenheit} = \text{°Celsius} \times \dfrac{9}{5} + 32 [/latex]
Step 2: Plug the numbers into the formula.[latex]\Large \text{°Fahrenheit} = \text{°Celsius} \times \dfrac{9}{5} + 32 [/latex]
[latex]\Large \text{°Fahrenheit} = 55 \times \dfrac{9}{5} + 32 [/latex]
[latex]\Large \text{°F}= 131 [/latex]
Example
[latex]\Large \text{°Fahrenheit} = \text{°Celsius} \times 1.8 + 32 [/latex]
Step 2: Plug the numbers into the formula.[latex]\Large \text{°Fahrenheit} = \text{°Celsius} \times 1.8 + 32 [/latex]
[latex]\Large \text{°F} = 1980 \times 1.8 + 32 [/latex]
[latex]\Large \text{°F} = 3596 [/latex]
Now we have to do the reverse. We have to turn Fahrenheit into Celsius and for that we once again need a formula.[latex]\Large \text{°Celsius} = (\text{°Fahrenheit} - 32) \times \dfrac{5}{9}[/latex]
Example
[latex]\Large \text{°Celsius} = (\text{°Fahrenheit} - 32) \times \dfrac{5}{9}[/latex]
Step 2: Plug the numbers into the formula[latex]\Large \text{°Celsius} = (\text{°Fahrenheit} - 32) \times \dfrac{5}{9}[/latex]
[latex]\Large \text{°C} = (425 - 32) \times \dfrac{5}{9} [/latex]
[latex]\Large \text{°C} = 218.3 [/latex]
[latex]\Large \text{°F} = \text{°C} \times \dfrac{9}{5} + 32[/latex]
Question 1
Question 2
Example
[latex]\Large \text{1 psi} = 6.895 \text{ kilopascals} [/latex]
Step 2: As usual build a ratio[latex]\Large \dfrac{\text{1 psi}}{\text{10 psi}} = \dfrac{6.895 \text{ kPa}}{\text{X kPa}} [/latex]
Step 3: Cross multiply[latex]\Large \begin{array}{c} \dfrac{\text{1 psi}}{\text{10 psi}} = \dfrac{6.895 \text{ kPa}}{\text{X kPa}} \\ 1 \times \text{X} = 10 \times 6.895 \\ \text{X} = 68.95 \\ \text{Answer} = 68.95 \text{ kPa} \end{array}[/latex]
Example
[latex]\Large \text{1 psi} = 6.895 \text{ kilopascals} [/latex]
Step 2: Build a ratio.[latex]\Large \dfrac{\text{1 psi}}{\text{X psi}} = \dfrac{6.895 \text{ kPa}}{\text{150 kPa}}[/latex]
Step 3: Cross multiply.[latex]\Large \begin{array}{c} \dfrac{\text{1 psi}}{\text{X psi}} = \dfrac{6.895 \text{ kPa}}{\text{150 kPa}} \\ 1 \times 150 = \text{X} \times 6.895 \\ \text{X} = \dfrac{150}{6.895} = 21.75 \\ \text{Answer} = 21.75 \text{ psi} \end{array} [/latex]
Example
[latex]\Large 1 \text{ psi} = 2.31 \text{ feet of head} [/latex]
Step 2: Build a ratio.[latex]\Large \dfrac{1 \text{ psi}}{14 \text{ psi}} = \dfrac{2.31 \text{ ft/hd}}{\text{X ft/hd}} [/latex]
Step 3: Cross multiply.[latex]\Large \begin{array}{c} \dfrac{1 \text{ psi}}{14 \text{ psi}} = \dfrac{2.31 \text{ ft/hd}}{\text{X ft/hd}} \\ 1 \times \text{X} = 14 \times 2.31 \\ \text{X} = 32.34 \\ \text{Answer} = 32.34 \text{ ft/hd}\end{array} [/latex]
Now change that feet of head into inches of water column. Step 1: Find a number that works between feet of head and inches of water column. Do you see a problem? The issue here is that we don’t have a number than translates between the two of them. Although we could figure this number out mathematically it would mean that in the end we would have a lot more numbers to remember. The solution here is to find a number that they both relate to and that number is pounds per square inch. What we can then do is go from feet of head to pounds per square inch and then from pounds per square inch to inches of water column. Step 2 (really step 1): We first need to change the feet of head to psi.[latex]\Large 1 \text{ psi} = 2.31 \text{ feet of head}[/latex]
Step 3: Build a ratio.[latex]\Large \dfrac{1 \text{ psi}}{\text{X psi}} = \dfrac{2.31 \text{ ft/hd}}{32.34 \text{ ft/hd}} [/latex]
Step 4: Cross Multiply.[latex]\Large \begin{array}{c}\dfrac{1 \text{ psi}}{\text{X psi}} = \dfrac{2.31 \text{ ft/hd}}{32.34 \text{ ft/hd}} \\ 1 \times 32.34 = \text{X} \times 2.31 \\ \text{X} = \dfrac{32.34}{2.31} = 14 \\ \text{Answer} = 14 \text{ psi}\end{array} [/latex]
What you’ll notice here is that we are back to where we started from. We actually just proved that our first calculation was correct. Step 5: Change the pounds per square inch into inches of water column. Find the number that translates between those two.[latex]\Large 1 \text{ psi} = 27.72 \text{ ″ w.c.} [/latex]
Step 6: Build a ratio.[latex]\Large \dfrac{\text{1 psi}}{\text{14 psi}} = \dfrac{27.72 \text{ ″ w.c.}}{\text{X ″ w.c.}} [/latex]
Step 7: Cross multiply.[latex]\Large\begin{array}{c} \dfrac{\text{1 psi}}{\text{14 psi}} = \dfrac{27.72 \text{ ″ w.c.}}{\text{X ″ w.c.}} \\ 1 \times \text{X} = 14 \times 27.72 \\ \text{X} = 388.08 \\ \text{Answer} = 388.08 \text{ ″ w.c.}\end{array} [/latex]
Question 1
Question 2
Version | Date | Change | Details |
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1.00 | June 21, 2021 | Book published. | |
1.01 | December 2, 2021 | Chapter removed. | "Heat Energy Measurements" chapter was removed from "Understanding and Working with Units" because the content contained errors that could not be fixed without moving away from the intent of the chapter. Related H5P questions were also deleted. |
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[latex]\Large 5 + 4 \times 3 = ? [/latex]
[latex]\Large \begin{array}{lrl}\text{Option A} & \text{Start with:} & 5 + 4 = 9 \\ & \text{Then:} & 9 \times 3 = 27 \end{array}[/latex]
[latex]\Large \begin{array}{lrl} \text{Option B}& \text{Start with:} & 4 \times 3 = 12 \\ & \text{Then:} & 12 + 5 = 17 \end{array} [/latex]
The idea here is that you can’t have two answers for the same question. That just won’t work in math. One of the two answers must be correct and in both options the actual math (by that I mean if you plugged the numbers into a calculator) is correct. No mistakes were made. The issue is that the order of operations in one of the options is wrong.And the correct answer is… Option B.
The next question then becomes "what are the rules to follow when dealing with equations?" This is where the term "BEDMAS" comes into play.[latex]\Large \text{Z} = 4 \times 2 \times (5 \times 9) [/latex]
This situation would indicate the first calculation we perform is 5 × 9. Then we would work through the rest of the equation. Exponents are the second order of operation in Bedmas. Have you ever looked at a math problem and seen the following?[latex]\Large \text{D} = 4 + 8 \times 5 - 3 + {9}^{3} [/latex]
Well the 9^{3} is what we are talking about when dealing with exponents. Specifically we are dealing with the 3 portion. Exponents tell you how many times you multiply a number by itself within an equation. In this case the 3 indicates that we multiply the 9 three times.[latex]\Large {9}^{3} = 9 \times 9 \times 9 [/latex]
We can look at another example to see how exponents work.[latex]\Large {5}^{6} = 5 \times 5 \times 5 \times 5 \times 5 \times 5[/latex]
After brackets and exponents we move onto dividing, multiplying, adding and subtracting. Let’s we go through a couple of examples to see if you get the idea.Example
[latex]\Large\text{A}=1+2\times 6\div 3[/latex]
Although we won’t include this in the steps to answer the question, the first thing you might want to do is write down Bedmas so you can refer to it visually.[latex]\Large \begin{array}{c}\text{A}=1+2\times \mathbf{6 \div 3} \\ 6 \div 3 = 2 \\ \text{So now we have:} \\ \text{A}=1+2 \times 2 \end{array}[/latex]
Step 2: Move on to the next step in Bedmas which is multiplication.[latex]\Large\begin{array}{c} \text{A} = 1+ \mathbf{2\times 2} \\ 2 \times 2 = 4 \\ \text{So now we have:} \\ \text{A} = 1 + 4 \end{array} [/latex]
Step 3: We only have one more operation to go so we are nearing the end. Just add the one and four and we have our answer.[latex]\Large \begin{array}{c} \text{A} = 1 + 4 \\ \text{A} = 5 \end{array} [/latex]
Example
[latex] \Large \text{Y} = (24+36) \times 2 + {4}^{2} [/latex]
Step 1: Refer to BEDMAS. Work through the brackets first.[latex]\Large \begin{array}{c} \text{Y}=\mathbf{(24+36)} \times 2 + {4}^{2} \\ 24 + 36 = 60 \\ \text{Y} = 60 \times 2 +{4}^{2} \end{array} [/latex]
Step 2: Deal with the exponents next.[latex]\Large \begin{array}{c} \text{Y} = 60 \times 2 + \mathbf{{4}^{2}} \\ {4}^{2} = 4 \times 4 = 16 \\ \text{Y} = 60 \times 2 + 16 \end{array}[/latex]
Step 3: Work through the multiplying.[latex]\Large \begin{array}{c} \text{Y}= \mathbf{60 \times 2} + 16 \\ 60 \times 2 = 120 \\ \text{Y} = 120 + 16 \end{array}[/latex]
Step 4: The last step in this question is to do the addition.[latex]\Large \begin{array}{c} \text{Y} = 120 + 16 \\ \text{Y} = 136 \\ \text{Final Answer: Y} = 36 \end{array} [/latex]
Example
[latex]\Large \text{M} = {17}^{2} \times 24 + 13 + 7 \times (45 \div 5)[/latex]
Step 1: Work through the brackets first.[latex]\Large \begin{array}{c} \text{M} = {17}^{2} \times 24 + 13 + 7 \times \mathbf{(45 \div 5) }\\ 45 \div 5 = 9 \\ \text{M} = {17}^{2} \times 24 + 13 + 7 \times 9 \end{array} [/latex]
Step 2: Deal with the exponents next.[latex]\Large \begin{array}{c} \text{M} = \mathbf{{17}^{2}} \times 24 + 13 + 7 \times (45 \div 5) \\ {17}^{2} = 17 \times 17 = 289 \\ \text{M} = 289 \times 24 + 13 + 7 \times 9 \end{array} [/latex]
Step 3: Work through the multiplying.[latex]\Large \begin{array}{c} \text{M} = \mathbf{289 \times 24} + 13 + \mathbf{7 \times 9} \\ 289 \times 24 = 6936 \\ 7 \times 9 = 63 \\ \text{M} = 6939 + 13 + 63 \end{array} [/latex]
Step 4: Complete the addition.[latex]\Large \begin{array}{c} \text{M} = 6939 + 13 + 63 \\ \text{M} = 7012 \\ \text{Final Answer: M} = 7012 \end{array}[/latex]
Question 1
[latex]\Large \text{D} = 5 + 6 \div 2 \times 7 + {4}^{3} \times (5+7)[/latex]
https://video.bccampus.ca/id/0_esv09nv5?width=608&height=402&playerId=23448552Question 2
[latex]\Large \text{M} = 17 + (6\times 3) + {5}^{2} \div 5 - 22[/latex]
https://video.bccampus.ca/id/0_dw741bcm?width=608&height=402&playerId=23448552[latex]\Large \text{A} = {\text{B}}^{2} \times 0.7854 \times \text{H} [/latex]
In a perfect world, you would like to solve for "A" and at the same time be given the values of both "B" and "H." But what if you were given "A" and you had to solve for "B"? How would you go about doing this? The idea here would be to move the variables around and isolate "B." What this means is that "B" is on one side of the equation by itself, and everything else is on the other side. Take a look at the equation again when this has been done.[latex]\Large \text{B} = \sqrt{\dfrac{\text{A}}{.7854 \times \text{H}}}[/latex]
Changing the formula around is referred to as "transposing" an equation. It’s not as simple as just moving stuff around though. There are rules to get to this point, and those rules and their application are what we are going to deal with in this part of the chapter. The most important thing to remember when transposing equations is that whatever is done to one side of the equation must also be done to the other side of the equation. If you look at it mathematically, this makes sense. We already determined that an equation is two mathematical expressions that are separated by an equal sign. What this means is the addition, subtraction, division, and multiplication variables and constants on one side of the equation are equal to all the addition, subtraction, division, and multiplication variables on the other side of the equation. So if you decide to add 5 to one side, you must add 5 to the other side. What this does is keep the equation equal. Take a look at the following example.[latex]\Large \begin{array}{c}10+7=9+8 \\ \text{This works out to be:} \\ 17=17\end{array}[/latex]
Here we have an equation that is true. Now add 5 to the left hand side of the equation, and you’ll see that, in order to keep the equation true, you’ll have to add 5 to the right hand side of the equation.[latex]\Large \begin{array}{c} 10 + 7 + \mathbf{5} = 9 + 8 + \textbf{?} \\ 22 = 9 + 8 + ? \\ 22 = 9 + 8 + \mathbf{5} \\ 22 = 22 \end{array}[/latex]
Keep in mind that this is an example where we added to one side. If we had subtracted, divided, or multiplied, things would be different. We would have to do the same thing to the other side.[latex]\Large \begin{array}{c}7+2=8+1 \\ \text{This works out to be:}\\ 9=9\end{array}[/latex]
Then add 4 to one side of the equation and solve.[latex]\Large 7+2+4=8+1+ \text{ ?} [/latex]
As the left hand side of the equation has had 4 added to it, the right hand side of the equation also has to have 4 added to it. We get:[latex]\Large \begin{array}{c}7+2+ \mathbf{4} = 8 + 1 + \textbf{?} \\ 13 = 9 + \text{ ?} \\ 13 = 9 + \mathbf{4} \\ 13 =13 \end{array} [/latex]
As stated before, the rule is that whatever you do to one side you must do to the other. In this case, if we add 4 to one side then we have to add 4 to the other side to keep it equal. Subtraction would work the same way. If we were to subtract 4 from one side we would have had to subtract 4 from the other side in order to keep it equal. Let’s try out this concept with an equation that has an unknown variable in it. Take a look at the following equation and solve for "J."[latex]\Large 7+ \text{J}=5+8[/latex]
In order to solve for "J" we must isolate "J" on one side of the equation, and it actually doesn’t matter which side we isolate "J" on. If we follow our rule, in order to isolate "J" we would have to get rid of the 7 on the left side. The question becomes, how is that done? What we essentially have to do is move the 7 from the left side to the right side. Once again, we always have to keep in mind that whatever we do to one side, we must do to the other. Start with this. Would you agree that 7 − 7 = 0? What if we subtracted 7 from the left hand side of the equation? What we would be left with is just "J" on the left side, which solves the problem of isolating "J."[latex]\Large \begin{array}{c} \mathbf{(7-7)} +\text{J} = 5+8 \\ 0 + \text{J} = 5+8 \\ \text{J}= 5+8 \end{array} [/latex]
Mathematically this is not correct yet as we only dealt with the left side of the equation. What we need to do now is to do the same thing to the right side and then solve the equation. So we end up subtracting 7 from the right side of the equation to make everything equal once again.[latex]\Large \begin{array}{c}(7-7)+\text{J} = 5+8-7 \\ 0 + \text{J} = 13-7 \\ \text{J} = 13 - 7 \\ \text{J}=6 \end{array}[/latex]
[latex]\Large \begin{array}{rl}\text{Replace J with 6} & \\ \downarrow & \\ 7 + \text{J} & = 5+8 \\ 7+6 & = 5+8 \\ 13 & = 13 \end{array} [/latex]
Example
[latex]\Large 27+ \text{G} = 43+49[/latex]
Step 1: Isolate the variable you are trying to find. In this case "G." In order to do this, we must remove the 27 from the left side of the equation. This is done by subtracting 27 from the left side and then also from the right side.[latex]\Large \begin{array}{c}27+\text{G} = 43+49 \\ \mathbf{(27-27)}+\text{G} = 43+49 - \mathbf{27}\end{array} [/latex]
Step 2: Work through the equation.[latex]\Large \begin{array}{c}(27-27) +\text{G} = 43+49-27 \\ 0 + \text{G} = 92-27 \\ \text{G}=65\end{array}[/latex]
Step 3: Check your answer.[latex]\Large \begin{array}{rl}\text{Replace G with 65} &\\ \downarrow & \\ 27 + \text{G} & = 43 + 49 \\ 27 + 65 & = 43 + 49 \\ 92 & =92\end{array}[/latex]
Example
[latex]\Large \text{H}-16=13+19[/latex]
Step 1: Isolate the variable you are trying to find. In this case "H." In order to do this we must remove the 16 from the left side of the equation. This is done by adding 16 to the left hand side of the equation and then also adding 16 to the right hand side of the equation.[latex]\Large \begin{array}{c}\text{H}-16 = 13+19 \\ \text{H} + \mathbf{(-16+16)}=13 + 19 \mathbf{ +16}\end{array}[/latex]
Step 2: Work through the equation.[latex]\Large \begin{array}{c}\text{H}+(-16+16)=13+19+16 \\ \text{H}+0=32+16 \\ \text{H}=48 \end{array}[/latex]
Step 3: Check your answer.[latex]\Large \begin{array}{l}\text{Replace H with 48} \\ \downarrow \\ \text{H}-16 = 13+19 \end{array}[/latex]
[latex]\Large \begin{array}{c} 48-16=13+19 \\32=32 \end{array}[/latex]
Question 1
[latex]\Large 142+\text{S}=198+257[/latex]
https://video.bccampus.ca/id/0_hwu98s5b?width=608&height=402&playerId=23448552Question 2
[latex]\Large \text{Y} - 22 = 51+53[/latex]
https://video.bccampus.ca/id/0_tmf6wncj?width=608&height=402&playerId=23448552[latex]\Large 4 \text{ and } \dfrac{1}{4}[/latex]
Don’t forget that when we write the number 4, we could also write it as:[latex]\Large\dfrac{4}{1}[/latex]
Do you remember what will happen if we multiplied those two together?[latex]\Large 4 \times \dfrac{1}{4} = \dfrac{4}{4} = 1 [/latex]
Reciprocals are numbers that when multiplied together equal 1. This is an important concept when transposing using multiplication and division. If we divide a number by its reciprocal and end up with one, we have essentially removed that number from the equation. Here is an example. Solve for K.[latex]\Large 8 \times \text{K} = 14 \times 17 [/latex]
Similar to transposing equations with addition and subtraction, the first thing to do is to isolate the variable we are trying to find. In this case, "K." This means that the 8 needs to be removed from the left hand side of the equation. Multiplying 8 by its reciprocal will give us a value of 1. Perfect![latex]\Large 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1[/latex]
Now adding this information to the equation, we would end up with this:[latex]\Large \begin{array}{c} 8 \times \dfrac{1}{8} = \dfrac{8}{8} = 1 \\ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K}= 14 \times 17 \\ \dfrac{8}{8} \times \text{K} = 14 \times 17\\ 1\times \text{K}=14\times 17 \\ \text{K}=14\times 17 \end{array}[/latex]
This would leave us with 1 × K on the left hand side of the equation which would end up being just "K" when multiplied together. This is exactly what we are looking for. We are not quite finished yet though. Now back to the golden rule. Whatever you do to one side of the equation you must do to the other. Therefore, as we multiplied the left hand side of the equation by ⅛ we need to multiply the right hand side of the equation by ⅛.[latex]\Large \dfrac{1}{8} \times 8 \times \text{K} = 14 \times 17 \times \dfrac{1}{8} [/latex]
So now if we followed this through we would end up with:[latex]\Large \begin{array}\\ \mathbf{\dfrac{1}{8}} \times 8 \times \text{K} = 14 \times 17 \times \mathbf{ \dfrac{1}{8}} \\ \dfrac{8}{8} \times \text{K} = \dfrac{238}{8} \\ \text{K} = 29.75 \end{array}[/latex]
We should do a check of the answer like we did previously to see if we are correct.[latex]\Large \begin{array}{rl}\text{Replace K with 29.75} & \\ \downarrow & \\ 8 \times \text{K} & = 14 \times 17 \\ 8 \times 29.75 & = 14 \times 17 \\ 238 & = 238\end{array}[/latex]
Example
[latex]\Large \dfrac{4}{9} \times \text{L} = 12 \times 12 [/latex]
Step 1: Isolate L. In this case, we have to move the 4/9 from one side to the other. In order to do this we have to multiply both sides by the reciprocal of 4/9. This will essentially remove 4/9 from the left hand side of the equation. The reciprocal is of 4/9 is 9/4.[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} = \dfrac{36}{36} = 1 [/latex]
Therefore we get:[latex]\Large \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4} [/latex]
Step 2: Solve the equation[latex]\Large \begin{array}{c} \dfrac{4}{9} \times \dfrac{9}{4} \times \text{L} = 12 \times 12 \times \dfrac{9}{4} \\ 1 \times \text{L} = 144 \times \dfrac{9}{4} \\ \text{L} = \dfrac{1296}{4} \\ \text{L}=324 \end{array}[/latex]
Step 3: Confirm your answer[latex]\Large \begin{array}{rl}\text{Replace L with 324} & \\ \downarrow & \\ \dfrac{4}{9} \times \text{L} & = 12 \times 12 \\ \dfrac{4}{9} \times 324 & = 12 \times 12 \\ \dfrac{1296}{9} & = 144 \\ 144 & = 144 \end{array}[/latex]
Example
[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]
Step 1: Determine which variable you must isolate. In this case it’s given for us and it’s "D."[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\textbf{D}}[/latex]
The challenge here is that "D" is in the denominator (bottom) of the fraction. If we were to isolate it but it was still in the denominator of a fraction we would not have something that we could work with. When "D" is isolated it must not only be by itself on one side of the equation but also appear as a whole number and not as the denominator of a fraction. Step 2: To make this process easier break the equation down to look like the following.[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}}[/latex]
Remember that:
[latex]\Large \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{A}}{\text{B}}[/latex]
We haven’t really changed anything mathematically we have just made the equation into something that is easier to work with. Step 3: Isolate "D." In order to do this we must remove "C" from the right hand side of the equation. Do this by multiplying both the right and left hand side by 1/C.[latex]\Large \mathbf{\dfrac{1}{\text{C}}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= \dfrac{\text{C}}{1} \times \dfrac{1}{\text{D}} \times \mathbf{\dfrac{1}{\text{C}}}[/latex]
Now this might start looking a little bit complicated but if you follow it through you’ll start to see the answer form. Look at the right hand side of the equation. It now has both and C/1 and a 1/C. Multiplying those reciprocals together you get 1 and effectively take out the "C" on the right hand side.[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{1}{\text{C}}= \dfrac{\text{C}}{\text{C}}=1[/latex]
So we end up with:[latex]\Large \dfrac{1}{\text{C}} \times \dfrac{\text{A}}{1} \times \dfrac{1}{\text{B}}= 1 \times \dfrac{1}{\text{D}}[/latex]
If we wanted to simplify this we could just do the following:[latex]\Large \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}}[/latex]
Step 4: Get D into the numerator of the equation. What we need to do will take a little bit of math and some patience. We need to follow the same rules that we have been following. The idea here is to multiply each side by D/1. This will make the right hand side of the equation equal to 1 but now the left hand side of the equation will have D in the numerator.[latex]\Large \begin{array}{c}\dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} = \dfrac{1}{\text{D}} \times \dfrac{\text{D}}{1} \\ \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C}\times \text{B}} = 1 \end{array}[/latex]
What you'll notice is that it actually creates more work but we have managed to get "D" into the numerator of the equation. The only problem is that we also have a bunch of other stuff on the same side as "D" and now our job becomes getting rid of all that stuff. Now all we have to do is follow the rules and get the A, B and C over to the right hand side and isolate D. I’ll do this simply and in one quick calculation.[latex]\Large \dfrac{\text{D}}{1} \times \dfrac{\text{A}}{\text{C} \times \text{B}} \times \dfrac{\text{C} \times \text{B}}{\text{A}} = \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]
What we end up with is:[latex]\Large \text{D}= \dfrac{\text{C} \times \text{B}}{\text{A}}[/latex]
There is a lot of math involved there but if you follow the rules and go one step at a time you’ll eventually get there. My suggestion here is that you go over what we just went through a couple times before moving on. Understanding the math involved here is quite important when it comes to transposing formulas and equations. At this point you might be wondering if there was a shortcut and as luck would have it there is for this procedure which I'll go through now.. We'll start with an example. Take the following equation:[latex]\Large \dfrac{3}{4}=\dfrac{3}{4}[/latex]
This equation is true. Now take each of the fractions and flip them around.[latex]\Large \dfrac{4}{3} = \dfrac{4}{3} [/latex]
It seems that if you flip both fractions around then the equation is still true. Mathematically you are doing the same thing to one side as you are doing to the other. There is a lot of math involved here but the main point is that if you were to do all the math you would end up with the same answer. How we can use this to help us simplify our question is to just do the following. Remember we started with:[latex]\Large \dfrac{\text{A}}{\text{B}}= \dfrac{\text{C}}{\text{D}}[/latex]
We needed to isolate D but the main problem (and the thing that causes us to do a lot of work) was that D was in the denominator and we needed it in the numerator. Through our math wizardry we can get D into the numerator by just flipping around the left hand fraction. Whatever we do to that side we then must do to the right side. What we end up with is:[latex]\Large \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}}[/latex]
All we need to do now is get C out of the left hand side by multiplying it by C/1 and then doing the same thing to the left hand side.[latex]\Large \dfrac{\text{C}}{1} \times \dfrac{\text{B}}{\text{A}}= \dfrac{\text{D}}{\text{C}} \times \dfrac{\text{D}}{1}[/latex]
We end up with…[latex]\Large \dfrac{\text{C} \times \text{B}}{\text{A}} = \text{D} [/latex]
Now to take all that in in one shot might be a little much so you may want to go back and reread that whole explanation. But always keep in mind the math that goes along with that.Example
[latex]\Large \begin{array}{c} \text{Area of a circle} \\ \text{A} = {\text{D}}^{2} \times 0.7854 \end{array}[/latex]
[latex]\Large \begin{array}{cl} \text{Where:} & \text{A = area of the circle} \\ & \text{D = diameter of the circle} \end{array}[/latex]
Solving for the area of the circle would be pretty straight forward but as we are dealing with transposing equations in this section what we’ll do is solve for the diameter (D). Step 1: Identify the variable you are trying to solve for.[latex]\Large \text{A} = \mathbf{{\text{D}}^{2}} \times 0.7854 [/latex]
From this you can see we have a couple problems we have to deal with. The first one is that we have to isolate D. The second one is that D has the exponent 2 attached to it so we have to somehow eliminate that. Step 2: Isolate D. For this we have to move the 0.7854 from the right hand side of the equation to the left hand side. We simply follow the rules we have used up to this point.[latex]\Large \begin{array}{c} \mathbf{\dfrac{1}{0.7854}} \times \text{A} = {\text{D}}^{2} \times 0.7854 \times \mathbf{\dfrac{1}{0.7854}} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times \dfrac{0.7854}{0.7854} \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \times 1 \\ \dfrac{\text{A}}{0.7854} = {\text{D}}^{2} \end{array}[/latex]
That takes care of the first problem. Now we have to tackle the D^{2} issue. Step 3: Remove the exponent from D. Once again we go back to our original rule. Whatever you do to one side you must to the same thing to the other side. We have to refer to the video that you watched earlier in this section regarding exponents and square roots. We’ll do a quick refresher before we solve for D. Remember that:[latex]\Large \begin{array}{c} {\text{D}}^{2} = \text{D} \times \text{D} \\ \text{and... } \\ \sqrt{\text{D} \times \text{D}} = \text{D} \end{array}[/latex]
So what we see is that if you square root a number that is squared (has an exponent of 2) you end up with the number itself. I’ll quickly show you with numbers.[latex]\Large \begin{array}{c} {4}^{2} = 4 \times 4 \\ {4}^{2} = 16 \\ \sqrt{16} = \sqrt{4 \times 4} \\ \sqrt{16} = 4 \end{array} [/latex]
Having gone through all that we can now solve the problem.[latex]\Large \begin{array}{c} \sqrt{\dfrac{\text{A}}{0.7854}}= \sqrt{{\text{D}}^{2}} \\ \sqrt{\dfrac{\text{A}}{0.7854}} = \text{D} \end{array}[/latex]
Question 1
[latex]\Large \dfrac{\text{A} \times \text{B}}{\text{C}}=\dfrac{\text{D} \times \text{E}}{\text{F}}[/latex]
https://video.bccampus.ca/id/0_91c0rn02?width=608&height=402&playerId=23448552Question 2
[latex]\Large \begin{array}{ll} \textbf{Square:} & \text{Perimeter}= \text{ side}+ \text{ side} + \text{ side} + \text{ side}\\ \textbf{Rectangle:} & \text{Perimeter} = \text{length} + \text{width} + \text{width} + \text{length}\end{array}[/latex]
Would there be any other way to express the two formulas? Take a guess.[latex]\Large \begin{array}{ll} \textbf{Square:} & \text{Perimeter}= \text{ side} \times 4 \\ \textbf{Rectangle:} & \text{Perimeter} = (\text{length}\times 2)+(\text{width} \times 2) \end{array}[/latex]
Both versions of each formula are correct and you can use whichever you choose to answer questions. This might be a good time to stop and discuss memorization. Memorization is a topic that comes up quite frequently in math. Students often ask if they have to memorize formulas and for the most part the answer is yes. Memorizing formulas can become time consuming and can also take up a lot of your brain power. The problem that occurs when students memorize formulas is they often forget what the formula represents. Students get really good at plugging numbers into formulas but they don't really understand what the formula does. If the numbers were to change or the question was asked in a different manner than normal the student might become lost. You might want to think about the following. Every time you get a formula, instead of memorizing that formula try to VISUALIZE that formula. What I mean by that is visualize what the formula represents. Hopefully it will become easier for you to work with the formula and easier for your brain to recall the formula when needed.Example
[latex]\Large \text{Perimeter}= \text{ side} + \text{ side} + \text{ side} + \text{ side}[/latex]
Step 2: Solve for perimeter.[latex]\Large \begin{array}{c} \text{Perimeter}=8+8+8+8 \\ \text{Perimeter} = 32 \text{ inches} \end{array}[/latex]
Example
[latex]\Large \text{Perimeter} = (\text{length}\times 2) + (\text{width} \times 2) [/latex]
Step 2: Solve for perimeter[latex]\Large \text{Perimeter of a polygon} = \text{side} + \text{ side} + \text{side and so on... }[/latex]
What the formula is saying is just add up all the sides to get the perimeter.Example
[latex] \Large \text{Perimeter of a polygon} = \text{side} + \text{ side} + \text{side and so on... }[/latex]
Step 2: Solve for perimeter[latex] \Large \begin{array}{c} \text{Perimeter of a polygon}= 5+10+6+4+7+8 \\ \text{Perimeter}=40\end{array}[/latex]
Example
[latex]\large \text{Perimeter of a polygon}= \text{side} + \text{ side} + \text{ side} + \text{side} + \text{ side} + \text{ side}[/latex]
Step 2: Solve for perimeter[latex]\Large \begin{array}{c}\text{Perimeter}=10+10+13+8+8+6 \\ \text{Perimeter} = 55 \end{array} [/latex]
Question 1
Jacques is a carpet layer who is going to re - carpet a room. In order to do this he has to nail a small piece of wood to the ground which runs the entire perimeter of the room. This wood is there to attach the carpet at the edges of the room.
The room itself is in the shape of a rectangle with the length being 12 feet 2 inches and the width being 10 feet 1 inch.
How much wood does Jacques need?
https://video.bccampus.ca/id/0_wyqknxvw?width=608&height=402&playerId=23448552Question 2
[latex]\Large \text{radius} = \dfrac{\text{diameter}}{2}[/latex]
Working with the equation we could also state that...[latex]\Large \text{diameter} = \text{radius} \times 2 [/latex]
No matter which way you work it both equations represent the relationship between the radius and diameter. Having gotten that out of the way we can look at the formula for finding the circumference of a circle. Before I go through the formula try and take a guess at what the formula is. Once again think about the relationship between the variables and how they might work together in the formula. Formula:[latex]\Large \begin{array}{c} \textbf{Circle} \\ \text{Circumference} = \pi \times \text{diameter} \end{array}[/latex]
We could also write it as…[latex]\Large \text{C} = \pi \times \text{D} [/latex]
Hang on a minute! What is that symbol? Have you ever seen or heard of this symbol before? Well it’s a constant. It’s referred to as "pi," and if you were to sound it out, it would sound like "pie." What "pi" represents is the relationship between the circumference of a circle and the diameter of that same circle. "Pi" is a constant and never changes. Having said that, the number "pi" is a bit of an anomaly. You would think that a constant would be something like 7 or maybe 12.64 or maybe even 0.00004. "Pi" is a bit trickier than that. The following is the value of "pi."[latex]\Large \begin{array}{c} \pi \\ = 3.141592653589793238462643383279 \end{array}[/latex]
And that’s just the first few digits. The constant "pi" goes on forever. It doesn’t actually stop. People have figured out "pi" to thousands of digits. The good news for us is that we don’t have to worry about all those digits that come after the decimal point. For our purpose we’ll use just the following...[latex]\Large \pi = 3.14 [/latex]
If you want to find out more about the number "pi" and how far people have calculated it, check out the following websites: Pi (Wikipedia) and 1 Million Digits of Pi (piday).Example
[latex]\Large \begin{array}{c} \text{circumference} = \pi \times \text{diameter} \\ \text{OR} \\ \text{C} = \pi \times \text{D}\end{array}[/latex]
Step 2: Solve for circumference.[latex]\Large \begin{array}{c} \text{C} = \pi \times \text{D} \\ \text{C} = 3.14 \times 24 \\ \text{C} = 75.36 \end{array}[/latex]
Example
[latex]\Large \text{C} = \pi \times \text{D}[/latex]
[latex]\Large \begin{array}{c} \text{diameter} = \text{radius} \times 2 \\ \text{diameter} = 8 \times 2 \\ \text{diameter} = 16 \end{array}[/latex]
Step 2: Solve for circumference.[latex]\Large \begin{array}{c} \text{C} = \pi \times \text{D} \\ \text{C} = 3.14 \times 16 \\ \text{C} = 50.24 \end{array}[/latex]
Example
[latex]\Large \text{circumference} = \pi \times \text{diameter} [/latex]
Step 2: Rework the formula to solve for diameter.[latex]\Large\text{diameter}= \dfrac{\text{circumference}}{\pi}[/latex]
Step 3: Solve for diameter.[latex]\Large \begin{array}{c} \text{diameter}= \dfrac{\text{circumference}}{\pi} \\ \text{diameter} = \dfrac{153}{3.14} \\ \text{diameter} = 48.73 \end{array}[/latex]
Question 1
Question 2
[latex]\Large \begin{array}{ll}\textbf{Square:} & \text{Area} = \text{side} \times \text{side}\\ \textbf{Rectangle:}&\text{Area}= \text{length}\times \text{width}\end{array}[/latex]
Take note here as there are a couple things different than dealing with perimeter. The first is that we are now multiplying rather than adding. This ends up leading to our second point. If we take a look at the formula for the square we see that we multiply a side by a side. As all the sides are the same it doesn’t really matter which two sides we multiply together. The issue becomes the units that we end up with. Remember that when dealing with perimeter we are only dealing with a one dimensional line. Our units end up being linear or essentially one dimensional. With area we end up with units that give us an answer using two dimensional units. The best way to understand this is to go through an example. Let’s say we have a square where each side is 5 inches long. Using the formula for area of a square we would get…[latex]\Large \begin{array}{c} \text{Area} = \text{side} \times \text{side} \\ \text{Area} = 5 \text{ inches} \times 5 \text{ inches}\end{array}[/latex]
We can conclude that 5 times 5 is equal to 25 but what happens when we multiply inches times inches.[latex]\Large \text{inches} \times \text{inches} = ?[/latex]
Well inches times inches ends up being inches squared or if you were to write it down it would look like this:[latex]\Large {\text{inches}}^{2} \qquad \text{OR}\qquad {\text{in}}^{2}[/latex]
So when you see measurements, such as feet, inches, miles, kilometres and so on with the square symbol after them what you are dealing with is a measurement of area. Put another way it is a measurement stated in two dimensions. If we go back to our example we would end up with this:[latex]\Large \begin{array}{c} \text{Area} = \text{side} \times \text{side} \\ \text{Area} = 5 \text{ inches} \times 5 \text{ inches} \\ \text{Area} = 25 {\text{ in}}^{2} \end{array}[/latex]
If we were to look at it visually here is what we would have:Example
[latex]\Large \text{Area} = \text{side} \times \text{side}[/latex]
Step 2: Solve for area.[latex]\Large \begin{array}{c} \text{Area} = \text{side} \times \text{side} \\ \text{Area} = 14 \text{ cm} \times 14 \text{ cm} \\ \text{Area} = 196 {\text{ cm}}^{2}\end{array}[/latex]
Example
[latex]\Large \text{Area} = \text{length} \times \text{width}[/latex]
Step 2: Solve for area.[latex]\Large \begin{array}{c} \text{Area} = \text{length} \times \text{width} \\ \text{Area} = 22 \text{ in} \times 15 \text{ in} \\ \text{Area} = 330 {\text{ in}}^{2} \end{array}[/latex]
[latex]\Large \text{foot} = 12 \text{ inches}[/latex]
Remember that’s a linear or one dimensional measurement. What we are looking to end up with is a two dimensional measurement.[latex]\Large \begin{array}{cc} \text{Area} = \text{side} \times \text{side} & \text{Area} = \text{side} \times \text{side} \\ \text{Area} = 1 \text{ ft} \times 1 \text{ ft} & \text{Area} = 12 \text{ in} \times 12 \text{ in} \\ \text{Area} = 1 {\text{ ft}}^{2} & \text{Area} = 144 {\text{ in}}^{2}\end{array} [/latex]
[latex]\Large \text{therefore} \longrightarrow 1 {\text{ ft}}^{2} = 144 {\text{ in}}^{2}[/latex]
Basically it takes 144 square inches to make one square foot.Example
[latex]\Large \text{Area} = \text{side} \times \text{side}[/latex]
Step 2: Solve for area.[latex]\Large \begin{array}{cc} \text{Area} = \text{side} \times \text{side} \\ \text{Area} =15 \text{ in} \times 15\text{ in} \\ \text{Area} = 225 {\text{ in}}^{2} \end{array} [/latex]
Step 3: Convert inches squared to feet squared.[latex]\Large \begin{array}{ccl}1{\text{ ft}}^{2}=144{\text{ in}}^{2} & & \\ \text{therefore} & \longrightarrow & {\text{ft}}^{2} = \dfrac{{\text{in}}^{2}}{144 {\text{/ft}}^{2}}\\ & & {\text{ft}}^{2} = \dfrac{225 {\text{ in}}^{2}}{144 {\text{ in/ft}}^{2}} \\ & & {\text{ft}}^{2} = 1.56 {\text{ ft}}^{2} \end{array}[/latex]
[latex]\Large \text{Area} = \dfrac{\text{base} \times \text{height}}{2} [/latex]
Let’s go through as example.Example
[latex]\Large \text{Area} = \dfrac{\text{base} \times \text{height}}{2} [/latex]
Step 2: Solve for area.[latex]\Large \begin{array}{c} \text{Area} = \dfrac{9 \text{ inches} \times 7 \text{ inches}}{2} \\ \text{Area} = \dfrac{63 {\text{ in}}^{2}}{2} \\ \text{Area} = {31.5 \text{ in}}^{2} \end{array}[/latex]
Examples
[latex]\Large \text{Area} = \dfrac{\text{base} \times \text{height}}{2} [/latex]
Step 2: Solve for area.[latex]\Large \begin{array}{c} \text{Area} = \dfrac{12 \text{ inches} \times 10 \text{ inches}}{2} \\ \text{Area} = \dfrac{120 {\text{ in}}^{2}}{2} \\ \text{Area} = 60 {\text{ in}}^{2} \end{array} [/latex]
We’ll try one more example but this time we’ll switch it up a bit and the area and base will be given and we’ll have to solve for height.Example
[latex]\Large \text{Area} = \dfrac{\text{base} \times \text{height}}{2} [/latex]
Step 2: Rearrange the formula so solve for height.[latex]\Large \text{Height} = \dfrac{\text{Area} \times 2}{\text{base}} [/latex]
Step 3: Solve for height.[latex]\Large \begin{array}{c} \text{Height} = \dfrac{\text{Area} \times 2}{\text{base}} \\ \text{Height} = \dfrac{200 {\text{ in}}^{2} \times 2}{17 \text{ in}} \\ \text{Height} = 23.53 \text{ in}\end{array} [/latex]
Question 1
Question 2