Chapter 9

The base metric unit for weight is Newtons (N), but usually you see the mass measured in grams (g). When you buy packages of food, for instance, they often have the weight in ounces and the mass in grams.

gram (g)

The base metric unit used for distance is meters (m).

meter (m)

The base metric unit for volume is liters (L).

liter (L)

If you are learning the meaning of the prefixes, it would be a good idea to write them on an index card or in your notebook.

To order the measures, look at the exponents in their magnitudes. Of the measures in the table, positive 3 is the largest power, so kilogram is the largest measure. Positive 2 is the next largest power, so hectogram is the next largest measure. Last comes decagram.

Ordered from largest to smallest: kilogram, hectogram, decagram

The lesson suggested a mnemonic for remembering the order of the prefixes: King Henry Died from Drinking Chocolate Milk (“from” does not have a prefix).

kilogram, hectogram, decagram

Think of a yardstick, which is roughly the length of a meter. A short hallway could be 2.5 meters long.

A kilometer is not a reasonable length for a hallway. Even one kilometer would be one thousand meters long. A 2.5-kilometer hallway would be 2,500 meters long.

The ruler you used in elementary school probably had 12 inches on one side and 30 centimeters on the other side. A 2.5-centimeter hallway would be shorter than your 12-inch ruler. This is not a reasonable length.

2.5 m

$2.45\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{100\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)=245\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

245 cm

$7.6\cancel{\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}}{10\cancel{\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=0.76\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

0.76 daL

$1,457\cancel{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}\left(\frac{1\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}}{10\cancel{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}}\right)=145.7\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}$

145.7 cg

Hint: You can make conversion factors out of any quantities that are equivalent, such as 1 dose being equivalent to 5 mL of amoxicillin.

$250\cancel{\mathrm{d}\mathrm{o}\mathrm{s}\mathrm{e}\mathrm{s}}\left(\frac{5\cancel{\mathrm{m}\mathrm{L}}}{1\cancel{\mathrm{d}\mathrm{o}\mathrm{s}\mathrm{e}}}\right)\left(\frac{1\mathrm{L}}{1000\cancel{\mathrm{m}\mathrm{L}}}\right)=1.25\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

1.25 L

Multiply the caffeine in each drink times the number of drinks you consume. Then add the results.

Twice the caffeine in coffee + three times the caffeine in soda = your daily intake

$2(95)+3(54)=190+162=352\mathrm{m}\mathrm{g}=\mathrm{y}\mathrm{o}\mathrm{u}\mathrm{r}\mathrm{d}\mathrm{a}\mathrm{i}\mathrm{l}\mathrm{y}\mathrm{i}\mathrm{n}\mathrm{t}\mathrm{a}\mathrm{k}\mathrm{e}$

$352\cancel{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}\left(\frac{1\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}}{1000\cancel{\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}}\right)=0.352\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}$

FDA recommended maximum = 0.5 grams

Your daily intake of 0.352 grams is less than the recommended maximum of 0.5 grams.

Yes, 352 mg of caffeine is equal to 0.352 g, which is less than 0.5 g.

$3,810\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{1000\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=3.810\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}$

Since 3,810 meters is 3.81 km, Kyrie’s skydive was a greater altitude than his friend’s jump at 3.2 km.

Since 3,810 m is 3.81 km, Kyrie’s skydive was at a greater altitude.

Both a centimeter and a millimeter are less than an inch, so square centimeters and square millimeters are not practical for measuring forests. Your best choice is the square kilometer.

km²

A meter is roughly the length of a yardstick. An area of 1,800 square meters would be larger than many homes. Now decide between the other measurements.

Suppose it was a square table. Find the measure of one side.

$\sqrt{1,800{\mathrm{m}\mathrm{m}}^2}\approx <!--\; \approx \; -->42.4\mathrm{m}\mathrm{m}=4.24\mathrm{c}\mathrm{m}$

This would be less than the width of your hand.

$\sqrt{1,800{\mathrm{c}\mathrm{m}}^2}\approx <!--\; \approx \; -->42.4\mathrm{c}\mathrm{m}$

The 42.4 cm table would work.

The 1,800 square centimeter table is the best choice.

1,800 cm²

A typical sheet of paper is 8.5 inches by 11 inches. They cut up four sheets of this paper to make their mosaic. Eleven inches is a little bit less than the length of a ruler, which is around 30 cm. So, you know the upper limit in 30 by 30 times 4, which is way overestimating since 8 is far less than 11.

$30(30)(4)=3600{\mathrm{c}\mathrm{m}}^2$

Since their calculation is 2,412.80 square centimeters, their calculation does make sense.

Yes. The unit of value, cm², is appropriate given the size of a sheet of paper. Because the total area of 2,412.89 cm² represents four sheets of paper, each sheet is approximately 600 cm², which seems reasonable for a sheet of letter-size paper.

$1,532,900{\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}^2{\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)}^2=\frac{1,532,900}{10,000}{\mathrm{m}}^2=153.29{\mathrm{m}}^2$

153.29 m²

$136.5{\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}^2{\left(\frac{100\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)}^2=136.5\times <!--\; \times \; -->10,000\mathrm{c}{\mathrm{m}}^2=1,365,000\mathrm{c}{\mathrm{m}}^2$

1,365,000 cm²

Convert the measures to meters.

$100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}$

$106\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=1.06\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

Use the area formula to determine the area of the fabric.

$A=L\times <!--\; \times \; -->W=\left(1\mathrm{m}\right)\left(1.06\mathrm{m}\right)=1.06{\mathrm{m}}^2$

1.06 m²

Step 1: Determine the area of the garden.

$A=L\times <!--\; \times \; -->W=\left(5\right)\left(8\right)=40{\mathrm{m}}^2$

Step 2: Determine the area of the sidewalk.

$A=L\times <!--\; \times \; -->W=\left(0.75\right)\left(8\right)=6{\mathrm{m}}^2$

Step 3: Subtract the area of the sidewalk from the garden to find how much topsoil you need.

$40{\mathrm{m}}^2-<!--\; -\; -->6{\mathrm{m}}^2=34{\mathrm{m}}^2$

You need to purchase 34 square meters of topsoil.

34 m²

Step 1: Convert 2,500 cm to meters.

$2,500\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=25\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

Step 2. Find the surface area of the pool.

$A=L\times <!--\; \times \; -->W=\left(50\mathrm{m}\right)\left(25\mathrm{m}\right)=1,250{\mathrm{m}}^2$

The surface area of the pool is 1,250 square meters.

1,250 m²

Step 1: Convert the dimensions of the living room to meters.

$600\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=6\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

$750\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=7.5\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

Step 2. Find the area of the living room.

$A=L\times <!--\; \times \; -->W=\left(6\mathrm{m}\right)\left(7.5\mathrm{m}\right)=45{\mathrm{m}}^2$

Step 3: Add the result to the provided area of the bedroom.

$25{\mathrm{m}}^2+45{\mathrm{m}}^2=70{\mathrm{m}}^2$

You need to buy 70 square meters of carpet.

70 m²

A cubic meter box would measure approximately one yard on each side. An 8 cubic meter space would make a reasonable storage area. The other measures would be much too small.

8 m³

$8{\cancel{\text{centimeters}}}^3{\left(\frac{1\cancel{\text{meter}}}{\text{100 }\cancel{\text{centimeters}}}\right)}^3{\left(\frac{\text{1,000 millimeters}}{\text{1 }\cancel{\text{meter}}}\right)}^3=\frac{8\times <!--\; \times \; -->1,000,\cancel{000,000}}{\cancel{1,000,000}}\text{ m}{\text{m}}^3\text{ = }8,000\text{ m}{\text{m}}^3$

8,000 mm³

$37.854{\cancel{\text{meters}}}^3{\left(\frac{100\text{ centimeters}}{\text{1 }\cancel{\text{meter}}}\right)}^3=37.854\times <!--\; \times \; -->1,000,000\text{ c}{\text{m}}^3\text{ = }37,854,000\text{ c}{\text{m}}^3$

37,854,000 cm³

$465,000,000{\cancel{\text{centimeters}}}^3{\left(\frac{1\text{ meter}}{\text{100 }\cancel{\text{centimeters}}}\right)}^3=\frac{465,000,000}{1,000,000}{\text{m}}^3\text{ = }465{\text{m}}^3$

465 m³

$19,000\cancel{\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{i}\mathrm{c}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}}{1,000\cancel{\mathrm{c}\mathrm{u}\mathrm{b}\mathrm{i}\mathrm{c}\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=19\mathrm{l}\mathrm{i}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

19 L

$37.854\cancel{\text{cubic meters}}{\left(\frac{100\cancel{\text{centimeters}}}{\text{1 }\cancel{\text{meter}}}\right)}^3\left(\frac{\text{ 1 liter}}{\text{ 1,000}\cancel{\text{cubic centimeters}}}\right)$

$=\frac{37.854\times <!--\; \times \; -->1,000,\cancel{000}}{\cancel{1,000}}=37,854\text{ liters}$

37,854 L

$800,000\cancel{\text{cubic millimeters}}\left(\frac{\text{ 0}\text{.001 milliliters}}{\text{ 1,000}\cancel{\text{cubic millimeters}}}\right)=800\text{ milliliters}$

800 mL

Step 1: Convert the dimensions of the box to meters.

$20\cancel{\text{centimeters}}\left(\frac{\text{1 meter}}{100\cancel{\text{centimeters}}}\right)=0.2\text{ meter}$

$5\cancel{\text{centimeters}}\left(\frac{\text{1 meter}}{100\cancel{\text{centimeters}}}\right)=0.05\text{ meter}$

$30\cancel{\text{centimeters}}\left(\frac{\text{1 meter}}{100\cancel{\text{centimeters}}}\right)=0.3\text{ meter}$

Step 2: Find the volume of one box.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{0}\text{.2 m}\right)\left(\text{0}\text{.05 m}\right)\left(0.3\text{ m}\right)=0.003{\text{m}}^3$

Step 3: Divide the flour milled by the volume of a box to find the number of boxes produced each day.

$300\cancel{{\text{m}}^3}\left(\frac{\text{1 box}}{0.003\cancel{{\text{m}}^3}}\right)=100,000\text{ boxes}$

The factory can produce 100,000 boxes per day.

100,000 boxes

Step 1: Find the volume of a box.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{25 cm}\right)\left(\text{8 cm}\right)\left(\text{35 cm}\right)=7,000\text{ c}{\text{m}}^3$

Step 2: Convert the volume of a box into liters.

$7,000\cancel{\text{cubic centimeters}}\left(\frac{\text{ 1 liter}}{\text{ 1,000}\cancel{\text{cubic centimeters}}}\right)=7\text{ liters}$

Step 3: Convert the factory production into boxes.

$\text{42,000 liters }\left(\frac{\text{1 box}}{\text{7 }\cancel{\text{liters}}}\right)=6,000\text{ boxes}$

They produce 6,000 boxes per day.

6,000 boxes

Step 1: Find the volume of the tank.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{75 cm}\right)\left(\text{20 cm}\right)\left(\text{25 cm}\right)=37,500\text{ c}{\text{m}}^3$

Step 2: Convert the volume into liters.

$37,500\cancel{\text{cubic centimeters}}\left(\frac{\text{ 1 liter}}{\text{ 1,000}\cancel{\text{cubic centimeters}}}\right)=37.5\text{ liters}$

Step 3: You only want half a tank of water, so divide the volume by two.

$37.5\text{ liters }÷<!--\; ÷\; -->2=18.75\text{ liters}$

You need 18.75 liters of water.

18.75 L

A gram is about the weight of a paper clip. A penny weighs about the same as 2.5 paperclips, so 2.5 grams is reasonable.

2.5 g

The average car weighs about 2 metric tons, or 2,000 kilograms. The best choice is 1,300 kilograms for a lightweight car.

1,300 kg

A gram is about the weight of a paper clip. A shrew that tiny can weigh the same as two paper clips, so 2 grams is reasonable.

A reasonable value for the weight of an Etruscan shrew is 2 g. A 2 L bottle of water weighs 2 kg, and since it is one of the smallest mammals, the shrew must weigh less than that, so 2 g seems reasonable.

$175,000\cancel{\text{milligrams}}\left(\frac{\text{1 }\cancel{\text{gram}}}{\text{ 1,000}\cancel{\text{milligrams}}}\right)\left(\frac{\text{1 kilogram}}{\text{1,000 }\cancel{\text{grams}}}\right)=\frac{175,000}{1,000,000}\text{ kilograms = }0.175\text{ kilograms}$

0.175 kg

$4.869\cancel{\text{kilograms}}\left(\frac{\text{1,000 grams}}{\text{ 1}\cancel{\text{kilogram}}}\right)=4,869\text{ grams}$

4,869 g

$1.23\cancel{\text{kilograms}}\left(\frac{1,000\cancel{\text{grams}}}{\text{1 }\cancel{\text{kilogram}}}\right)\left(\frac{1,000\text{ milligrams}}{1\cancel{\text{gram}}}\right)=1,230,000\text{ milligrams}$

1,230,000 mg

7.5 mL

Step 1: Convert the grams to kilograms.

$17,683\cancel{\text{grams}}\left(\frac{1\text{ kilogram}}{1,000\cancel{\text{grams}}}\right)=\text{ 17}\text{.863 kilograms}$

Step 2: Use the chart to identify the dosage.

The child fits within the 16 to 21 kg range.

The correct dosage is 7.5 mL.

Step 1: Convert the grams to kilograms.

$500\cancel{\text{grams}}\left(\frac{1\text{ kilogram}}{1,000\cancel{\text{grams}}}\right)=\text{ 0}\text{.5 kilograms}$

Step 2: Multiply by 10 to find the amount of flour needed.

$\text{10(0}\text{.5 kg) = 5 kg}$

Five kilograms of flour are needed to make 10 batches of croissants.

5 kg

Step 1: Convert 225 grams to kilograms to find the weight of one potato in kilograms.

$225\cancel{\text{grams}}\left(\frac{1\text{ kilogram}}{1,000\cancel{\text{grams}}}\right)=\text{ 0}\text{.225 kilograms}$

Step 2: Use equivalences to create conversion factors. Writing the units helps you understand.

$5,000\cancel{\text{bags}}\left(\frac{\text{5 }\cancel{\text{kilograms}}}{\text{1 }\cancel{\text{bag}}}\right)\left(\frac{1\text{ potato}}{\text{0}\text{.225 }\cancel{\text{kilograms}}}\right)=\frac{5,000\times <!--\; \times \; -->5}{0.225}\text{ potatoes }\approx <!--\; \approx \; -->111,111\text{ potatoes}$

111,111 potatoes

$C=\frac{5}{9}\left(F-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(482-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(450\right)$

${250}^{\circ <!--\; \circ \; -->}\mathrm{C}$

250 °C

$F=\frac{9}{5}C+32$

$F=\frac{9}{5}\left(75\right)+32$

$F=135+32$

${167}^{\circ <!--\; \circ \; -->}\mathrm{F}$

167 °F

Step 1: Convert Toronto’s Fahrenheit temperature to Celsius.

$C=\frac{5}{9}\left(F-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(57.4-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(25.4\right)$

Approximately ${14.1}^{\circ <!--\; \circ \; -->}\mathrm{C}$

Step 2: Subtract the two temperatures.

$16.7-<!--\; -\; -->14.1={2.6}^{\circ <!--\; \circ \; -->}\mathrm{C}$

Madrid has the higher temperature by approximately ${2.6}^{\circ <!--\; \circ \; -->}\mathrm{C}$.

Since Toronto has a temperature of 14.1 °C, Madrid has the higher temperature by 2.6 °C.

The lesson mentioned that a sample temperature in a hot desert was 104°F or 49°C. Of these choices, 37°C is the most reasonable.

37 °C

The temperature of boiling water is 100°C. While chili is not likely to be that hot, a temperature of 60°C is reasonable for cooking chili.

60 °C

A fall in the temperature means it is lower. Subtract 6° from 4°C.

$4-<!--\; -\; -->6=-<!--\; -\; -->2^{\circ <!--\; \circ \; -->}\mathrm{C}$

The temperature at 6 AM was −2°C.

–2 °C

An increase in the temperature means it is higher. Add 59° to 4°C.

$4+59={63}^{\circ <!--\; \circ \; -->}\mathrm{C}$

The temperature of the cooked hot dog was 63°C.

63 °C

Step 1: Convert the Fahrenheit temperature to Celsius.

$C=\frac{5}{9}\left(F-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(302-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(270\right)$

${150}^{\circ <!--\; \circ \; -->}\mathrm{C}$

Step 2: Find the temperature difference between ${150}^{\circ <!--\; \circ \; -->}\mathrm{C}$ and $0^{\circ <!--\; \circ \; -->}\mathrm{C}$.

$150-<!--\; -\; -->0={150}^{\circ <!--\; \circ \; -->}\mathrm{C}$

Step 3: Use the fact that the temperature decreases $5^{\circ <!--\; \circ \; -->}\mathrm{C}$ every 6 minutes.

${150}^{\circ <!--\; \circ \; -->}\mathrm{C}\left(\frac{6\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}}{5^{\circ <!--\; \circ \; -->}\mathrm{C}}\right)=180\mathrm{m}\mathrm{i}\mathrm{n}\mathrm{u}\mathrm{t}\mathrm{e}\mathrm{s}$

It will take 180 minutes for the temperature to decrease the required amount.

Note: You could also have expressed your answer as three hours.

180 minutes, or 1 hour and 30 minutes

The base metric unit used for distance is meters (m).

meter

The base metric unit for weight is Newtons, but usually you see the mass measured in grams. When you buy packages of food, for instance, they often have the weight in ounces and the mass in grams.

grams

Since hecto- has a magnitude of 2 and centi- has a magnitude of −2, you know that 12 hectoliters have a greater volume than 12 centiliters.

12 hectoliters

$1,520\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=15.2\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

15.20 m or 15.2 m

It can be easier sometimes to make conversion factors to base units instead of trying to figure out how many factors of 10 to use.

$1.34\cancel{\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1000\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}{1\cancel{\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)\left(\frac{1\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{10\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=134\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}$

134 dam

It can be easier sometimes to make conversion factors to base units instead of trying to figure out how many factors of 10 to use.

$12,700\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}\left(\frac{1\cancel{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}}\right)\left(\frac{1\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}}{100\cancel{\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}}}\right)=1.27\mathrm{h}\mathrm{e}\mathrm{c}\mathrm{t}\mathrm{o}\mathrm{g}\mathrm{r}\mathrm{a}\mathrm{m}\mathrm{s}$

1.27 hg

$750\cancel{\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1000\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}{1\cancel{\mathrm{k}\mathrm{i}\mathrm{l}\mathrm{o}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)\left(\frac{1,000\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)=750,000,000\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

750,000,000 mm

Convert them both to meters.

$0.04\cancel{\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{10\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{d}\mathrm{e}\mathrm{c}\mathrm{a}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)=0.4\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

$40\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)=0.4\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}$

They are equal in size.

They are equal in size.

A kilometer is a little farther than half a mile. Square kilometers are not practical for measuring walls.

Square centimeters are around the size of your fingerprint. This is too small for measuring your wall. Square millimeters are even smaller.

A meter is roughly the length of a yardstick. 12 square meters is a good measure for your wall.

12 m²

Square centimeters are around the size of your fingerprint. This is too small for measuring your park. Square millimeters are even smaller.

A meter is roughly the length of a yardstick. 1,200 square meters would be good for a city park.

A kilometer is a little farther than half a mile. 1,200 square kilometers is too large for a city park.

1,200 m²

Square centimeters are around the size of your fingerprint. This is too small for measuring your table. Square millimeters are even smaller.

A meter is roughly the length of a yardstick. 2.5 square meters would be good for a table.

A kilometer is a little farther than half a mile. 2.5 square kilometers is far too large for a table.

2.5 m²

$20,000{\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}^2{\left(\frac{1\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)}^2=\frac{20,000}{10,000}{\mathrm{m}}^2=2{\mathrm{m}}^2$

2

$5.7{\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}^2{\left(\frac{100\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)}^2=5.7\times <!--\; \times \; -->10,000\mathrm{c}{\mathrm{m}}^2=57,000\mathrm{c}{\mathrm{m}}^2$

57,000

$217{\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}^2{\left(\frac{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}{100\cancel{\mathrm{c}\mathrm{e}\mathrm{n}\mathrm{t}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}}\right)}^2{\left(\frac{1,000\mathrm{m}\mathrm{i}\mathrm{l}\mathrm{l}\mathrm{i}\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}\mathrm{s}}{1\cancel{\mathrm{m}\mathrm{e}\mathrm{t}\mathrm{e}\mathrm{r}}}\right)}^2=\frac{217\times <!--\; \times \; -->1,000,000}{10,000}\mathrm{m}{\mathrm{m}}^2=21,700\mathrm{m}{\mathrm{m}}^2$

21,700

A 50,000 cubic centimeter terrarium is roughly equivalent to a rectangular prism that is 37 centimeters on each side. This is a reasonable value.

50,000 cm³

A 236-milliliter carton of milk is close to a half-pint carton of milk. This is a reasonable value.

236 mL

A 1,500-cubic centimeter box of crackers, if it were a rectangular prism, would be approximately 11.4 cm on each side. This is a reasonable value.

1,500 cm³

42.5

$42,500{\cancel{\text{millimeters}}}^3{\left(\frac{1\cancel{\text{meter}}}{\text{1,000 }\cancel{\text{millimeter}}}\right)}^3{\left(\frac{\text{100 centimeters}}{\text{1 }\cancel{\text{meter}}}\right)}^3$

$=\frac{42,500\times <!--\; \times \; -->\cancel{\text{1,000,000}}}{1,000,\cancel{\text{000,000}}}\text{ c}{\text{m}}^3\text{ = }42.5\text{ c}{\text{m}}^3$

1,500

$1.5\cancel{\text{cubic decimeters}}\left(\frac{\text{ 1,000 milliliters}}{\text{ 1}\cancel{\text{cubic decimeter}}}\right)=1,500\text{ milliliters}$

Note: You often have more than one conversion from which to choose. You might have chosen to convert one cubic decimeter conversion to one liter, and then convert liters to milliliters.

$6.75{\cancel{\text{centimeters}}}^3{\left(\frac{1\cancel{\text{meter}}}{\text{100 }\cancel{\text{centimeters}}}\right)}^3{\left(\frac{\text{1,000 millimeters}}{\text{1 }\cancel{\text{meter}}}\right)}^3=\frac{6.75\times <!--\; \times \; -->\text{1,000,}\cancel{\text{000,000}}}{\cancel{\text{1,000,000}}}\text{ m}{\text{m}}^3\text{ = }6,750\text{ m}{\text{m}}^3$

6,750

800

Step 1: Convert 20 meters to centimeters.

$20\cancel{\text{meters}}\left(\frac{100\text{ centimeters}}{1\cancel{\text{meter}}}\right)=2,000\text{ centimeters}$

Step 2: Find the volume of a box.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{20 cm}\right)\left(\text{2,000 cm}\right)\left(\text{20 cm}\right)=800,000\text{ c}{\text{m}}^3$

Step 3: Convert the volume of a box into liters.

$800,000\cancel{\text{cubic centimeters}}\left(\frac{\text{ 1 liter}}{\text{ 1,000}\cancel{\text{cubic centimeters}}}\right)=800\text{ liters}$

Step 1: Convert all the units to centimeters.

$12\cancel{\text{millimeters}}\left(\frac{1\cancel{\text{meter}}}{\text{1,0}\cancel{\text{00}}\cancel{\text{millimeter}}}\right)\left(\frac{\cancel{100}\text{ centimeters}}{1\cancel{\text{meter}}}\right)=1.2\text{ centimeters}$

Step 2: Find the volume of a box.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{1}\text{.2 cm}\right)\left(\text{5 cm}\right)\left(\text{1}\text{.7 cm}\right)=10.2\text{ c}{\text{m}}^3$

Step 3: Convert the volume of a box into milliliters.

$10.2\cancel{\text{cubic centimeters}}\left(\frac{\text{ 1 milliliter}}{\text{ 1}\cancel{\text{cubic centimeter}}}\right)=10.2\text{ milliliters}$

10.2

163.52

Step 1: Find the volume of a box.

$V=L\times <!--\; \times \; -->W\times <!--\; \times \; -->H=\left(\text{7}\text{.3 m}\right)\left(\text{3}\text{.2 m}\right)\left(\text{7 m}\right)=163.52{\text{m}}^3$

A gram is about the weight of a paper clip. A candy bar weighs about the same as 50 paper clips, so 50 grams is reasonable.

50 g

A liter of water weighs about the same as one kilogram, so 180 kilograms is reasonable for the lion.

180 kg

A gram is about the weight of a paper clip. A basketball weighs about the same as 624 paper clips, so 624 grams is reasonable.

624 g

$8,900\cancel{\text{grams}}\left(\frac{1\text{ kilogram}}{1,000\cancel{\text{grams}}}\right)=\text{ 8}\text{.9 kilograms}$

8.9

$17\cancel{\text{grams}}\left(\frac{1,000\text{ milligrams}}{1\cancel{\text{gram}}}\right)=17,000\text{ milligrams}$

17,000

$0.07\cancel{\text{kilograms}}\left(\frac{\text{1000 grams}}{\text{ 1}\cancel{\text{kilogram}}}\right)=70\text{ grams}$

70

Step 1: Convert the grams to kilograms.

$48\cancel{\text{grams}}\left(\frac{1\text{ kilogram}}{1,000\cancel{\text{grams}}}\right)=\text{ 0}\text{.048 kilograms}$

Step 2: Multiply by 3.

$\text{3(0}\text{.048 kg) = 0}\text{.144 kg}$

0.144

Step 1: Convert the grams to milligrams.

$28\cancel{\text{grams}}\left(\frac{1,000\text{ milligrams}}{1\cancel{\text{gram}}}\right)=28,000\text{ milligrams}$

Step 2: Multiply by 7.

$\text{7(28,000 mg) = 196,000 milligrams}$

196,000

Step 1: Convert the milligrams to grams.

$15\cancel{\text{milligrams}}\left(\frac{\text{ 1 gram}}{\text{ 1,000}\cancel{\text{milligrams}}}\right)=0.015\text{ grams}$

Step 2: Multiply by 6.

$\text{6(0}\text{.015 g) = 0}\text{.09 grams}$

0.09

Water freezes at 0°C, which is 32°F. The reasonable temperature is 28.5°F.

28.5 °F

Water freezes at 0°C, which is 32°F. A room temperature of 20°F would be below the freezing temperature of water. Most people would prefer the 20°C room temperature.

20 °C

The temperature of boiling water is 100°C or 212°F. As a cup of coffee cools, the more reasonable temperature is 71.1°C. This temperature in Fahrenheit degrees would be very lukewarm.

71.1 °C

$C=\frac{F-<!--\; -\; -->30}{2}$

$C=\frac{450-<!--\; -\; -->30}{2}$

$C=\frac{420}{2}$

${210}^{\circ <!--\; \circ \; -->}\mathrm{C}$

210

$F=2C+30$

$F=2\left(35\right)+30$

$F=70+30$

${100}^{\circ <!--\; \circ \; -->}\mathrm{F}$

100

$C=\frac{F-<!--\; -\; -->30}{2}$

$C=\frac{100-<!--\; -\; -->30}{2}$

$C=\frac{70}{2}$

${35}^{\circ <!--\; \circ \; -->}\mathrm{C}$

35

$C=\frac{5}{9}\left(F-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(225-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(193\right)$

${107}^{\circ <!--\; \circ \; -->}\mathrm{C}$

107

$F=\frac{9}{5}C+32$

$F=\frac{9}{5}\left(27\right)+32$

$F=48.6+32$

${81}^{\circ <!--\; \circ \; -->}\mathrm{F}$

81

$C=\frac{5}{9}\left(F-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(750-<!--\; -\; -->32\right)$

$C=\frac{5}{9}\left(718\right)$

${399}^{\circ <!--\; \circ \; -->}\mathrm{C}$

399