# 9.3 Simplification of denominate numbers  (Page 2/2)

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## Practice set b

Perform each operation. Simplify when possible.

Add 4 gal 3 qt to 1 gal 2 qt.

6 gal 1 qt

Add 9 hr 48 min to 4 hr 26 min.

14 hr 14 min

Subtract 2 ft 5 in. from 8 ft 7 in.

6 ft 2in.

Subtract 15 km 460 m from 27 km 800 m.

12 km 340 m

Subtract 8 min 35 sec from 12 min 10 sec.

3 min 35 sec

Add 4 yd 2 ft 7 in. to 9 yd 2 ft 8 in.

14 yd 2 ft 3 in

Subtract 11 min 55 sec from 25 min 8 sec.

13 min 13 sec

## Multiplying a denominate number by a whole number

Let's examine the repeated sum

$\underset{\text{3 times}}{\underbrace{\text{4 ft 9 in.}+\text{4 ft 9 in.}+\text{4 ft 9 in.}}}=\text{12 ft 27 in.}$

Recalling that multiplication is a description of repeated addition, by the distribu­tive property we have

$\begin{array}{cccc}\hfill 3\left(\text{4 ft 9 in}\text{.}\right)& =& 3\left(\text{4 ft}+\text{9 in.}\right)\hfill & \\ & =& 3\cdot \text{4 ft}+3\cdot \text{9 in.}\hfill & \\ & =& \text{12 ft}+\text{27 in.}\hfill & \text{Now, 27 in.}=\text{2 ft 3 in.}\hfill \\ & =& \text{12 ft}+\text{2 ft}+\text{3 in.}\hfill & \\ & =& \text{14 ft}+\text{3 in.}\hfill & \\ & =& \text{14 ft 3 in.}\hfill & \end{array}$

From these observations, we can suggest the following rule.

## Multiplying a denominate number by a whole number

To multiply a denominate number by a whole number, multiply the number part of each unit by the whole number and affix the unit to this product.

## Sample set c

Perform the following multiplications. Simplify if necessary.

$\begin{array}{ccc}\hfill 6\cdot \left(\text{2 ft 4 in.}\right)& =& 6\cdot \text{2 ft}+6\cdot \text{4 in.}\hfill \\ & =& \text{12 ft}+\text{24 in.}\hfill \end{array}$

Since $\text{3 ft}=\text{1 yd}$ and ,

$\begin{array}{ccc}\hfill \text{12 ft}+\text{24 in.}& =& \text{4 yd}+\text{2 ft}\hfill \\ & =& \text{4 yd 2 ft}\hfill \end{array}$

$\begin{array}{ccc}\hfill 8\cdot \left(\text{5 hr 21 min 55 sec}\right)& =& 8\cdot 5\text{hr}+8\cdot \text{21 min}+8\cdot \text{55 sec}\hfill \\ & =& \text{40 hr}+\text{168 min}+\text{440sec}\hfill \\ & =& \text{40 hr}+\text{168 min}+\text{7 min}+\text{20 sec}\hfill \\ & =& \text{40 hr}+\text{175 min}+\text{20 sec}\hfill \\ & =& \text{40 hr}+\text{2 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{42 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{24hr}+\text{18hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{1 da}+\text{18 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{1 da 18 hr 55 min 20 sec}\hfill \end{array}$

## Practice set c

Perform the following multiplications. Simplify.

$2\cdot \left(\text{10 min}\right)$

20 min

$5\cdot \left(\text{3 qt}\right)$

$\text{15 qt}=\text{3 gal 3 qt}$

$4\cdot \left(5\text{ft 8 in}\text{.}\right)$

$\text{20 ft 32 in}\text{.}=\text{7 yd 1 ft 8 in}\text{.}$

$\text{10}\cdot \left(2\text{hr 15 min 40 sec}\right)$

$\text{20 hr 150 min 400 sec}=\text{22 hr 36 min 40 sec}$

## Dividing a denominate number by a whole number

To divide a denominate number by a whole number, divide the number part of each unit by the whole number beginning with the largest unit. Affix the unit to this quotient. Carry any remainder to the next unit.

## Sample set d

Perform the following divisions. Simplify if necessary.

$\left(\text{12 min 40 sec}\right)÷4$

Thus $\left(\text{12 min 40 sec}\right)÷4=3\text{min 10 sec}$

$\left(\text{5 yd 2 ft 9 in}\text{.}\right)÷3$

$\begin{array}{}\text{Convert to feet: 2 yd 2 ft}=\text{8 ft}\text{.}\\ \text{Convert to inches: 2 ft 9 in}\text{.}=\text{33 in}\text{.}\end{array}$

Thus $\left(\text{5 yd 2 ft 9 in}\text{.}\right)÷3=1\text{yd 2 ft 11 in}\text{.}$

## Practice set d

Perform the following divisions. Simplify if necessary.

$\left(\text{18 hr 36 min}\right)÷9$

2 hr 4 min

$\left(\text{34 hr 8 min}\text{.}\right)÷8$

4 hr 16 min

$\left(\text{13 yd 7 in}\text{.}\right)÷5$

2 yd 1 ft 11 in

$\left(\text{47 gal 2 qt 1 pt}\right)÷3$

15 gal 3 qt 1 pt

## Exercises

For the following 15 problems, simplify the denominate numbers­.

16 in.

1 foot 4 inches

19 ft

85 min

1 hour 25 minutes

90 min

17 da

2 weeks 3 days

25 oz

240 oz

15 pounds

3,500 lb

26 qt

6 gallons 2 quarts

300 sec

135 oz

8 pounds 7 ounces

14 tsp

18 pt

2 gallons 1 quart

3,500 m

16,300 mL

16 liters 300 milliliters (or 1daL 6 L 3dL)

For the following 15 problems, perform the indicated opera­tions and simplify the answers if possible.

Add 6 min 12 sec to 5 min 15 sec.

Add 14 da 6 hr to 1 da 5 hr.

15 days 11 hours

Add 9 gal 3 qt to 2 gal 3 qt.

Add 16 lb 10 oz to 42 lb 15 oz.

59 pounds 9 ounces

Subtract 3 gal 1 qt from 8 gal 3 qt.

Subtract 3 ft 10 in. from 5 ft 8 in.

1 foot 10 inches

Subtract 5 lb 9 oz from 12 lb 5 oz.

Subtract 10 hr 10 min from 11 hr 28 min.

1 hour 18 minutes

Add 3 fl oz 1 tbsp 2 tsp to 5 fl oz 1 tbsp 2 tsp.

Add 4 da 7 hr 12 min to 1 da 8 hr 53 min.

5 days 16 hours 5 minutes

Subtract 5 hr 21 sec from 11 hr 2 min 14 sec.

Subtract 6 T 1,300 lb 10 oz from 8 T 400 lb 10 oz.

1 ton 1,100 pounds (or 1T 1,100 lb)

Subtract 15 mi 10 in. from 27 mi 800 ft 7 in.

Subtract 3 wk 5 da 50 min 12 sec from 5 wk 6 da 20 min 5 sec.

2 weeks 23 hours 29 minutes 53 seconds

Subtract 3 gal 3 qt 1 pt 1 oz from 10 gal 2 qt 2 oz.

## Exercises for review

( [link] ) Find the value: ${\left(\frac{5}{8}\right)}^{2}+\frac{\text{39}}{\text{64}}$ .

1

( [link] ) Find the sum: $8+6\frac{3}{5}$ .

( [link] ) Convert $2\text{.}\text{05}\frac{1}{\text{11}}$ to a fraction.

$2\frac{\text{14}}{\text{275}}$

( [link] ) An acid solution is composed of 3 parts acid to 7 parts water. How many parts of acid are there in a solution that contains 126 parts water?

( [link] ) Convert 126 kg to grams.

126,000 g

#### Questions & Answers

can someone help me with some logarithmic and exponential equations.
sure. what is your question?
20/(×-6^2)
Salomon
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I don't understand what the A with approx sign and the boxed x mean
it think it's written 20/(X-6)^2 so it's 20 divided by X-6 squared
Salomon
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Salomon
I got X =-6
Salomon
ok. so take the square root of both sides, now you have plus or minus the square root of 20= x-6
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