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Perform each operation. Simplify when possible.
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
$\begin{array}{cccc}\hfill 3(\text{4 ft 9 in}\text{.})& =& 3(\text{4 ft}+\text{9 in.})\hfill & \\ & =& 3\cdot \text{4 ft}+3\cdot \text{9 in.}\hfill & \\ & =& \text{12 ft}+\text{27 in.}\hfill & \text{Now, 27 in.}=\text{2 ft 3 in.}\hfill \\ & =& \text{12 ft}+\text{2 ft}+\text{3 in.}\hfill & \\ & =& \text{14 ft}+\text{3 in.}\hfill & \\ & =& \text{14 ft 3 in.}\hfill & \end{array}$
From these observations, we can suggest the following rule.
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 $\text{12}\text{in}\text{.}=\text{1 ft}$ ,
$\begin{array}{ccc}\hfill \text{12 ft}+\text{24 in.}& =& \text{4 yd}+\text{2 ft}\hfill \\ & =& \text{4 yd 2 ft}\hfill \end{array}$
$\begin{array}{ccc}\hfill 8\cdot \left(\text{5 hr 21 min 55 sec}\right)& =& 8\cdot 5\text{hr}+8\cdot \text{21 min}+8\cdot \text{55 sec}\hfill \\ & =& \text{40 hr}+\text{168 min}+\text{440sec}\hfill \\ & =& \text{40 hr}+\text{168 min}+\text{7 min}+\text{20 sec}\hfill \\ & =& \text{40 hr}+\text{175 min}+\text{20 sec}\hfill \\ & =& \text{40 hr}+\text{2 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{42 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{24hr}+\text{18hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{1 da}+\text{18 hr}+\text{55 min}+\text{20 sec}\hfill \\ & =& \text{1 da 18 hr 55 min 20 sec}\hfill \end{array}$
Perform the following multiplications. Simplify.
$5\cdot \left(\text{3 qt}\right)$
$\text{15 qt}=\text{3 gal 3 qt}$
$4\cdot \left(5\text{ft 8 in}\text{.}\right)$
$\text{20 ft 32 in}\text{.}=\text{7 yd 1 ft 8 in}\text{.}$
$\text{10}\cdot \left(2\text{hr 15 min 40 sec}\right)$
$\text{20 hr 150 min 400 sec}=\text{22 hr 36 min 40 sec}$
Perform the following divisions. Simplify if necessary.
$\left(\text{12 min 40 sec}\right)\xf74$
Thus $\left(\text{12 min 40 sec}\right)\xf74=3\text{min 10 sec}$
$\left(\text{5 yd 2 ft 9 in}\text{.}\right)\xf73$
$\begin{array}{}\text{Convert to feet: 2 yd 2 ft}=\text{8 ft}\text{.}\\ \text{Convert to inches: 2 ft 9 in}\text{.}=\text{33 in}\text{.}\end{array}$
Thus $\left(\text{5 yd 2 ft 9 in}\text{.}\right)\xf73=1\text{yd 2 ft 11 in}\text{.}$
Perform the following divisions. Simplify if necessary.
$\left(\text{13 yd 7 in}\text{.}\right)\xf75$
2 yd 1 ft 11 in
$\left(\text{47 gal 2 qt 1 pt}\right)\xf73$
15 gal 3 qt 1 pt
For the following 15 problems, simplify the denominate numbers.
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 9 gal 3 qt to 2 gal 3 qt.
Subtract 3 gal 1 qt from 8 gal 3 qt.
Subtract 5 lb 9 oz from 12 lb 5 oz.
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.
( [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?
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