D dlawrence New member Joined Jan 30, 2012 Messages 2 Jan 30, 2012 #1 Here is the problem: (c-d)/((1/d)-(1/c)) Possible answers are: a. (c-d)/dc b. dc/(c-d) c. dc d. -dc e. 1/dc
Here is the problem: (c-d)/((1/d)-(1/c)) Possible answers are: a. (c-d)/dc b. dc/(c-d) c. dc d. -dc e. 1/dc
D dlawrence New member Joined Jan 30, 2012 Messages 2 Jan 30, 2012 #2 (c-d) * [(cd)/(c-d)] = cd or dc Thanks so much, do you think you could tell me the name of that rule?
(c-d) * [(cd)/(c-d)] = cd or dc Thanks so much, do you think you could tell me the name of that rule?
S soroban Elite Member Joined Jan 28, 2005 Messages 5,584 Jan 30, 2012 #3 Hello, dlawrence! \(\displaystyle \dfrac{c-d}{\frac{1}{d} - \frac{1}{c}}\) . . \(\displaystyle (a)\;\dfrac{c-d}{cd} \qquad (b)\;\dfrac{cd}{c-d} \qquad (c)\;cd \qquad (d)\;-cd \qquad (e)\;\dfrac{1}{cd}\) Why can't they write varaibles in alphabetical order? Click to expand... Multiply the fraction by \(\displaystyle \frac{cd}{cd}\!:\;\;\dfrac{cd(c-d)}{cd\left(\frac{1}{d} - \frac{1}{c}\right)}\;=\;\dfrac{cd(c-d)}{c - d} \;=\;cd \quad\text{ . . . answer (c )} \)
Hello, dlawrence! \(\displaystyle \dfrac{c-d}{\frac{1}{d} - \frac{1}{c}}\) . . \(\displaystyle (a)\;\dfrac{c-d}{cd} \qquad (b)\;\dfrac{cd}{c-d} \qquad (c)\;cd \qquad (d)\;-cd \qquad (e)\;\dfrac{1}{cd}\) Why can't they write varaibles in alphabetical order? Click to expand... Multiply the fraction by \(\displaystyle \frac{cd}{cd}\!:\;\;\dfrac{cd(c-d)}{cd\left(\frac{1}{d} - \frac{1}{c}\right)}\;=\;\dfrac{cd(c-d)}{c - d} \;=\;cd \quad\text{ . . . answer (c )} \)
