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Suppose that $a$, $b$, and $c$ are real numbers such that $\frac{a}{b} = \frac{\sqrt{10}}{\sqrt{21}}$ and $\frac{b}{c} = \frac{\sqrt{135}}{\sqrt{8}}$. Find $\frac{a}{c}$. Completely simplify and rationalize the denominator.

Guest Jan 13, 2018

#1
+5924
+1

$$\frac{a}{b}\,=\,\frac{\sqrt{10}}{\sqrt{21}} \qquad\text{so}\qquad a\,=\,\frac{b\sqrt{10}}{\sqrt{21}}\\~\\ \frac{b}{c}\,=\,\frac{\sqrt{135}}{\sqrt8}\qquad\text{so}\qquad c\,=\,\frac{b\sqrt{8}}{\sqrt{135}} \\~\\ \ \\~\\\frac{a}{c}\,=\,(\frac{b\sqrt{10}}{\sqrt{21}})\,/\,(\frac{b\sqrt{8}}{\sqrt{135}}) \,=\,(\frac{b\sqrt{10}}{\sqrt{21}})(\frac{\sqrt{135}}{b\sqrt{8}}) \,=\,(\frac{\sqrt{10}}{\sqrt{21}})(\frac{\sqrt{135}}{\sqrt{8}}) \\~\\ \frac{a}{c}\,=\,\frac{\sqrt{10\,\cdot\,135}}{\sqrt{21\,\cdot\,8}} \,=\,\frac{\sqrt{2\cdot5\cdot5\cdot3\cdot3\cdot3}}{\sqrt{3\cdot7\cdot2\cdot2\cdot2}} \,=\,\frac{\sqrt{2}\cdot\sqrt{5\cdot5}\cdot\sqrt{3\cdot3}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{7}\cdot\sqrt{2\cdot2}\cdot\sqrt{2}} \,=\,\frac{\sqrt{5\cdot5}\cdot\sqrt{3\cdot3}}{\sqrt{7}\cdot\sqrt{2\cdot2}} \\~\\ \frac{a}{c}\,=\,\frac{5\,\cdot\,3}{2\sqrt{7}} \,=\,\frac{15\sqrt7}{14}$$

hectictar  Jan 13, 2018
edited by hectictar  Jan 13, 2018
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#1
+5924
+1

$$\frac{a}{b}\,=\,\frac{\sqrt{10}}{\sqrt{21}} \qquad\text{so}\qquad a\,=\,\frac{b\sqrt{10}}{\sqrt{21}}\\~\\ \frac{b}{c}\,=\,\frac{\sqrt{135}}{\sqrt8}\qquad\text{so}\qquad c\,=\,\frac{b\sqrt{8}}{\sqrt{135}} \\~\\ \ \\~\\\frac{a}{c}\,=\,(\frac{b\sqrt{10}}{\sqrt{21}})\,/\,(\frac{b\sqrt{8}}{\sqrt{135}}) \,=\,(\frac{b\sqrt{10}}{\sqrt{21}})(\frac{\sqrt{135}}{b\sqrt{8}}) \,=\,(\frac{\sqrt{10}}{\sqrt{21}})(\frac{\sqrt{135}}{\sqrt{8}}) \\~\\ \frac{a}{c}\,=\,\frac{\sqrt{10\,\cdot\,135}}{\sqrt{21\,\cdot\,8}} \,=\,\frac{\sqrt{2\cdot5\cdot5\cdot3\cdot3\cdot3}}{\sqrt{3\cdot7\cdot2\cdot2\cdot2}} \,=\,\frac{\sqrt{2}\cdot\sqrt{5\cdot5}\cdot\sqrt{3\cdot3}\cdot\sqrt{3}}{\sqrt{3}\cdot\sqrt{7}\cdot\sqrt{2\cdot2}\cdot\sqrt{2}} \,=\,\frac{\sqrt{5\cdot5}\cdot\sqrt{3\cdot3}}{\sqrt{7}\cdot\sqrt{2\cdot2}} \\~\\ \frac{a}{c}\,=\,\frac{5\,\cdot\,3}{2\sqrt{7}} \,=\,\frac{15\sqrt7}{14}$$

hectictar  Jan 13, 2018
edited by hectictar  Jan 13, 2018
#3
+91445
+1

None of this is LaTex displaying for me at all.

Melody  Jan 13, 2018
#2
+80978
+2

Find  $$\frac{a}{c}$$

Given :  $$\frac{a}{b} = \frac{\sqrt{10}}{\sqrt{21}}$$         $$\frac{b}{c} = \frac{\sqrt{135}}{\sqrt{8}}$$

b  =  √(21/10) a

b =  √(135/ 8)  c

Which implies  that

√(21/10) a  =   √(135/ 8)  c      ⇒

a / c   =   √(135/ 8)  /   √(21/10)

a / c  =  √   [ ( 135 * 10)  / (21 * 8) ]  =  √ [ (45 * 5)  / (7 * 4) ]   =  15 / [ 2√7 ]  =

15√7 / 14

CPhill  Jan 13, 2018

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