I need help in solving this problem (sorry I didn't know how to write it on here).
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1I think the question is evaluating $(5/3)^3(-3/5)^2$ into simplest terms. – Kaj Hansen Jun 17 '14 at 00:10
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Usually an equation needs an equals sign. Do you mean $\frac 53 x^3 = - \left(\frac{3}{5}\right)^2$? – Sten Jun 17 '14 at 00:10
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1404 calculus/linear algebra. – Shahar Jun 17 '14 at 00:10
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Sorry It's not an equation – LuisDavis Jun 17 '14 at 00:18
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Do you know what powers are? Do you know how to multiply rational numbers together? Do you know that negative times negative is positive? If the answer to all three of these questions is yes, then you have all the tools you need to to simplify the expression. All you have to do is use them. – anon Jun 17 '14 at 00:22
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Yeah thanks i figured it out – LuisDavis Jun 17 '14 at 00:27
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Why all the dowmvotes? Seems like a legitimate question. – Cheerful Parsnip Jun 17 '14 at 00:42
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This is a question about skill, not a mathematical question. – Hans Engler Jun 17 '14 at 01:03
3 Answers
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Hint:
$\left(\dfrac{5}{3}\right)^3*\left(-\dfrac{3}{5}\right)^2$ = $\left(\dfrac{5^3}{3^3}\right)*\left(\dfrac{(-3)^2}{5^2}\right)$
Another hint:
To multiply two fractions, you can multiply the two numerators by each other, and the two denominators by each other to get the new numerator and denominator.
RogueBaneling
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**Hint:**$$\left(\frac 53 \right)^3 \left(-\frac 35 \right)^2=\left(\frac 53 \right)\left(\frac 53 \right)^2 \left(-\frac 35 \right)^2=\left(\frac 53 \right)\left(\frac 53 \cdot-\frac 35 \right)^2$$ Can you simplify $\displaystyle \frac 53 \cdot -\frac 35$?
Cookie
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$\begin{align}\left(\dfrac{5}{3}\right)^3\left(\dfrac{-3}{5}\right)^2 & = \left(\dfrac{5^3}{3^3}\right)\cdot\left(\dfrac{(-1)^2 3^2}{5^2}\right) & \text{by commutativity of exponents} \\ ~ & = \dfrac{(-1)^2 5^1}{3^1} & \text{by associativity of exponents} \\ ~ & = \dfrac{5}{3} & \text{by }(-1)^2=1, a^1 = a \end{align}$
Graham Kemp
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