What is 4s over 3t to the negative 2nd power times 2s over 6t to the 2nd power? The format is a bit weird .

Question

${(\frac{4 s}{3 t})}^{- 2} \times {(\frac{2 s}{6 t})}^{2}$ $((4s)/(3t))^(-2) xx ((2s)/(6t))^2$

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Answer 1 · 2018-04-18T22:35:08

$\frac{1}{16}$ $1/16$

${(\frac{4 s}{3 t})}^{- 2} \cdot {(\frac{2 s}{6 t})}^{2}$ $((4s)/(3t))^(-2)*((2s)/(6t))^2$

First when given a negative exponent, I reciprocate the expression and make the exponent positive, so:

${(\frac{3 t}{4 s})}^{2} \cdot {(\frac{2 s}{6 t})}^{2}$ $((3t)/(4s))^(2)*((2s)/(6t))^2$

$\frac{{(3 t)}^{2}}{{(4 s)}^{2}} \cdot \frac{{(2 s)}^{2}}{{(6 t)}^{2}}$ $(3t)^2/(4s)^2*(2s)^2/(6t)^2$

$\frac{(3 t) (3 t)}{(4 s) (4 s)} \cdot \frac{(2 s) (2 s)}{(6 t) (6 t)}$ $((3t)(3t))/((4s)(4s))*((2s)(2s))/((6t)(6t))$

$\frac{9 t^{2}}{16 s^{2}} \cdot \frac{4 s^{2}}{36 t^{2}}$ $(9t^2)/(16s^2)*(4s^2)/(36t^2)$

Cross simplify:

$(cancel(9t^2))/(cancel(16s^2)4)*(cancel(4s^2))/(cancel(36t^2)4)=$

$1/16$