Question

MEDAL!!!


Problem will be posted below...

Answer 1

If \(k\), \(n\), \(x\), and \(y\) are positive numbers satisfying \(x^{-\frac{4}{3}} = k^{-2}\) and \(y\frac{4}{3} = n^2\), and is \((xy)^{-2\frac{2}{3}}\) in terms of \(n\) and \(k\) ?

Answer 2

The second \(\rm and\) is supposed to be \(\rm what\).

Answer 3

we can write this:
\[\Large {y^{ - 4/3}} = \frac{1}{{{n^2}}}\]

Answer 4

(kn)^2

Answer 5

Isn't it supposed to be
\[\large y^{-4/3} = \frac{1}{\color \red{k}^2}\]

Answer 6

therefore:
\[\Large {\left( {xy} \right)^{ - 4/3}} = {x^{ - 4/3}}{y^{ - 4/3}} = ...?\]

Answer 7

I don't understand that

Answer 8

I have applied a property of powers

Answer 9

where are u stuck?
what is your approach ?
how do u think we can do this?

Answer 10

@Michele_Laino i think -2(2/3) is a mixed fraction

Answer 11

I don't know how to evaluate something that has a fraction as its exponent.

Answer 12

\[-2\frac{ 2}{ 3 }=\frac{ (-2 \times 3)+2 }{ 3}\]

Answer 13

Is that supposed to help me evaluate something that has a fraction as its exponent? @mom.

Answer 14

yes! you are right! @baru

Answer 15

first we convert the mixed fraction 2&2/3 to 8/3

Answer 16

next step is:
\[{\left( {xy} \right)^{ - 2\;2/3}} = {\left( {xy} \right)^{ - 8/3}} = {\left( {{{\left( {xy} \right)}^{ - 4/3}}} \right)^2}\]

Answer 17

then express x in terms of a power of k

Answer 18

yes

Answer 19

& y as a power of n

Answer 20

and substitute into micheles new equation

Answer 21

just use the normal rules of exponents on fractions just like we do for whole numbers

Answer 22

@Michele_Laino Why did you raise the last part to the second power?

Answer 23

there's no real need to do that

Answer 24

have you worked out the new expressions for x & y ?

Answer 25

since we have to compute this:
\[\Large {\left( {xy} \right)^{ - 2\;2/3}}\]

Answer 26

I still don't understand.

Can we go through this step-by-step?

Answer 27

michele do you want to help him out?

Answer 28

we have the subsequent steps:
\[\Large 2\frac{2}{3} = 2 + \frac{2}{3} = \frac{8}{3} = 2 \cdot \left( {\frac{4}{3}} \right)\]

Answer 29

ok! @alekos

Answer 30

therefore:
\[\Large {\left( {xy} \right)^{ - 2\;2/3}} = {\left( {xy} \right)^{ - 8/3}} = {\left( {xy} \right)^{ - 2 \cdot \left( {4/3} \right)}} = {\left( {{{\left( {xy} \right)}^{ - 4/3}}} \right)^2}\]

Answer 31

Oh okay I understand now! Thanks @Michele_Laino

Answer 32

:)