rass22706 rass22706

29-11-2019
Mathematics

contestada

If y>x>1, and (x^1/3 y^1/6) = 256 what is the value of xy?

Respuesta :

PollyP52 PollyP52

29-11-2019

Answer:

xy = 2^39.

Step-by-step explanation:

8 *32 = 256

so x could be 8^3 and y could be 32^6

So xy = 8^3 * 32^6

= (2^3)^3 * (2^5)^6

= 2^9 * 2^30

= 2^39.

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