Y varies jointly as the cube of x and the square root of w, and Y = 128 when x = 2 and w = 16. Find Y when x = 1/2 and w = 64? P.S. Thank you for helping me for this problem.

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Answer 1 · 2017-03-14T15:44:47

Given that y varies jointly as the cube of x and the square root of w,

$y=ax^3xxsqrtw.....(1)$ ,

where $a$ variation constant

Again inserting

y = 128 when x = 2 and w = 16 in equation (1)

$128=axx2^3xxsqrt16$

$=>128=axx8xx4$

$=>a=4$

Now the equation (1) becomes

$y=4x^3xxsqrtw$

Inserting x = 1/2 and w = 64 we get
$y=4(1/2)^3xxsqrt64$

$=>y=4xx1/8xx8=4$