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How do I simplify this algebra equation?

0 votes

h = 16/pi(4/pi)^(2/3)

I was asked by my professor to not solve and give an approximation but to simplify for h and leave the (pi) alone

asked Apr 3, 2013 in ALGEBRA 1 by anonymous Apprentice

3 Answers

+1 vote

h = 16 / pi(4 / pi)2/3

Apply power rule (a / b)x = ax / ay

h = 16 / pi(42/3 / pi2/3)

h = 16 / (pi / pi2/3(42/3)

Apply power rule (a / a2/3 = a1-2/3

h = 16 / (pi1-2/3)(42/3)

h = 16 / (pi1/3)(161/3)

h = 161-1/3 / (pi)1/3

h = 162/3 / pi1/3

h = 2561/3 / pi1/3

Apply power rule (a / b)1/3 = a1/3 / b1/3

Therefore h = (256 / pi)1/3.

answered Apr 3, 2013 by diane Scholar
0 votes

h = (16/π)(4/π)^2/3

= 16* π^(-1)*4^(2/3) * π^(2/3)

= (16 * ∛4² ) π^(-1/3)

= 32 ∛2 / ∛π

that's 32 times cube root of 2 over cube root of pi.

Source: http://answers.yahoo.com/

answered Apr 4, 2013 by Naren Answers Apprentice

Simplify for h, the equation h = (16/π)(4/π)^2/3

h = (32 ∛2)  (π)^(-5/3).

0 votes

h = (16/π)(4/π)^2/3

= 16* π^(-1)*4^(2/3) * π^(-2/3)

= (16 * ∛4² ) π^(-1-2/3)

= (16 * ∛4² ) π^(-1-2/3)

= (32 ∛2)  (π)^(-5/3)

answered Apr 4, 2013 by steve Scholar

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