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Posted: 4/2/2002 1:53:15 AM EDT
Okay, i must have forgot my friggin algebra, but the equation is: [((1/2)^(2/3))(a^(1/3))(b^(2/3))(y)]+[(2^(1/3))(a^(1/3))(b^(2/3))(y)] It might not be clear, so I'll say: The first quantity consists of: One-half raised to the 2/3, times a raised to the 1/3 times b raised to the 2/3 times y. Add this to the second quantity: 2 raised to the 1/3 times a raised to the 1/3 times b raised to the 2/3 times y. The answer says: 3[(ab^2)/4]^(1/3)(y) 3 times the quantity: a times b squared all over 4 Quantity raised to the 1/3 All of this times "y". Please write if you need any clarification. If the above is the answer, what are the steps to get to it? If not, what is the answer and the steps involved? Thanks!
Link Posted: 4/2/2002 2:36:51 AM EDT
Sorry man, wish I could help. College was way too many years ago. I have dig out my old books to help my daughters with their homework these days. Good luck. Somebody here [b]will[/b] be able to help you though. Try instant messaging DPeacher.....
Link Posted: 4/2/2002 6:22:20 AM EDT
Your answer looks right to me. Here's how I get it: (1/2)^(2/3)a^(1/3)b^(2/3)y + 2^(1/3)a^(1/3)b^(2/3)y = [(1/2)^(2/3) + 2^(1/3)] a^(1/3)b^(2/3)y = [(1/2)^(2/3) + 2^(1/3)] (ab^2)^(1/3)y = [(1/4)^(1/3) + 2^(1/3)] (ab^2)^(1/3)y = [(1/4)^(1/3) + (8/4)^(1/3)] (ab^2)^(1/3)y = [(1/4)^(1/3) + 2*(1/4)^(1/3)] (ab^2)^(1/3)y = 3*[(1/4)^(1/3)] (ab^2)^(1/3)y = 3*[(1/4)(ab^2)]^(1/3)y
Link Posted: 4/3/2002 3:25:46 PM EDT
Originally Posted By SoCalGunner: Okay, i must have forgot my friggin algebra, but the equation is: [((1/2)^(2/3))(a^(1/3))(b^(2/3))(y)]+[(2^(1/3))(a^(1/3))(b^(2/3))(y)] It might not be clear, so I'll say: The first quantity consists of: One-half raised to the 2/3, times a raised to the 1/3 times b raised to the 2/3 times y. Add this to the second quantity: 2 raised to the 1/3 times a raised to the 1/3 times b raised to the 2/3 times y. The answer says: 3[(ab^2)/4]^(1/3)(y) 3 times the quantity: a times b squared all over 4 Quantity raised to the 1/3 All of this times "y". Please write if you need any clarification. If the above is the answer, what are the steps to get to it? If not, what is the answer and the steps involved? Thanks!
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Your equation would equal the following algebraic collection of terms The orginal equation would go on to ... =[(1/2)^(2/3)+(2^1/2)(a^1/3)(b^2/3)(y) =(3/(2^2/3)(a^1/3)(b^2/3)(y) Then the answer is....!! =3[ab^2/4]^(1/3)(y)
Link Posted: 4/3/2002 3:29:19 PM EDT
Link Posted: 4/3/2002 3:42:22 PM EDT
I'm glad I don't use that in real life. Home Depot scenario: I'll take 1/2 over 2/3rds of those bolts, plus, 2 over 1/3'rd nuts, plus, yb over 2/3rd washers please.
Link Posted: 4/3/2002 4:17:58 PM EDT
Originally Posted By Wolfpack: Why did I feel like an instant idiot 2 seconds after clicking this thread? [:E]
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You are not alone! Fred
Link Posted: 4/3/2002 4:19:55 PM EDT
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