x^(3/2) (x+x^(5/2) - x^2)
Multiply through..
[x^(3/2) * x] + [x^(3/2) * x^(5/2)] - [x^(3/2) * x^2]
= x^(3/2 + 1) + x^(3/2 + 5/2) - x^(3/2 + 2)
= x^(5/2) + x^(4) - x^(7/2)
That's all you can do.
2007-06-20 08:59:22
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answer #1
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answered by Mathematica 7
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This problem uses a combination of distributive property A(B + C) = AB + AC. And the exponent addition rule X^a * X^b = X^(a+b).
So for this particular problem, you have x^(3/2 + 1) + x^(3/2 + 5/2) + x^(3/2 + 2)
= x^5/2 + x^4 + x^7/2, and that's as simplified as you can get.
2007-06-20 08:59:04
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answer #2
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answered by Vangorn2000 6
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x^(3/2)[ x^1 + x^(5/2) - x^2 ]
By the distributive law and (a^m)(a^n) = a^(m+n):
= x^(3/2 + 1) + x^(3/2 + 5/2) - x^(3/2 + 2)
= x^(3/2 + 2/2) + x^(8/2) - x^(3/2 + 4/2)
= x^(5/2) + x^4 - x^(7/2)
2007-06-20 08:59:40
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answer #3
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answered by mathjoe 3
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add the powers
x^3/2 (x+x^5/2-x^2)
x^3/2(x^1+x^5/2-x^2)
=> x^5/2 + x^8/2 - x^7/2
=> x^5/2 + x^4 - x^7/2
2007-06-20 09:00:13
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answer #4
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answered by harry m 6
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= x^(5/2) + x^4 - x^(7/2)
2007-06-20 10:57:10
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answer #5
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answered by Como 7
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x^3/2(x+x^5/2-x^2) get rid of denominators by multiplying by 2.
x^3*2/2(2(x+x^5/2-x^2)
2x^3(2x+x^5-2x^2) Now multiply the 2x^3 to what is parenthesis--remember when you multiply exponents you just add the exponents.
4x^4+2x^8-4x^5 Now reduce by 2x^4
2x^4(2+x^4-x)
2007-06-20 08:59:20
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answer #6
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answered by txmama423 3
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(2x^3-x^5) / (2x+2x^5)
i don't know if this is what you're looking for, but i think it's right
2007-06-20 08:58:38
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answer #7
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answered by 0826 2
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5x/(x-x^2)=5x/x(a million-x)=(5x/x(a million-x)) situations (a million-x)= 5x/x=5 ingredient out an x from the (x-x2) to get x(a million-x). then multiply precise and backside by utilising (a million-x); and it cancles out. u r left with 5x/x. divide out the x to get 5.
2016-11-07 01:21:33
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answer #8
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answered by ? 4
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34f4
2014-08-08 02:53:24
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answer #9
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answered by Anonymous
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