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hey guys!
this is a silly question but can someone help me simplify:

x^3/2(x+x^5/2-x^2)
ur help will be appreciated greatly...
thanks!!!

2007-06-20 08:52:43 · 9 answers · asked by Anonymous in Science & Mathematics Mathematics

there has been some confusion:
the problem is::

x^(3/2) (x+x^(5/2) - x^2)

2007-06-20 09:01:51 · update #1

9 answers

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 · answer #1 · answered by Mathematica 7 · 0 0

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 · answer #2 · answered by Vangorn2000 6 · 0 0

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 · answer #3 · answered by mathjoe 3 · 0 0

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 · answer #4 · answered by harry m 6 · 0 0

= x^(5/2) + x^4 - x^(7/2)

2007-06-20 10:57:10 · answer #5 · answered by Como 7 · 0 0

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 · answer #6 · answered by txmama423 3 · 0 0

(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 · answer #7 · answered by 0826 2 · 0 0

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 · answer #8 · answered by ? 4 · 0 0

34f4

2014-08-08 02:53:24 · answer #9 · answered by Anonymous · 0 0

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