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Problem Solving Help!!

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by phoenix9801 » Thu Jun 03, 2010 8:43 pm
Hi Guys, I was wondering if anyone can help me please. Would appreciate if you can explain step-by-step instructions (detials) on how to solve it.


1- (2x^0 y^2)^-3 * 2yx^3 =


2- (2mp^-1 q^0)^-4 * 2m^-1 p^3 / 2pq^2 =


3- (2 h j^2 k^-2 * h^4 j^-1 k^4)^0 / 2 h^-3 j^-4 k^-2 =
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by papgust » Thu Jun 03, 2010 9:22 pm
1- (2x^0 y^2)^-3 * 2yx^3

Anything to the power 0 is 1. So, x^0 is 1.

(2 * y^2)^-3 * 2yx^3

[1 / (2 * y^2) ^3 ] * 2yx^3

[1 / 8 * y^6] * 2yx^3

= x^3 / 4y^3


2- [(2mp^-1 q^0)^-4] * [2m^-1 p^3 / 2pq^2]

[1 / (2m p^-1)^4] * [p^3 / m^1 * q^2]

[p^4 / 16 * m^4] * [p^3 / m^1 * q^2]

= p^6 / (16 * m^5 * q^2)


3- (2 h j^2 k^-2 * h^4 j^-1 k^4)^0 / 2 h^-3 j^-4 k^-2

The whole numerator is 1 since it is raised to the power 0.

1 / 2 h^-3 j^-4 k^-2

1/2 * h^3 * j^4 * k^2
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by uwhusky » Thu Jun 03, 2010 10:41 pm
Going to bed, so I can only do the first one =).

1- (2x^0 y^2)^-3 * 2yx^3

This of it as (2x^0)^-3 * (y^2)^-3 * 2 * y * x^3 = 1 * y^-6 * 2 * y * x^3.

Combine like terms, you get 1 * y^-5 * 2 * x^3.
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by phoenix9801 » Fri Jun 04, 2010 10:03 am
papgust wrote:1- (2x^0 y^2)^-3 * 2yx^3

Anything to the power 0 is 1. So, x^0 is 1.

(2 * y^2)^-3 * 2yx^3

[1 / (2 * y^2) ^3 ] * 2yx^3

[1 / 8 * y^6] * 2yx^3

= x^3 / 4y^3


2- [(2mp^-1 q^0)^-4] * [2m^-1 p^3 / 2pq^2]

[1 / (2m p^-1)^4] * [p^3 / m^1 * q^2]

[p^4 / 16 * m^4] * [p^3 / m^1 * q^2]

= p^6 / (16 * m^5 * q^2)


3- (2 h j^2 k^-2 * h^4 j^-1 k^4)^0 / 2 h^-3 j^-4 k^-2

The whole numerator is 1 since it is raised to the power 0.

1 / 2 h^-3 j^-4 k^-2

1/2 * h^3 * j^4 * k^2

I understand number 1, but not 2 and 3. I thought the answer for number 2 is m^3 / 16p^2 q^2 and for number 3 is H^3j^4k^2 / 2 ?????????????? Please explain
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by uwhusky » Fri Jun 04, 2010 10:23 am
The way you're writing the question is not very concise either.

2- (2mp^-1 q^0)^-4 * 2m^-1 p^3 / 2pq^2

I'll just go with what I think you are writing.

(2mp^-1)^-4 = 2mp^4

(q^0)^-4 = 1

So far it is 2mp^4 * 1 * 2m^-1 * p^3 / 2pq^2

2mp^4 = 2 * m^1 * p^4

Now to combine like terms:

2 * 2 = 4

m^1 * m^-1 = m^0 = 1

p^4 * p^3 = p^7

4 * p^7 is what's left.

4p^7 / 2pq^2

4 / 2 = 2

p^7 / p^1 = p^6

Nothing else left to reduce and you end up with:

2p^6 / q^2
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