need help

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need help

b^2(1/2b^-1)^2 (3b^3)^-2/(9b^4)-1

The answer is suppose to be 1/4b^2
I can't seem to get it though.
Amyyy offline

Re: need help

Given b^2(1/2b^-1)^2 (3b^3)^-2/(9b^4)-1

After looking over this question, I think it might write like this
b^2(1/2 b^-1)^2 (3b^3)^-2/(9b^4)^-1

In this problem we have a negative exponent to start with. That means we need to take the reciprocal of the base. Note that we DO NOT take the reciprocal of the exponent, only the base.

b^2 * (1/2b)^2 * (1/3b^3)^2 / (1/9b^4)
b^2 * (1/4b^2) * (1/9b^6) / (1/9b^4)
--- [remember when multiplying exponents to add them b^3 * b^3= b^6]
b^2/36b^8 / (1/9b^4)
simplify the numerator
1/36b^6 / ( 1/9b^4)
invert and multiply when dividing fractions.
1 9b^4
------ * ------
36b^6 1

[Typing it this way makes it easier to read, but can also be typed
1/36b^6 * 9b^4/1 ]
9b^4/36b^6
Simplify
1/4 b^2

Of course this is assuming that there is a typo in the expression.
Hope this helps, and good luck.
Ivy Orchid offline

Re: need help

When I typed to make it look more clear, it didn't save my spaces so it just looks weird there now
1/36b^6 * 9b^4/1

Just wanted to say I did put spaces in there, but they didn't stay. Sorry about that.
Ivy offline

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