Uhh.. maybe.
\[(-3a^0)(-2a ^{-2}b^3)\]
By the way, as a general rule, anything to to the zero power equals 1. So as a hint: \(a^{0} = \dfrac{z}{z}\)
So would I add it again or cancel out?
There's nothing to "add". \(a^{0}\) simply becomes 1 or you can just remove it from the expression.
Alright.
This looks sooooo wrong. \[(6-3a^63b ^{-9}\]
Well basically, what you do once again is pair together like terms: \((-3)\cdot(-2)(a^{-2})(b^3)\)
Would I distribute the -3 to the rest?
Distribution implies that two terms are being added or subtracted which is not the case here.
Remember the distributive property is a(b PLUS c) = ab PLUS ac. In the problem you're given, there is no addition or subtraction element
The minus signs strictly apply to the negative numbers and not subtraction
For example if you had (-3)(-3a - bc) now you have to distribute because the minus sign separates two terms -3a and bc. but with (-5)(-3abc) there is no addition or subtraction sign separating two terms within parentheses.
I gotcha
Awesome.
Do I need to turn it into a fraction now?
Yes correct. If you see a negative exponent, automatically convert the expression to fraction form.
\[\frac{ b^3 }{ 6a^2 }\]
Why did you put the 6 in the denominator?
Because I’m a idiot..
Not true.
The 3 and 2 were both negative so I thought they wen to the demoninator so I multiplied them.
They were negative, but not negative exponents.
So do they stay where they are?
Yes, they remain in the numerator. Only the variable with the negative exponent gets converted to the denominator. If the coefficients are in the numerator they stay in the numerator unless their exponents are negative. If the coefficients in the denominator, they stay in the denominator unless their exponents are negative. In this case (-3) and (-2) have positive exponents.
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