help please!
\[\frac{ 7ab ^{-2} }{ 3w }\]
First, Nina, what's your goal in solving this problem?
simplify
In this type of problem, what would "simplify" mean?
I need and want to know how YOU see this.
simplify expression
see what?
What it means to "simplify" an expression such as the one you've presented here.
to reduce an equation to a simpler form
In other words, Nina, what will the correct result look like, in comparison to the expression you're starting with?
OK, let's start with what you've just said. What part of the expression (7ab^(-2))/(3W) needs simplifying, and why?
that is, which part of the following expression needs simplification, and why? \[\frac{ 7ab ^{-2} }{ 3w }\]
Hint: don't worry about the 7, the a, the 3 or the w. They're fine as is. What's "wrong" with the remainder of this expression?
the top part 7ab^-2
^-2 exponent turns to a positive so b^2
Just the b^(-2) part. Sorry to have been so roundabout. My point is that we need to understand what actions to take to simplify an expression that contains a negative exponent. Perhaps you know already how to re-write the following WITHOUT a negative exponent: \[b ^{-2}=?\]
yes
Building on what you have already said, where are y ou going to write that \[b ^{2} \] which you got a moment ago?
Hint: rewrite \[b ^{-2}\] as a fraction.
next to the 7a?
Yes, but for now please just focus on re-writing that \[b ^{-2}.\]
Let me drop another hint: In this problem we need to simplify the given expression by getting rid of negative exponents. That's why I keep asking you how you'd do this re-writing.
i think i got it
the 7a is left at the top write so
Yes, the 7a stays in the numerator. And the 3w stays in the denominator. All you have to do is to rewrite that \[b ^{-2}.\]
\[\frac{ 7a }{ 3b ^{2}w }\]
ok is this right
Yes, that's absolutely perfect. Congrats. You're really showing your stuff now. What I'd encourage you to do
Thanks ^^
is to go back and re-write the original problem statement to read, "simplify the following expression by eliminating the negative exponent."
\(\bf \cfrac{7ab^{-2}}{3w}\qquad \textit{recall}\quad a^{-n} = \cfrac{1}{a^n}\qquad thus\\ \quad \\ \cfrac{7ab^{-2}}{3w}\implies \cfrac{7a}{3w}\cdot b^{-2}\implies \cfrac{7a}{3w}\cdot \cfrac{1}{b^{2}}\)
I will practice more
but i get it now and it's easy
I'm so glad you see it that way! Good going, Nina. Take care. H. N. Y.
@mathmale can you help me with another problem?
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