if a,b,c are positive and a^-2b^2c^2 > a^-3b^3c^2 which of the following must be true? 1. a>b 2. a>c 3. b>c I'd really appreciate the help :')
\[a^{-2}b^2c^2 > a^{-3}b^3c^2 \]\[\frac{b^2c^2}{a^2} > \frac{b^3c^2}{a^3} \]Multiply a^3 to both sides, then divide both sides by b^2c^2 to get your answer.
So you get b^3 c^2?
How do I know which one is greater ? 1,2, or 3 though?
Hello ):
I'm sorry that I was away. When you multiply a^3 to both sides of the question, what do you get?
a^3b^2C^2/a^2 > b^3c^2
its fine!
For the left side, can you simplify the expression?
Yeah and then you get a^3b^2c^2/a^2 = 1/b^2c^2 and then you can cancel the equation out
That doesn't look right to me ... \[a^{-2}b^2c^2 > a^{-3}b^3c^2\]\[\frac{b^2c^2}{a^2} > \frac{b^3c^2}{a^3}\]Multiply both sides by a^3\[a^3 \times \frac{b^2c^2}{a^2} > a^3 \times\frac{b^3c^2}{a^3}\]Cancel the common factor on both sides, what do you get?
Omg can you just tell me the answer I hate math! but you a^3 X b^2c^2/a^2 = b^3C^2
I can't, I'm sorry! \[a^3 \times \frac{b^2c^2}{a^2} > \cancel{a^3} \times\frac{b^3c^2}{\cancel{a^3}}\]\[a^3 \times \frac{b^2c^2}{a^2} > b^3c^2\]That's what you got. For the left, you can do something similar. \[a(a^2) \times \frac{b^2c^2}{a^2} > b^3c^2\] Can you try again? Cancel the common factor on the left.
This would be easier if I could type like that but okay a^3b^2c^2 >b^3c^2
You can use the equation button to type that :D Still... it's not right :| \[a(a^2) \times \frac{b^2c^2}{a^2} > \cancel{a^3} \times\frac{b^3c^2}{\cancel{a^3}}\]\[a\cancel{(a^2)} \times \frac{b^2c^2}{\cancel{a^2}} > \cancel{a^3} \times\frac{b^3c^2}{\cancel{a^3}}\]What's left?
Omg lol im so dumb \[ab^2c^2 > b^3c^2 ? \]
It looks right now :) Now, divide both side by c^2. What do you get?
\[ab^2> b^3\]
Yup! Now, divide both sides by b^2, what do you get?
a>b ??
Yup :)
You did it !! :)
Omg thankyou so much!! :')
Welcome :D
you're great aahh thanks <3333
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