(x-²y-³)-³ answer is requested with a positive exponent. I just finished my first round of algebra, passing but whatever, and now onto the next level. It is all confusing to me, especially when not everyone (myself included) cannot get the expressions to appear in text fashion...any help on how to solve this would be great.
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algebra
- Thread starter amelia
- Start date
Hello, amelia!
I must assume you are familiar with the Laws of Exponents.
\(\displaystyle \left(x^{\text{-}2}y^{\text{-}3}\right)^{\text{-}3} \;=\;\left(x^{\text{-}2}\right)^{\text{-}3}\left(y^{\text{-}3}\right)^{\text{-}3} \;=\;x^{(\text{-}2)(\text{-}3)}y^{(\text{-}3)(\text{-}3)} \;=\; x^6y^9\)
\(\displaystyle \text{Express in positive exponents: }\:\left(x^{\text{-}2}y^{\text{-}3}\right)^{\text{-}3}\)
I must assume you are familiar with the Laws of Exponents.
\(\displaystyle \left(x^{\text{-}2}y^{\text{-}3}\right)^{\text{-}3} \;=\;\left(x^{\text{-}2}\right)^{\text{-}3}\left(y^{\text{-}3}\right)^{\text{-}3} \;=\;x^{(\text{-}2)(\text{-}3)}y^{(\text{-}3)(\text{-}3)} \;=\; x^6y^9\)
I am learning the Law of Exponents. I am not liking the text, and I am learning this online of all things, and I feel that it is self taught, thus my apprehension...traditional classroom may have been better for me.
What I did notice is that there is not a / in the paranthesis...so that means nothing? I was thinking this was a fraction and I did get muddled at it. I see how you worked it and I was close but I was confused on the / mark, so I guess I will (should have) ask the presenter first...which is not always easy when you are not as smart as they are...thanks for your help. I will be back I am sure before then end of this class.
What I did notice is that there is not a / in the paranthesis...so that means nothing? I was thinking this was a fraction and I did get muddled at it. I see how you worked it and I was close but I was confused on the / mark, so I guess I will (should have) ask the presenter first...which is not always easy when you are not as smart as they are...thanks for your help. I will be back I am sure before then end of this class.
J
JeffM
Guest
There are frequently many ways to do things. You can solve your problem using the / operator.amelia said:I am learning the Law of Exponents. I am not liking the text, and I am learning this online of all things, and I feel that it is self taught, thus my apprehension...traditional classroom may have been better for me.
What I did notice is that there is not a / in the paranthesis...so that means nothing? I was thinking this was a fraction and I did get muddled at it. I see how you worked it and I was close but I was confused on the / mark, so I guess I will (should have) ask the presenter first...which is not always easy when you are not as smart as they are...thanks for your help. I will be back I am sure before then end of this class.
(x[sup:2rrlzlel]-2[/sup:2rrlzlel]y[sup:2rrlzlel]-3[/sup:2rrlzlel])[sup:2rrlzlel]-3[/sup:2rrlzlel] = 1/[(x[sup:2rrlzlel]-2[/sup:2rrlzlel]y[sup:2rrlzlel]-3[/sup:2rrlzlel])[sup:2rrlzlel]3[/sup:2rrlzlel]] = 1/[x[sup:2rrlzlel]-6[/sup:2rrlzlel]y[sup:2rrlzlel]-9[/sup:2rrlzlel]] = (1/1)/[(1/x[sup:2rrlzlel]6[/sup:2rrlzlel])(1/y[sup:2rrlzlel]9[/sup:2rrlzlel])] = (1/1)(x[sup:2rrlzlel]6[/sup:2rrlzlel]/1)(y[sup:2rrlzlel]9[/sup:2rrlzlel]/1) = x[sup:2rrlzlel]6[/sup:2rrlzlel]y[sup:2rrlzlel]9[/sup:2rrlzlel]
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Soroban's way is quicker and more elgant, but both ways end in the same spot.
I know you think you are doing my homework, but may I ask a question that has been presented and I cannot find an answer in my textbook, (it) covers this topic, but I cannot find a definitive answer....searched the Internet first, but have had no results. If you would please, the question is
Address the issue of a missing term in the dividend. How is this issue resolved? We are working on polynomial division when the divisor is a monomial, BUT when the divisor IS NOT a monomial?
Thank you.
Amelia
Address the issue of a missing term in the dividend. How is this issue resolved? We are working on polynomial division when the divisor is a monomial, BUT when the divisor IS NOT a monomial?
Thank you.
Amelia