dragonstar57 Posted January 31, 2011 Posted January 31, 2011 in math class today the teacher wrote in ()'s 3x^3y^4 over 2xy^2 and outside of the ()"S was a ^3 the teacher then said that anyone that finished the problem before the end of the hour would get extra points i answered that it was 3^3*x^9*y^12 over (2x)^3*y^6 is that right? if not explain?
timo Posted January 31, 2011 Posted January 31, 2011 No offense meant, but I don't see a problem in the first place. Your re-arrangement [math]\left( \frac{3x^3y^4}{2xy^2} \right)^3 = \frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] is a mathematically correct step. But what is it supposed to be (good for)?
mooeypoo Posted January 31, 2011 Posted January 31, 2011 I assume the exercise means simplify since there's no equation to solve (?). Also, just to make sure, this is the exercise? [math]\big( \frac{3x^{3y^{4}}}{2xy^2} \big)^3[/math] Okay, yeah it looks like a simplification problem. You start from: [math]\left( \frac{3x^3y^4}{2xy^2} \right)^3 = \frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] And open all brackets, then simplify: [math]\frac{27x^9y^{12}}{8x^3y^6} = [/math] [math]\frac{27x^9}{8x^3}*\frac{y^{12}}{y^6}=[/math] [math]\frac{27}{8}x^6 * 12y^6=(3/2)*27x^6y^6[/math]
Janus Posted January 31, 2011 Posted January 31, 2011 in math class today the teacher wrote in ()'s 3x^3y^4 over 2xy^2 and outside of the ()"S was a ^3 the teacher then said that anyone that finished the problem before the end of the hour would get extra points i answered that it was 3^3*x^9*y^12 over (2x)^3*y^6 is that right? if not explain? Try reducing [math]\frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] Further. Hint: what does [math]\frac{y^{12}}{y^6}[/math] reduce to?
timo Posted January 31, 2011 Posted January 31, 2011 @moo: I know and understand that not everyone is such a pedant as me who wants (their) students to understand why they are doing what and to be able to communicate this to others (of course I am not completely clueless what the task might have been). But actually giving the answer with all steps is not exactly the idea of the homework section, and a bit too much - see Janus' post for what I guess everyone would consider a reasonable reply. 1
dragonstar57 Posted January 31, 2011 Author Posted January 31, 2011 (edited) @moo: I know and understand that not everyone is such a pedant as me who wants (their) students to understand why they are doing what and to be able to communicate this to others (of course I am not completely clueless what the task might have been). But actually giving the answer with all steps is not exactly the idea of the homework section, and a bit too much - see Janus' post for what I guess everyone would consider a reasonable reply. the teacher then said that anyone that finished the problem before the end of the hour would get extra points the answer is of little use to me now what matters is how the answer is found. it was a bonus problem that we have not been taught to do yet and now that the hour is over it can't help me (i got the points with my answer but the teacher said that it was close but "not quite") Also, just to make sure, this is the exercise? [math]\big( \frac{3x^{3y^{4}}}{2xy^2} \big)^3[/math] Okay, yeah it looks like a simplification problem. You start from: [math]\left( \frac{3x^3y^4}{2xy^2} \right)^3 = \frac{3^3x^9y^{12}}{(2x)^3y^6}[/math] yeah that is what i was tiring to say the problem was but i don't know how to enter the problems like that (yeah it asks to simplify) Edited January 31, 2011 by dragonstar57
imatfaal Posted February 2, 2011 Posted February 2, 2011 Dragonstar and Mooeypoo Just to set my mind to rest - there are two different questions represented in Mooey's post. The first after "Also, just to make sure, this is the exercise?" treats 3y^4 as the exponent of 3x^3 [math]\big( \frac{3x^{3y^{4}}}{2xy^2} \big)^3[/math] The second after "You start from.." treats y^4 as a multiplicand of 3x^3 [math] \big(\frac{3x^3y^4}{2xy^2} \big)^3[/math] You did say you had removed some brackets - so I am not sure which question was the correct one. everyone has answered the second one; perversely when I read your question I had assumed the first was the question (it's got a bit more meat to it). Shows the benefit of being dull and putting in all the brackets - or using laTex. There are some really easy to use generators on the web if you cannot remember the codes. I assume the exercise means simplify since there's no equation to solve (?). Also, just to make sure, this is the exercise? [math]\big( \frac{3x^{3y^{4}}}{2xy^2} \big)^3[/math] Okay, yeah it looks like a simplification problem. You start from: [math]\left( \frac{3x^3y^4}{2xy^2} \right)^3 = \frac{3^3x^9y^{12}}{(2x)^3y^6}[/math]
timo Posted February 2, 2011 Posted February 2, 2011 I would not expect that "3^3*x^9*y^12 over (2x)^3*y^6" is a consistent rearrangement of "(3x^3y^4 over 2xy^2)^3" in the alternative reading - that's how I came up with my interpretation, at least. I agree that unambiguous presentation of terms would be cool, though (but still think "not saying what's actually asked for" is a more serious mistake).
dragonstar57 Posted February 2, 2011 Author Posted February 2, 2011 [math] \big(\frac{3x^3y^4}{2xy^2} \big)^3[/math] that is right 3x^3*y^4 not 3x^3y^4
dragonstar57 Posted February 3, 2011 Author Posted February 3, 2011 (edited) 15^5*y^6 over 7xy^4 all this was in ()'s and its to the power of 4 i think that it is 22*x^16*y^8 is that right? Edited February 3, 2011 by dragonstar57
imatfaal Posted February 3, 2011 Posted February 3, 2011 Dragon I am sure that others who actively teach would echo this point, you need to be rigorous and careful in your maths! The first question in this post had a potential problem - and this one is just plain wrong. There is no x component in the numerator! Maths is subtle and abstract, it can require a strange sort of weirdness in the brain; but it also requires boring exactness and strict book-keeping. Will repeat - learn to use laTex - or be obsessive with your brackets and double checking
dragonstar57 Posted February 4, 2011 Author Posted February 4, 2011 (edited) (-2x^2)*(5x^6) I'm not sure what I should do here it is supposed to be simplified but I don't think that I can just combine the x^2 and the x^6 because of the ()'s Dragon I am sure that others who actively teach would echo this point, you need to be rigorous and careful in your maths! The first question in this post had a potential problem - and this one is just plain wrong. There is no x component in the numerator! Maths is subtle and abstract, it can require a strange sort of weirdness in the brain; but it also requires boring exactness and strict book-keeping. Will repeat - learn to use laTex - or be obsessive with your brackets and double checking I'm sorry i copied the problem down incorrectly the question in my previous post 15^5*y^6 over 7xy^4 all this was in ()'s and its to the power of 4 i think that it is 22*x^16*y^8 is that right? should have been 15x^5*y^6 Edited February 4, 2011 by dragonstar57
hypervalent_iodine Posted February 4, 2011 Posted February 4, 2011 (-2x^2)*(5x^6) I'm not sure what I should do here it is supposed to be simplified but I don't think that I can just combine the x^2 and the x^6 because of the ()'s The brackets are there by manner of convention. As you have written it, you can remove the brackets and it won't change anything. Remember you also have to multiply the coefficients.
mooeypoo Posted February 4, 2011 Posted February 4, 2011 @moo: I know and understand that not everyone is such a pedant as me who wants (their) students to understand why they are doing what and to be able to communicate this to others (of course I am not completely clueless what the task might have been). But actually giving the answer with all steps is not exactly the idea of the homework section, and a bit too much - see Janus' post for what I guess everyone would consider a reasonable reply. You're right, and I am usually the one lecturing others on that issue of not giving away the answer. I thought dragonstar wanted a verification, and so I put up the info, and later realized that was a mistake. Totally my bad. 1
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