This Blog exists for the collective benefit of all algebra students. While the posts are specific to Mr. Chamberlain's class, any and all "algebra-ticians" are welcome. The more specific your question (including your own attempts to answer it) the better.
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Friday, March 4, 2011
hw #7-4 POWERS TO THE PEOPLE!!
hw #7-4
pg 436-437
#11-37 Odd
#51-55 Odd
& Problem Set 7A
I'm not sure I like the way you are using the "ysub" terminology. To say "ysub" usually refers to a subscripted variable, like back in our slope days when we said that m=(ysub2 - ysub1)/(xsub2-xsub1)... I don't think that's what you meant, right?
For blogging purposes, y raised to the 10th should be written as y^10 and y raised to the negative 10th should be written as y^(-10)... can you re-phrase the question now?
I updated MathChamber Unit 7 with video tutors and assessment dates, so you might want to take a look! The video tutors in this section are very good, I recommend watching them for reinforcement even for those of you who GOTSK it!
just throwing something out there....
ReplyDeleteif you simplify an equation to this:
ysub10 times 1/4 time ysub6
could u multiply the six and the ten to get ysub60 times 1/4?
I'm not sure I like the way you are using the "ysub" terminology. To say "ysub" usually refers to a subscripted variable, like back in our slope days when we said that m=(ysub2 - ysub1)/(xsub2-xsub1)... I don't think that's what you meant, right?
ReplyDeleteFor blogging purposes, y raised to the 10th should be written as y^10 and y raised to the negative 10th should be written as y^(-10)... can you re-phrase the question now?
For #33 on pg 437 I'm lost.. You have to distribute the p & the ^-2?
ReplyDeleteIf the equation is 4j^2 * 2j^33 you would add the exponents 33+2 right?
ReplyDeleteFor #4 in the Problem set, I don't know what to do. Would it be 1/(3/4)^3?
ReplyDeleteAARRGGHH!! Don't use the word distribute unless you see multiplication "over" addition... like 5(x+7) or p(7-x).
ReplyDeleteIn class, I believe we agreed to use the word "applied"...
and remember, blob^-2 = 1/blob^2, right??
So, in #33, your first step could be
p/[p^(-7)*q^3)^2 * q^3)]
Do you see, that means p OVER "everything"??
Then, you can apply the "squared" so that you have:
p/[p^(-14)*q^6) * q^3)]
then, after you send the little piggy home, you have:
p^15/q^9, right?
... OR ANOTHER WAY, just "apply" the -2 right from the start:
p * [p^(-7)*q^3)]^(-2) * q^(-3)
p * p^14 * q^(-6) * q^(-3)
Then you just add the exponents (for powers of the same base):
p^15 * q^(-9)
... and then finish with positive exponents only (sending the little piggy home)
p^15/q^9
Ca-peesh?
Sorry I meant #6..
ReplyDeleteSORRY!!!!!!!
ReplyDeleteAnd yep! I gotsk it.
Is the solution to #6 64/27
ReplyDeleteI went through the long TWISTED process with this one.. I got 4x^2/9y^4 for #7?? Is that right?
ReplyDeleteThis is actually kind of fun :P
ReplyDeleteThe problems in the problem set :)
Another LONG problem.. #8... is the solution
ReplyDeletew^12 x^3/y^6
A bunch of lopsided big fat ones there...
And #10.. is it -1?
ReplyDeleteFor PS 7A
ReplyDeleteIn #4, the coefficients "stay home" and the variables "go to market"
In #6, yup, it's a flip-flop, 64/27!
In #7, yup & yup, but I don't understand what was so long & twisted... let's look at it in class...
#10, yup!
I updated MathChamber Unit 7 with video tutors and assessment dates, so you might want to take a look! The video tutors in this section are very good, I recommend watching them for reinforcement even for those of you who GOTSK it!
ReplyDelete