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.
At this point, we employ our tried and true properties of algebra (all bow) such as associative (grouping doesn't matter for mult'pln or add'tn) and commutative (order doesn't matter for mult'pln or add'tn).
So,
(2xy)^3=(2xy)(2xy)(2xy)=(2^3)(x^3)(y^3) =8x^3y^3
Please do not call it "distribution"... let's reserve that word for the distributive property. The exponent is "applied" to each factor in the term, ok?
For 781#10, I would set up for the elimination method by multiplying the first equation by 5 and then subtracting the second from my modified first, yes?
Would a way to begin solving #24 on pg 782 be multiply both terms by a big fat one to get to the common denominator of 6, and then just subtract across the numerator?
[4,5) is interval notation from inequalities... your example is equivalent to all real numbers >=4 and <5... the bracket includes the number the paren excludes, kinda like filled dot, open dot... yes?
When you "Distribute" the exponent (in a power to a power) do you "Distribute" the exponent to the coefficients too?
ReplyDeleteex. (2x^2)^2
Does it become (2^2)*x^4?
I don't understand how to solve #22 on pg 710..
ReplyDelete#20 on pg 710 is confusing me.
ReplyDeleteAs for your first question,
ReplyDelete(2xy)^3 is simply SHORTHAND for
(2xy)(2xy)(2xy)
At this point, we employ our tried and true properties of algebra (all bow) such as associative (grouping doesn't matter for mult'pln or add'tn) and commutative (order doesn't matter for mult'pln or add'tn).
So,
(2xy)^3=(2xy)(2xy)(2xy)=(2^3)(x^3)(y^3)
=8x^3y^3
Please do not call it "distribution"... let's reserve that word for the distributive property. The exponent is "applied" to each factor in the term, ok?
Ok! More questions coming soon!
ReplyDeleteTo quote "The Monkees"... It's the Last Train to Clarksville" (youtube it!).
ReplyDeleteWait, do we have to memorize a bunch of formulas again? Or will we have a reference sheet/the formulas will be in the test?
ReplyDeleteTrouble with #3 on pg 780..
ReplyDeleteI need help solving #9 on pg 781.
ReplyDeleteI can't figure out the best way to solve #10 on ph 781.
ReplyDeleteAnd in Vertex Form (Or whatever it was called), is the last integer the vertex?
ReplyDeleteWill he have to know vertex form and stuff from unit 12 on the final exam?
ReplyDeleteFor #14 on pg 781, how do you figure out which way the inequality sign goes from the shading?
ReplyDeleteThe Final Exam Review link on Math Chamber doesn't work :(
ReplyDeleteHow do you find the best measure of Central Tendency? (From a Pearson Quiz)
ReplyDeleteDon't forget operations (mult & div) with scientific notation!
ReplyDeleteNo formula sheets... so you need to know slope, slope-intercept, point-slope, quad formula, etc... NO heavy geometry at all (wait til next year!).
ReplyDeleteOk! And Alge-Chat, please?
ReplyDeleteDear Shady,
ReplyDeleteUse test points to determine where THE TRUTH is!
Ca-peesh?
Don't worry about central tendencies for this test...
ReplyDeleteVertex form YES
ReplyDeletey=(x-h)^2+k where (h,k) is the vertex of the parabola... of course, this requires you to complete the square :) YES YES YES!!
Unit 12 NO NO NO!!
ReplyDeleteWhat about Unit 12? What do you mean "NO NO NO!!"
ReplyDeleteOh, right, nevermind! I just saw Karel's comment.
ReplyDeleteFor 781#10, I would set up for the elimination method by multiplying the first equation by 5 and then subtracting the second from my modified first, yes?
ReplyDeleteBut after finding test points, how do you know which inequality symbol to choose?
ReplyDelete-Shady
;)
For 781#10, oh right! I forgot that you can multiply the equations by an integer!
ReplyDeleteHow can the "last integer" be the vertex. The vertex is an ordered pair, yes? (oxygen, folks, oxygen!)
ReplyDeleteThen what is the vertex in...
ReplyDeleteex.
y= (x-3)^2-4
How can you figure it out from that??
I don't know how to do #15 on pg 781...??
ReplyDeleteFor 781#9, you should be able to FACTOR FIRST (sing it for me, "Factors cancel, terms don't"). You should find a common factor of (2x+3), yes?
ReplyDeleteDivide out and conquer!!
I am conquering!!
ReplyDeleteNo probability on the test, but fo 780#3, the prob of an odd number is 1/2 yes? So the prob of two odd numbers would be (1/2)(1/2) or (1/2)^2, yes?
ReplyDeleteThe prob of n odd numbers would be (1/2)^n, yes?
Oh, gotsk it for #3!
ReplyDeleteNo questions from Chapter 12...
ReplyDeletealge-chat at or around 7pm...
ReplyDeleteSorry, I'm just doing the "End of Course Assessment" in the book, so it has some Chap 12 questions!
ReplyDeleteOk! I'll be here for Alge-Chat!
ReplyDeleteI was just remembering something.. What was it when something had solutions like this [4,5)??
ReplyDeleteI just remembered that.. With the different brackets???
How do you answer #37 on ph 784?
ReplyDeleteWould a way to begin solving #24 on pg 782 be multiply both terms by a big fat one to get to the common denominator of 6, and then just subtract across the numerator?
ReplyDeleteI'm having trouble figuring out #33 on pg 783..
ReplyDeletealge-chat delayed until 7:30, sorry
ReplyDelete[4,5) is interval notation from inequalities... your example is equivalent to all real numbers >=4 and <5... the bracket includes the number the paren excludes, kinda like filled dot, open dot... yes?
Alright, it's okay!
ReplyDeleteAnd gotsk it!
For example, #47 in pg 785, will histograms be on the test? Because I don't really understand them!!
ReplyDeleteDo we have to know the compound interest formula?
ReplyDeleteI don't know if I'll be online for Alge-chat.. Might be dinner time! I'll try my best to be on around that time!
ReplyDeleteYou should understand compound interest...
ReplyDeleteNo histograms... no chap 12
Ok!!
ReplyDeleteCan you answer my other questions, like #33 & #37?
Alge-chat now??
ReplyDeletelet's make algechat at 8pm...
ReplyDeleteIf you divide a polynomial by a binomial, do you just use the 1st term of the binomial as the divisor?
ReplyDelete