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, December 9, 2011
Unit 3 Test "Recovery"
Do you have questions on the test corrections?... ask them here... you will get a second chance on Thursday to STRUT YOUR STUFF... make the most of it.
Looking back at my test I realize that it was a series of stupid mistakes that I should have payed more attention to, but could you explain #8 | 6q + 9|<= 9. The first thing I did was subtract 9 from both sides, my next step (now left with |6q|<= 0 is to divide each side by 6, but I know that's wrong could you explain the next step? Do I divide 6q or leave the problem as is?
Those darn stupid mistakes... REMEMBER, SOLVE FOR THE BLOB... this one IS solved for the BLOB, so your first step should have been to split into two inequalities
6q+9≥-9 AND 6q+9≤9
THEN and ONLY THEN, you can start the solving process. NOTICE, that after the "split" the abs val brackets are gone... that's because they performed their function of splitting into a compound inequality.
Thanks for your input it is helpful,so the answer would be |q| >= -3 AND |q|<=0/6 yes?? And why did you switch the sign,I don't see any dividing by a negative happening.....
I did not SWITCH the symbol, the two "opposing" symbols were created on the "split"...
Please re-play MathChamber Academy Unit 3 from 3-7g on... you still need to understand that Absolute Value gives us the distance from zero, whether the value is on the negative or positive side of zero.
The expression |6q+9|≤9 can be re-stated as follows:
The quantity 6q+9 is within 9 units of zero. That means that the quantity 6q+9 can be as little as -9 or as much as +9, since all of those values in between -9 and 9 are solutions. So, this information is what causes the "split"... 6q+9 must be ≥ -9 AND it must be ≤ +9... so that is why we must solve both inequalities. The single absolute value inequality is actually a compound inequality in disguise!
Re-play the videos... I think you are close to "getting" this... yes, it takes a little struggle and some steam coming out of the ears.
Ask Julia... her head just exploded while doing a proof in geometry... what a mess!!
IT'S CLICKING! The distance between the two is the same as a graph, that's why they're connected! I got it before but now I understand it, if that makes sense. Its least and greatest -9 thorough 9 so anything below -9 or above 9 isn't included in the equation, video 3-6a (Writing Compound inequalities) made it easier to understand, thanks :) Another question for you.... On Lesson Check Do you know how number 4 it asks to tell whether each set of ordered pairs in Exercise 1 represents a function and justify my answer, can you further explain? What does it mean by represent a function? A function of what?
Sometimes it just takes a little longer to understand... thanks for sharing your thoughts, it helps me "understand what students don't understand" which can be difficult for me at times.
Your test grade has nudged up quite a bit already... from your dialogue I can tell that you do understand... so sharing has its benefits!
I transferred the other question to the 4-2 blog post, since that's where it belongs.
Hey Mr. Chamberlain, Do you think I could come in early tomorrow morning and do my test corrections with you? I think that if i do then alone I will just do them to get it done and not really understand everything fully.
Could you explain number 17b AUB when A= {xIx is an integer greater than 7} and B={xlx is and integer less than or equal to 12}? I thought U meant to include everything from A and B.
I have 3 questions Mr.C or anyone that knows the answer....
1) If |d|<-3 that wouldn't be a "No solution" Right...because |d| is LESS THAN -3 not equal to??
2) 17b A U B I was confused on how I would put the statement A{x|x is an integer greater than 7} would that be....
{8-∞,12,11,10,9,8,7,6,5,4,3,2,1}
or just
{ ∞, 12,11,10,9,8,7,6,5,4,3,2,1}
3) On number 23 -4>= x OR 0> x I'm confused on the interval notation, on the test I wrote (0,-4] should I start with -4 since its to the left of 0 and everything less than it is x??
I'm in a soc right now "Algebra State Of Confusion" and need some guidance. Thanks to anyone who replys:)
"D" was correct... Set B is a subset of the set specified.
If you answered "B", you assumed that the Universal Set was all real numbers. Set A was x>-1... if A' would have to be x≤-1, not x<-1... see the difference?
Brandon - U are right, U means include everything. So, if Set A gives me every integer > 7 and Set B gives me every integer ≤ 12, I thinks I gotsk every integer there is on the planet (I gotsk 8,9,10,11, and 12 twice but I'll ditch the duplicates). Can you thinks of an integer I DON'T have??
Dear SOX, 1) |d|<-3 yields no solution because the absval function always returns a positive#, so it can NEVER be less than a neg#
2) I do not understand what U are saying... do U understand my response to Brandon above?
3) An OR is an OR is an OR. OR's cannot be abbreviated into a single interval.
The solution you gave me: -4>= x OR 0> x ...is slightly wrong... I would have written it as: x≤-4 OR x>0 If you look at the graph, there are two "intervals" (aka sections)
For #14 you leave the "blob" alone until everything around it is gone and you gotta work back words in gemdas so subtract 1 first then divide both sides by 2 and the your left with d + 5 < 3 hopefully you can do that part hope this helps
I go to bed early... I need my beauty sleep... it's not easy lookin' as good as I do every day, I might seem like a natural to you guys, but it actually takes a lot of work each morning.
This is a tad late, but before the recovery test, can you explain 25 on the test or maybe give me a problem like it for practice. It would be very much appreciated.
Looking back at my test I realize that it was a series of stupid mistakes that I should have payed more attention to, but could you explain #8 | 6q + 9|<= 9. The first thing I did was subtract 9 from both sides, my next step (now left with
ReplyDelete|6q|<= 0 is to divide each side by 6, but I know that's wrong could you explain the next step? Do I divide 6q or leave the problem as is?
Those darn stupid mistakes... REMEMBER, SOLVE FOR THE BLOB... this one IS solved for the BLOB, so your first step should have been to split into two inequalities
ReplyDelete6q+9≥-9 AND 6q+9≤9
THEN and ONLY THEN, you can start the solving process. NOTICE, that after the "split" the abs val brackets are gone... that's because they performed their function of splitting into a compound inequality.
Watch a video again and see.
Mr. C.
Thanks for your input it is helpful,so the answer would be |q| >= -3 AND |q|<=0/6 yes?? And why did you switch the sign,I don't see any dividing by a negative happening.....
ReplyDeleteYes, that is the answer.
ReplyDeleteI did not SWITCH the symbol, the two "opposing" symbols were created on the "split"...
Please re-play MathChamber Academy Unit 3 from 3-7g on... you still need to understand that Absolute Value gives us the distance from zero, whether the value is on the negative or positive side of zero.
The expression |6q+9|≤9 can be re-stated as follows:
The quantity 6q+9 is within 9 units of zero. That means that the quantity 6q+9 can be as little as -9 or as much as +9, since all of those values in between -9 and 9 are solutions. So, this information is what causes the "split"... 6q+9 must be ≥ -9 AND it must be ≤ +9... so that is why we must solve both inequalities. The single absolute value inequality is actually a compound inequality in disguise!
Re-play the videos... I think you are close to "getting" this... yes, it takes a little struggle and some steam coming out of the ears.
Ask Julia... her head just exploded while doing a proof in geometry... what a mess!!
IT'S CLICKING! The distance between the two is the same as a graph, that's why they're connected! I got it before but now I understand it, if that makes sense. Its least and greatest -9 thorough 9 so anything below -9 or above 9 isn't included in the equation, video 3-6a (Writing Compound inequalities) made it easier to understand, thanks :) Another question for you.... On Lesson Check Do you know how number 4 it asks to tell whether each set of ordered pairs in Exercise 1 represents a function and justify my answer, can you further explain? What does it mean by represent a function? A function of what?
ReplyDeleteOrdered Pairs are.....
a. (0,0) (1,1) (2,2) (3,3) (4,4)
b.(0,8) (1,6) (2,4) (3,2) (4,0)
c. (3,0) (3,1) (3,2) (3,3) (3,4)
Sometimes it just takes a little longer to understand... thanks for sharing your thoughts, it helps me "understand what students don't understand" which can be difficult for me at times.
ReplyDeleteYour test grade has nudged up quite a bit already... from your dialogue I can tell that you do understand... so sharing has its benefits!
I transferred the other question to the 4-2 blog post, since that's where it belongs.
Alright Mr.C Thanks Again!
ReplyDeleteHey Mr. Chamberlain,
ReplyDeleteDo you think I could come in early tomorrow morning and do my test corrections with you? I think that if i do then alone I will just do them to get it done and not really understand everything fully.
-Ethan C.
On the test, the "stupid" question, why is D the answer and not B Im a little confuzzled on why D is right
ReplyDeleteCould you explain number 17b AUB when A= {xIx is an integer greater than 7} and B={xlx is and integer less than or equal to 12}? I thought U meant to include everything from A and B.
ReplyDeleteHELP WANTED!!
ReplyDeleteI have 3 questions Mr.C or anyone that knows the answer....
1) If |d|<-3 that wouldn't be a "No solution" Right...because |d| is LESS THAN -3 not equal to??
2) 17b A U B I was confused on how I would put the statement A{x|x is an integer greater than 7} would that be....
{8-∞,12,11,10,9,8,7,6,5,4,3,2,1}
or just
{ ∞, 12,11,10,9,8,7,6,5,4,3,2,1}
3) On number 23 -4>= x OR 0> x I'm confused on the interval notation, on the test I wrote (0,-4] should I start with -4 since its to the left of 0 and everything less than it is x??
I'm in a soc right now "Algebra State Of Confusion" and need some guidance. Thanks to anyone who replys:)
I will be in by 7:15 on Tues & Weds morning.
ReplyDelete"D" was correct... Set B is a subset of the set specified.
ReplyDeleteIf you answered "B", you assumed that the Universal Set was all real numbers. Set A was x>-1... if A' would have to be x≤-1, not x<-1... see the difference?
Brandon - U are right, U means include everything. So, if Set A gives me every integer > 7 and Set B gives me every integer ≤ 12, I thinks I gotsk every integer there is on the planet (I gotsk 8,9,10,11, and 12 twice but I'll ditch the duplicates). Can you thinks of an integer I DON'T have??
ReplyDeleteDear SOX,
ReplyDelete1) |d|<-3 yields no solution because the absval function always returns a positive#, so it can NEVER be less than a neg#
2) I do not understand what U are saying... do U understand my response to Brandon above?
3) An OR is an OR is an OR. OR's cannot be abbreviated into a single interval.
The solution you gave me:
-4>= x OR 0> x
...is slightly wrong... I would have written it as:
x≤-4 OR x>0
If you look at the graph, there are two "intervals" (aka sections)
(-∞,-4] OR (0,∞)
Ca-peesh??
K Mr. C Ill see you at 7:15 with my test corrections in hand.
ReplyDelete-Ethan c.
Thanks mr c.
ReplyDeleteMr.C Is it BYOD tomorrow morning?
ReplyDeleteI getz it
ReplyDeleteHeyy Mr.C I'm not sure how to begin #14 for the test corrections. Do I distribute the 2? Or do I subtract 1 from both sides?
ReplyDeleteFor #14 you leave the "blob" alone until everything around it is gone and you gotta work back words in gemdas so subtract 1 first then divide both sides by 2 and the your left with d + 5 < 3 hopefully you can do that part hope this helps
ReplyDelete#14 GREAT COLLABORATION, the answer is CORRECT!
ReplyDeleteI go to bed early... I need my beauty sleep... it's not easy lookin' as good as I do every day, I might seem like a natural to you guys, but it actually takes a lot of work each morning.
mr.c im reviewing for our make up quiz tom. can you show me the way to answer #19 on the chapter test
ReplyDeleteI'm confused by#38 in the chapter review can some one help me
ReplyDeleteThis is a tad late, but before the recovery test, can you explain 25 on the test or maybe give me a problem like it for practice. It would be very much appreciated.
ReplyDelete