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.
Just generally, I'm still confused on finding unions & intersections.. Even though we spent a lot of time on that in class. :( I don't understand how to find unions & intersections in inequalities & equations.. -Union-tersection Trouble
For #33, how would you solve it if it were an equation? You use the same techniques on an inequality... you should know the only difference by now. Let me know if that helped. Hint: Don't let x/2 fool you... it can be written as (1/2)x, right?
For #37, the salesperson earns $200 whether she sells anything or not. She wants to earn no less than (i.e. at least) $450. You don't know the total sales number, so assign it a variable s. How do you show 4% of s? The sum of "what plus what" must be "<=>"(??) $450?? Did that help?
I'm gonna let you struggle with #60 for awhile... I think you'll get it!
For pg227#18, we started that in class the other day. First of all, I don't like the question because if a question is going to deal with complements of sets it needs to explicitly define the universal set. If I were to ask this question, I would lead into it with "Assume that the universal set is ALL REAL NUMBERS." So let's assume that statement is there.
(A) is false because A U B is clearly not the empty (null) set. The only way a UNION of two sets can be the empty set is if both sets are also empty.
(B) is false because the number 1 is left out of the mix. A' is {x|x<=-1}
(C) is true (I hope you know why)
(D) is false. A subset MAY NOT contain any elements that are not in the set. Can you name an element that is in B and that is not a member of the set {x|x<2}? (Hint: What is 1+1?)
Would #60 be... For every WHOLE multiple of 6, there is a multiple of 2, because the number 2 goes into any even number, and all WHOLE multiples of 6 are even. ????
I think you are over-thinking #60: The members of N are 2,4,6,8,... The members of P are 6,12,18,24...
The intersection is the members that both sets have in common. Every member of P is represented in N, right?? So, the intersection of N and P can be stated as "P" or "multiples of 6" or a variety of other ways, right??
Just generally, I'm still confused on finding unions & intersections.. Even though we spent a lot of time on that in class. :( I don't understand how to find unions & intersections in inequalities & equations..
ReplyDelete-Union-tersection Trouble
Well, I can try to help you with specific questions, or I am available for extra help by appt. I will be around tomorrow after school.
ReplyDeleteHow would you solve #33 in the review?
ReplyDelete4+x/2>2x
D:
Also, I don't understand how to write an inequality for #37..
ReplyDeleteLastly, for #60, I don't understand how to "explain" the intersection between the 2 :o
ReplyDeleteconfused abotu 18 on pg 227
ReplyDeleteFor #33, how would you solve it if it were an equation? You use the same techniques on an inequality... you should know the only difference by now. Let me know if that helped. Hint: Don't let x/2 fool you... it can be written as (1/2)x, right?
ReplyDeleteFor #37, the salesperson earns $200 whether she sells anything or not. She wants to earn no less than (i.e. at least) $450. You don't know the total sales number, so assign it a variable s. How do you show 4% of s? The sum of "what plus what" must be "<=>"(??) $450?? Did that help?
I'm gonna let you struggle with #60 for awhile... I think you'll get it!
For pg227#18, we started that in class the other day. First of all, I don't like the question because if a question is going to deal with complements of sets it needs to explicitly define the universal set. If I were to ask this question, I would lead into it with "Assume that the universal set is ALL REAL NUMBERS." So let's assume that statement is there.
(A) is false because A U B is clearly not the empty (null) set. The only way a UNION of two sets can be the empty set is if both sets are also empty.
(B) is false because the number 1 is left out of the mix. A' is {x|x<=-1}
(C) is true (I hope you know why)
(D) is false. A subset MAY NOT contain any elements that are not in the set. Can you name an element that is in B and that is not a member of the set {x|x<2}? (Hint: What is 1+1?)
I understand what you're saying 33, but I still don't quite understand #37..
ReplyDeleteWould it be
200+0.04s>+ 450
????
Oops above I meant ">="
ReplyDeleteWould #60 be...
ReplyDeleteFor every WHOLE multiple of 6, there is a multiple of 2, because the number 2 goes into any even number, and all WHOLE multiples of 6 are even.
????
For #37, YUP.. now can you solve it?
ReplyDeleteI think you are over-thinking #60:
The members of N are 2,4,6,8,...
The members of P are 6,12,18,24...
The intersection is the members that both sets have in common. Every member of P is represented in N, right?? So, the intersection of N and P can be stated as "P" or "multiples of 6" or a variety of other ways, right??
Oh I get it now :)
ReplyDeleteI hope I got a good grade on the test!
And luckily there wasn't a question similar to 60! :)