Hello All,
I looked at the Chapter Review (pg 408-410) #1-32 and all problems look reasonable to me. As for word problems, if I were you I would make sure that I focus on our Problem Sets - the style and the wording of the problems on our test will mirror the Problem Sets much more so than the text book. The word problems on the Chapter Test (pg 411) are written in a style that doesn't actually ask a question, just that you "model the situation." I would rather that you answer a question, so don't focus on the word problems on the Chapter Test. Problems #1-12 on the Chapter Test are all reasonable in my view.
To reiterate, you shouldn't feel obligated to do every problem... do every third problem or so as a first pass and see how you do. Are you happy with your results? You can probably stop there.
I'll try to post a few videos this weekend.
See you on Monday,
Mr. C.
I was going over the chapter review and i had one question on question number 18 on page 409, I would like some assistance please
ReplyDeleteTry a relate-define-write...
ReplyDelete[the total earnings from haircuts] is [earnings from owner] plus [earnings from assistant]...
you should be able to "break down" the two earnings "boxes" into component parts that make sense.
Identify/define your variables and move forward. Let me know if this helps.
Could you help me with the combining problems. I'm doing the extra work problems on Mathchamber and i dont get one.
ReplyDelete"A scientist has a container of 2% acid solution and a container of 5% acid solution. How many fl. oz. of each concentration should be combined to make 25 fluid ounces of 3.2% acid solution?"
I dont even know where to begin..
That's a toughie... and I decided to forgo the "liquid % mixture" problems, so you can rest assured that it will not be on the test... but since you asked...
DeleteRELATE DEFINE WRITE!!!
What relationships can we set up?
One simple relationship is that we will be combining Solution-A (the 2% mixture) with Solution-B (the 5% mixture) for a total of 25 ounces.
So, let's define:
a is the # of ounces of 2%-solution in the 25-ounce mixture
b is the # of ounces of 5%-solution in the 25-ounce mixture
As a constant, we know that the target solution will be 25 ounces at 3.2%, or .032 concentration.
Therefore we can write:
a + b = 25 (DUH!)
Here's the tough part (read it a couple of times and it'll make sense):
The percent of acid of a certain number of ounces of Solution-A when mixed with the percent of acid in a certain number of ounces of Solution-B will yield a blended solution of a specified number of ounces of a specified percentage (in between the low and high %). That's a real mouthful... can you see why I skipped these??
So, we can write:
.02a + .05b = .032(25)
I'm sure that you don't like the decimals (either do I), so let's ditch 'em by multiplying by 1000 (yup, 1000!)
That will set up a system:
a + b = 25
20a + 50b = 32(25)
or
a = 25 - b
20a + 50b = 800
Solve!
20(25-b) + 50b = 800 (substitute)
500 - 20b + 50b = 800 (distribute)
30b + 500 = 800 (CLT)
30b = 300 (Subtract POE)
b = 10
Clearly, a=15
So we need 15 ounces of the 2% solution combined with 10 ounces of the 5% solution to create a 25-ounce mixture at 3.2% concentration of acid.
Does this solution for the "solution" (get it? the solution for the solution?!... I crack myself up!) make sense. Well, since .32 is a bit closer to 2% than 5%, it does makes sense that we needed more of the 2% solution that the 5% solution.
Too easy, right?
this is kinda like in the first video tutor under 6-4 !!
Deletefyi, the 7th graders had a great discussion about various problems, so you should check their blog for some answers to problem sets 6F and 6R... no sense repeating the same info here. If you had trouble with 6F or 6R it should help you BIG TIME.
ReplyDeleteDon't forget to review my videos on the "President's Weekend" blog post (on this blog)... I know that many of you did NOT view them... the feedback I got from those that DID VIEW the videos was that they were helpful.
USE YOUR RESOURCES, FOLKS!!
... and you better be using the TEST POINT technique to graph the shaded areas of your inequalities... it you don't know what I'm talking about, see the text book on pages 394-96 where it talks about the "half-plane" (i.e. the shaded region) testing points on either side of the line (i.e. the TEST POINTS - DUH!).
ReplyDeleteIm having a bit of trouble with number 8 in Problem set R please help
ReplyDeleteThere are a few "ugly" intersections (i.e. messy fractions) so I'll pick one and YOU LET ME KNOW if it helps (not all of them are this messy). You should not have had any difficulty with the graphs, and you should have been as neat & precise as you could so that you could check your algebraic solutions against the graph with a reasonable degree of comfort.
DeleteThe system for problem #8/#3 (which combines equations A&D) in a system is:
2x - y = -2
x + 2y = -4
As a first step, I will choose to multiply the 1st equation by 2, yielding:
4x - 2y = -4
x + 2y = -4
Adding the equations yields:
5x = -8
x= -8/5
Substitute this x value into the 2nd equation and you get a solution for y:
x + 2y = -4
-8/5 + 2y = -4 (add 8/5 tbs)
2y = -12/5 (divide bs by 2)
y = -12/10 (REDUCE!!)
y = -6/5
Substitute x&y into the second equation, you should get a TRUE statement:
2x - y = -2
2(-8/5) - (-6/5)= -2
-16/5 + 6/5 = 2
-10/5 = -2
-2 = -2 TRUE!!
I'll admit it's messy, BUT:
a) you know how to combine/solve a system of equations AND
b) you know how to add positives and negatives AND
c) you know how to deal with fractions (this is only 1/5'ths guys, you've been doing this since 4th or 5th grade) DON'T LET FRACTIONS INTIMIDATE YOU!!!!!
Capeesh?
Mr. C I'm having a lot of trouble with #8 on pg. 409. I read your comment about setting up the relate step but i am still having difficulty solving for it! Can you please explain it? Thanks.
ReplyDelete#8 on 409 is a "raw" system of equations... we only use the "relate" step as part of a relate-define-write for a word/story problem.
DeleteA good first step for this problem is to substitute for one of the 'y' variables:
-2x - 21 = x - 7
Was this your question?
Oops!! I meant #18 on page 409!! Sorry.
ReplyDeleteim having trouble on #1 on the problem set 6r
ReplyDeletesee below...
DeleteOk, so we know that the owner gave 6 haircuts and the assistant gave 12 haircuts. The owner charges $20 more per haircut. On the day in question they earned $750. How much does the owner charge for a haircut?
ReplyDeleteRELATE-DEFINE-WRITE
RELATIONSHIP A: [the total earned] eq [the amt earned by the asst] plus [the amt earned by the owner]
RELATIONSHIP B: [amt charged by ownr] eq [amt chrgd by asst] + [the difference amount] (Note from Mr. C. most people probably wouldn't write this because it is so obvious)
[the amt earned by the asst] is [the # of hc's]*[asst's charge]
[the amt earned by the ownr] is [the # of hc's]*[ownr's charge]
total earned: $750
# of hc's by asst: 12
# of hc's by the owner: 6
difference in charge: $20
x is the amount charged by the asst
y is the amount charged by the owner
Can you set up the equations and solve now? lmk
Hello MR. C. I am confused about question 1 on problem set 6R, how can you find variables from a question like that?
ReplyDeleteThe simplest of problems fool everyone.
DeleteUh, how about x is the first unknown number and y is the second unknown number. Of course, you need to know what 'sum' and 'difference' mean... if you do, we can say:
x + y = 16
x - y = 20
We gotsk ourselves a system! Can you solve that?
Thank you MR. C
DeleteMr. C,
ReplyDeleteHi, I am doing the chapter review and I got stuck on #14 on page 409 in the book. I keep trying to drop out one of the variables, but i keep getting 0x+0y=12. How is this possible? Anyway, i also tried adding them together and got -y=2x+42, and if that IS right, where do i go from there?
Does 0=12? "NO"
DeleteWill it ever equal 12 depending on what x and y are? "NO"
So there is "NO SOLUTION"... this was uncovered in ProbSet 6D Special Cases "Weird Things that can happen"... you are referencing your resources, right??
What do you think would happen if you graphed the lines? They would have NO POINTS in common, so the lines must be _________.?
Thanks Mr. C. I set up the system of equations for #18 (with a little difficulty to my dismay!)and ended up getting the right answer! On the test, will their be a problem similar to this one?
ReplyDeletemore or less... yup!
DeleteMr.C can you help me with number 4 in the packet
ReplyDeleteMr c
ReplyDeleteU marked number 11 wrong on my test that we received today. I showed it to my mom and she can't figure out why it's wrong please post the correct answer or create a video of it. Thank you
Danie