TEST on Wednesday, Jan 19th!!
If you are looking for MORE review problems, pgs 294-95 #1-21 offer good stuff, especially the word problems, #12,13,21
for TEST review
pg 270 #1-17 Odd
for Variables on Both Sides lesson
pg 278 #7-19 Odd
im a lil confused on problem #1 on page 270. i keep getting 4.5. Help!
ReplyDelete-RJ Meyer
im also confused on problem 7 on the same page
ReplyDeletestill rj
on page 278 #17 i know he catches him before 10 but how do i solve formula
ReplyDeleteluke
i dont get number 11 on page 278 how do you solve the problem????????????????????????????????????????
ReplyDeleteDear RJ,
ReplyDeleteFor #1, I'll ask you 2 questions.
1) What is 4+7?
2) What is 11/2?
Hopefully, you know the answers AND can tell me why I asked you the questions.
For #7, many find a problem like that a bit on the tough side. Tell me what happens if you multiply both sides by 'p'? Then let me know if you still have a question.
btw, if I gave you a problem like #7 on a test, I would allow the use of a calculator.
For #17, there is a way to solve it precisely (algebraically), but your answer is good enough if you can explain why (which I think you probably can).
ReplyDeleteOne way to answer this question would be with a timeline sketch. That would be as good or better that a narrative explanation.
Now that I look at it more closely, I don't like this question. They should have asked you to ESTIMATE the time and explain or draw a picture supporting your estimate. The "algebra" to do this problem is something called solving "systems" of equations, which you will learn about next year.
Dear #11 (pg 278)
ReplyDelete-k = 9(k-10)
This is a tough, yet fair, problem.
Talk me through it... from our work in class, you should have a first step (or two or three)?
i dont get number 15.
ReplyDeleteDear #15,
ReplyDeleteA lovely question, indeed!!!
Articulate, Articulate, Articulate!!
Here's the problem:
At movie shack, movie rentals cost $3.99 each. The cost of renting three movies and one video game is $.11 less than renting five movies. How much does renting a video game cost?
Well-zers, if we can set this up as an equation in one variable, we should be able to solve it, right?
We don't know the cost of a video game, so let's assign a variable to it, ok? How about 'v'?
We KNOW the cost of a movie rental, so we DO NOT need a variable for that, it's simply $3.99, right?
So, here is a sample "articulation" which is an attempt to translate words into algebraic expressions and or equations.
The cost of three movie rentals and a video game rental is 3 times $3.99 plus 'v', right?
The cost of 5 movie rentals is 5 times $3.99, right?
If we add $.11 to the cost of the 3 movie/1 video game combo, it is the same as (hint: I smell an equal sign) the 5 movie rental.
So NOW, can you write an equation and then solve it? Another hint, when you write out the algebra, lose the '$' signs... they're not needed.
lmk if this helped...
Mr. C.
Dearest Bloggers,
ReplyDeleteI will be unavailable today, but I will be back tomorrow (Monday). I would love to see some of you helping each other with questions and hints. Remember, answers don't help, suggestions and hints do.
Recall your blogging etiquette, and lmk if my comments/hints above helped you or not.
See my suggested optional review problems above and don't forget your mathchamber video tutors!
Mr. C.
I get #7 on pg 270, by multiplying p on each side, but I did the check step and din't get the right answer.
ReplyDelete-Maggie
Well, when you mbs by p, you should get:
ReplyDelete-3.4 = -0.06p
then you divide both sides by -.06, right?
What was your problem?
Oh I added, I get it now dividing makes more sense.
ReplyDelete-Maggie
Ask yourself... what's happening to the variable?... if it's being multiplied by anything, even a negative decimal number, you must UNDO the operation by using the inverse operation.
ReplyDeleteDo we have the use of calculators on the test????
ReplyDelete-RJ Meyer
You may use calculators for this test.
ReplyDeleteOh... and #15 & #20 on page 278 (using different numbers) will be on the test. Make sure you know how to do them.
Feel free to "pass it on" to your buddies. MAYBE NOW some folks will join the blog... or not!
OOPS! I copied pg 278 #15 out of the book incorrectly, so I'll redo it here. The logic of how you attack the problem is basically the same.
ReplyDeleteHere's the problem:
At movie shack, movie rentals cost $3.99 each. The cost of renting three movies and one video game is $.11 less than renting five video games. How much does renting a video game cost?
Well-zers, if we can set this up as an equation in one variable, we should be able to solve it, right?
We don't know the cost of a video game, so let's assign a variable to it, ok? How about 'v'?
We KNOW the cost of a movie rental, so we DO NOT need a variable for that, it's simply $3.99, right?
So, here is a sample "articulation" which is an attempt to translate words into algebraic expressions and or equations.
The cost of three movie rentals and a video game rental is 3 times $3.99 plus 'v', right?
The cost of 5 video games is 5 times 'v', right?
If we add $.11 to the cost of the 3 movie/1 video game combo, it is the same as (hint: I smell an equal sign) the cost of 5 video game rentals.
So NOW, can you write an equation and then solve it? Another hint, when you write out the algebra, lose the '$' signs... they're not needed.
Sorry about that folks!!!
I was thinking of maybe sending you another Unit Test hint, but since NO ONE has bothered to say "thank you, Mr. Chamberlain, your the best-est teacher in the whole world and stuff like that there" I guess I won't bother.
ReplyDeleteThe MathChamber Unit 6 Pearson OnLine Quizzes would be a good way to get some extra practice. The OnLine Test would be good, too. Just remember, we didn't do INEQUALITIES, so you can skip those.
Last chance for questions. I'm going to bed early tonight!
Thank you Mr. Chamberlain, your the best teacher in the whole world. Any more hints?
ReplyDelete-Maggie
Oh sure, NOW you say it!!!!!!!!!!! (maybe, stay tuned)... and that's best-EST... look it up!
ReplyDeleteMake sure you can do problems like pg 270 #13 which involve "distributing the negative"
ReplyDelete2h - 4(h-5)
2h - 4h + 20
-2h + 20
remember our "language"
You are REMOVING 4 sets of "h and the loss of 5"
you end up removing 4 h's (i.e. -4h) and removing 4 -5's essentially (-4)(-5)=+20
ca-peesh?
Ca-peesh. Hopefully I ca-peesh tomorrow on the test.
ReplyDelete-Maggie
Center Cut!
ReplyDelete