Anybody here good at Math?

[maycocksean]

So I'm teaching Algebra II this year, which I haven't taught in seven years (and then only for one year). It's tough going compared to Pre-Algebra and Algebra I which I've teaching for years.

I hit this problem I just cannot figure out and I was wondering if any Interference math whizzes could help me out (so that I can, in turn, help my students out).

"Most of todays languages are thought to be descended from a few common ancestral languages. The longer the time since the languages split from the ancestral language the fewer common words there are in the descendant languages. Let c = the number of centuries since two languages split from an ancestral language. Let w = the percentage of words from the ancestral language which are common to the the two descendant languages. In linguistics, the equation below (in which r = .86 is the index of retention) has been used to relate c and w.

10/c = 2 log r/log w

If about 15% of the words in an ancestral language are common to two different languages, about how many centuries ago did they split from the ancestral language?"

The book provides the answer of about 63 centuries. Try as I might I cannot seem to get this answer or anything close.

Can anyone help me? Thanks!

 

[Angela Harlem]

I think I'd rather have a urinary tract infection than do maths again any time soon.

 

[DrTeeth]

10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89

 

[achtung_girl]

^agreed! I would do it the same way.

 

[~BrightestStar~]

10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89

QFT


That's how I would have worked it out.



On an unrelated note, as it is not true in this case, surely after teaching for so many years you are familiar with how often textbooks get the answers wrong?
I think they do it on purpose to mess with people's minds.

 

[redhotswami]

10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89

 

[anitram]

I used to be good at math. Then I went to law school. Now I only need to know how to bill. In 6 minute increments. Which really, is an art unto itself.

 

[Laura M]

how often textbooks get the answers wrong?

I've found that. I prefer to do everything myself first anyway. What's annoying is when something in the pupils' book is wrong.

 

[Diemen]

I haven't taken math since high school. Dr. Teeth's answer looks like an alien language to me.

 

[Lila64]

^

Maddy had algebra homework the other night, and needed help (she's in 8th grade). Her math teacher has been out 2 weeks due to illness, so she's had 2 substitutes during that time. The 2nd sub wasn't very helpful she said, and I thought I could try & offer my assistance. She asked if I knew something (monomial? - a word I've never heard of). So I said, let me look at it to see if anything made sense. I made a suggestion, which turned out to be helpful to her. But I haven't had math since high school, which was over 20 years ago I was shocked, and pleased with myself

 

[maycocksean]

10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89

Okay, but how do you end up dividing 10 by 0.159? Shouldn't it be the other way around, dividing 0.159 by 10 (i.e. multiplying by one tenth?).

by the way, thanks for the help and commentary everyone! I teach my Algebra II class in about 15 minutes. We'll see how it goes.

 

[indra]

I haven't taken math since high school. Dr. Teeth's answer looks like an alien language to me.

Me too! I definitely had the face going when I read his answer.

 

[DrTeeth]

It's easier to figure out the mechanism if you substitute the numbers for numbers you're more familiar with.

Instead of 10/c = 0.159, you could write 6/3 =2. It's easy now to see how what you have to do with the 6 and the 2, to get 3 as an answer. Once you've figured that out, you only have to change back those numbers to the numbers in your formula.

 

[MsPurrl]

Shoot me now!

 

[~BrightestStar~]

Okay, but how do you end up dividing 10 by 0.159? Shouldn't it be the other way around, dividing 0.159 by 10 (i.e. multiplying by one tenth?).

I take it today has been one of those horrible days where the longer you look at something that should be simple, the more confusing it gets?


I suppose even math teachers can have their bad days



And I mean no offense.


Anyway, just remember it's a simple equation of:

number1 = number2
variable

(or alternatively)
1 = 2
x


how would you solve that on any other day?

 

[cobl04]

There's an Algebra 2???

I could juuust do normal algebra and basic calculus, and I thought I was ok at maths. I think I passed Year 11 Maths Methods with about a D+/C average or something. Soooooo glad I didn't go on with Yr 12 Methods.

 

[maycocksean]

It's easier to figure out the mechanism if you substitute the numbers for numbers you're more familiar with.

Instead of 10/c = 0.159, you could write 6/3 =2. It's easy now to see how what you have to do with the 6 and the 2, to get 3 as an answer. Once you've figured that out, you only have to change back those numbers to the numbers in your formula.

Yeah, this is how I actually figured it out eventually. It makes pretty logical sense. I guess I was overthinking it. And I'm STILL wondering why the "rules" seem to be different here--in other words normally you're looking to solve for c by multiplying by 1/10 but here you're multiplying each side by c first and THEN dividing by .159.

Maybe I'm there's just a big gap in my understanding of the basic principles here.

 

[maycocksean]

I take it today has been one of those horrible days where the longer you look at something that should be simple, the more confusing it gets?


I suppose even math teachers can have their bad days



And I mean no offense.




how would you solve that on any other day?

No offense taken. I think part of the problem is that I'm really not a math teacher. My certification areas are English and Behavioral Sciences and I'm quite knowledgable in History. However, I've ended up having to teach Math for the past several years and I've gotten much better at it (I hated Math as a kid and always tried to take the absolute minimum of math throughout my education) but I think there are still gaps in my understanding that wouldn't be there with a "real" Math teacher.

 

[~BrightestStar~]

No offense taken. I think part of the problem is that I'm really not a math teacher. My certification areas are English and Behavioral Sciences and I'm quite knowledgable in History. However, I've ended up having to teach Math for the past several years and I've gotten much better at it (I hated Math as a kid and always tried to take the absolute minimum of math throughout my education) but I think there are still gaps in my understanding that wouldn't be there with a "real" Math teacher.

Ahhhh, I see.
This explains a lot.

I know when I've tutored people that manipulation of the formulas (like what we're talking about here, with the 10/c) seems to be something that catches those with a weak background in math.
I'm not sure why exactly this is, it must be a conceptual thing. Cause for me, shifting this stuff around is almost automatic.

I assume there is some shortage of available teachers or funding as a reason why they have thrown you into the pool of math?
And kudos to you for learning as you go. It's tough to do. And hell, you hate math put are teaching algebra?


Anyway, there are a couple ways you can look at it.
I usually just think about doing opposites.
Due to the equal sign, you have to keep both sides of the equation balanced. Thus whatever you do one one side, you do the reciprocal of on the other.
So there, you are dividing by c, to oppose this you would multiply by c on the other side.
It cancels the process out you see, cause you are doing the same things to both sides of the equation, but lets you get the variable where you want it.
c(1/c) = c (2) (random numbers used for simplicity)
When the c's cancel on the left you have 1 = 2c

That's how I end up thinking about it and explaining it to others. It would prob come across better in person I'm afraid.



And, technically, you are right, you could simply divide by 10. But then your answer would be equal to 1/c not c.
It just adds extra steps...

eg

10/c = 2
1/c =2/10
1/c = 0.2
( 1 = 0.2c)
1/0.2 = c
5 = c

vs

10/c = 2
10 = 2c
10/2 =c
5=c




Anyway, I'm rambling now.

Good luck with your class!

 

[Vincent Vega]

^Why didn't any teacher before tell me that was possible?

Thank you.

 

[maycocksean]

The light bulb went on!

1/c! Of course! I forgot that multiplying by 1/10 would leave me with 1/c NOT c on the other side of the equation.

thanks, BrightestStar! Your screen name is quite apt!

 

[Got Philk?]

There's a reason I want to teach kindergarten

 

[BluRmGrl]

Aaaah! Aaahhh! Aaaahhhh! Higher math in the Crack House -- oh dear God, make it stop!!!



Seriously - my utmost respect to you, Dr. Teeth. I marvel at your intellect.

 

[joyfulgirl]

I used to be brilliant at algebra. But I forgot it all as soon as I aced the last test, like a normal person. What is wrong with you people? You're not to supposed to actually retain this stuff.

 

[gvox]

how would you solve that on any other day?

I'd shoot myself in the head first, then give it some serious thought.

 

[GibsonGirl]

I'm an English major!



I used to be good at maths...but I haven't had to bother with it for quite some time now. The only thing I can remember about calculus is "cutting off Britney Spears' head." My Taiwanese professor had some rather unconventional ways of explaining derivatives.

 

[theFly51]

Wow...i just finished taking high school pre-cal
and that looks like WTF to me

 

[Salome]

10/c = 2 log r/log w

r = 0.86
w = 15%. which is the same as 0.15 (which is probably where you go wrong)

10/c = 2 log 0.86/log 0.15

10/c = 0.159

10/0.159 = approx. 62.89

i've always loved this sort of stuff

pyramids, cubes etc though