Well, folks, you might be in for a surprise, but when I was thinking what to name my math program.... You know, first of all, English isn't my native language. Secondly, I'm not super creative when it comes to naming math programs.
I chose "mammoth" because it sort of rhymes with "math". I thought people would be able to REMEMBER it easily! You know, let's say a person stumbles on my website, and days later they try to remember what was it called? Maybe, just maybe, the woolly animal would have made a connection, even if an amusing one, in their mind.
So there you have it. There are no hidden implications. It's not ancient math, nor "humongous" in any sense. The math in "Math Mammoth" is pretty normal and logical.
All this was spurred by a really enjoyable and fun review of my books by Mary Grace at Books and Bairns. She really has a knack for writing!
Selasa, 30 Desember 2008
Jumat, 26 Desember 2008
The New Year 2009
I hope you all have had some happy family time (or otherwise) during these Christmas days! Now, I'm already going to turn your thoughts towards the changing of calendar year.
We're about to change from 2008 to 2009. If you'd like to have some mathematical fun with a new year's theme, check out MathNotation's Get Ready for Happy 41*7^2.
Basically, what you do for this "game" is try to find something special about the number 2009. Like the title of Dave's post tells us, 2009 factored is 41 × 72. So one thing you can do is ask the kids to factorize 2009.
Then, it's just up to you - or the students - to find anything interesting or special about the number 2009. Maybe they can explore the remainders when 2009 is divided by various numbers and find something that sounds "special". Maybe they can explore what kinds of sums they can make with it (it's 1004 + 1005 for example). Or, how about this sum: 2009 = 777 + 29 + 92 + 209 + 902. Or, you could simply ask students to write 2009 as a sum of four (or five or seven) whole numbers that are as close together as possible.
Read also Dave's post.
Yet another way to play a game with 2009 is explained at Math Forum's 2009 Year Game. here you will use the digits from 2009 to form all counting numbers from 1 to 100.
But, most importantly I want to wish you a prosperous & good year 2009! Mathematically and otherwise.
We're about to change from 2008 to 2009. If you'd like to have some mathematical fun with a new year's theme, check out MathNotation's Get Ready for Happy 41*7^2.
Basically, what you do for this "game" is try to find something special about the number 2009. Like the title of Dave's post tells us, 2009 factored is 41 × 72. So one thing you can do is ask the kids to factorize 2009.
Then, it's just up to you - or the students - to find anything interesting or special about the number 2009. Maybe they can explore the remainders when 2009 is divided by various numbers and find something that sounds "special". Maybe they can explore what kinds of sums they can make with it (it's 1004 + 1005 for example). Or, how about this sum: 2009 = 777 + 29 + 92 + 209 + 902. Or, you could simply ask students to write 2009 as a sum of four (or five or seven) whole numbers that are as close together as possible.
Read also Dave's post.
Yet another way to play a game with 2009 is explained at Math Forum's 2009 Year Game. here you will use the digits from 2009 to form all counting numbers from 1 to 100.
But, most importantly I want to wish you a prosperous & good year 2009! Mathematically and otherwise.
Minggu, 14 Desember 2008
Solving direct and inverse variations in chart form
Dave Marain recently featured my blog on his, and now it just so happens I get to promote his, because I really liked his post about learning direct and inverse variations.
He uses a beagle problem with interesting numbers:
I want to go one step further with this. Let's try a little more awkward numbers:
In each step on our chart, we change ONE variable (either the number of beagles, the number of holes, or the number of days), keep ONE variable unchanged, and figure out how the third variable changes. You need to carefully think if that third variable is multiplied or divided — if it is in direct or inverse variation.
For example: if the number of beagles is halved, and there are the same amount of holes, how will the number of days change?
Or: if the number of holes is quadrupled, and there are the same amount of beagles, how will the number of days change?
Let's start with the situation given in the problem.
We will want to find out how many days it takes ONE beagle to dig ONE whole, and then use that as a "stepping stone" to find how many days it takes 5 beagles to dig 9 holes .
So, we first change the chart so we have only ONE beagle, digging the same amount of holes. This, of course, TRIPLES the amount of days.
Then, to the question. How many days will it take 5 beagles to dig 9 holes? First, let's increase the amount of beagles to 5, digging the one hole. That will slash the amount of days by 5:
Lastly, we increase the number of holes from 1 to 9, so that the amount of days will also increase 9-fold:
You could also go through this in some other order. But the beauty of this chart approach is that it will work for any numbers. So, kids who have hard time with joint variation formulas might be able to use such chart approach successfully for all these types of work problems.
d = k * (h/b)
To solve for k, we plug in the values from the original situation (Three beagles can dig 7 holes in eight days).
8 = k * (7/3) from which k = 24/7.
Then we have our formula ready to use:
d = (24/7) * (h/b)
The question was: how many days will it take 5 beagles to dig 9 holes? So b is 5 and h is 9:
d = (24/7) * (9/5) = 216/35 ≈ 6.17 days.
He uses a beagle problem with interesting numbers:
"Three beagles can dig 4 holes in five days. How many days will it take 6 beagles to dig 8 holes?"The solution is actually quite easy — just think how there are exactly twice as many beagles and also twice as many holes. Dave shows a chart that can help youngsters grasp the solution.
I want to go one step further with this. Let's try a little more awkward numbers:
"Three beagles can dig 7 holes in eight days. How many days will it take 5 beagles to dig 9 holes?"I want to show you how a "chart" approach will still work. This situation describes joint variation, because there is both inverse and direct variation involved: the number of days it takes to dig the holes varies inversely with the number of beagles (the more beagles, the less days it takes), and varies directly with the number of holes (the more holes, the more days it takes).
In each step on our chart, we change ONE variable (either the number of beagles, the number of holes, or the number of days), keep ONE variable unchanged, and figure out how the third variable changes. You need to carefully think if that third variable is multiplied or divided — if it is in direct or inverse variation.
For example: if the number of beagles is halved, and there are the same amount of holes, how will the number of days change?
Or: if the number of holes is quadrupled, and there are the same amount of beagles, how will the number of days change?
Let's start with the situation given in the problem.
Beagles Holes Days
----------------------------------
3 7 8
We will want to find out how many days it takes ONE beagle to dig ONE whole, and then use that as a "stepping stone" to find how many days it takes 5 beagles to dig 9 holes .
So, we first change the chart so we have only ONE beagle, digging the same amount of holes. This, of course, TRIPLES the amount of days.
Next, we find out how long it takes this one beagle to dig just ONE hole. Now, the amount of days is divided by 7.
Beagles Holes Days
----------------------------------
3 7 8
1 7 24
Beagles Holes Days
----------------------------------
3 7 8
1 7 24
1 1 24/7 = 3 3/7
Then, to the question. How many days will it take 5 beagles to dig 9 holes? First, let's increase the amount of beagles to 5, digging the one hole. That will slash the amount of days by 5:
Beagles Holes Days
----------------------------------
3 7 8
1 7 24
1 1 24/7
5 1 (24/7)/5 = 24/35
Lastly, we increase the number of holes from 1 to 9, so that the amount of days will also increase 9-fold:
Beagles Holes Days
----------------------------------
3 7 8
1 7 24
1 1 24/7
5 1 (24/7)/5 = 24/35
5 9 9 * 24/35 ≈ 6.17
You could also go through this in some other order. But the beauty of this chart approach is that it will work for any numbers. So, kids who have hard time with joint variation formulas might be able to use such chart approach successfully for all these types of work problems.
Joint Variation: The formula
For completeness sake, I'll also solve this problem of mine using the formula. Let d be the number of days, h be the number of holes, and b be the number of beagles. And, k is the constant of variation. We know that d varies directly with the number of holes (the more holes, the more days it takes). Also, d varies inversely with the number of beagles (the more beagles, the less days). So:d = k * (h/b)
To solve for k, we plug in the values from the original situation (Three beagles can dig 7 holes in eight days).
8 = k * (7/3) from which k = 24/7.
Then we have our formula ready to use:
d = (24/7) * (h/b)
The question was: how many days will it take 5 beagles to dig 9 holes? So b is 5 and h is 9:
d = (24/7) * (9/5) = 216/35 ≈ 6.17 days.
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