Helloooo, today in math class, we did a test. It wasnt too hard, but I kept messing up in my formula for the last question, and so I eventually just counted spaces on a graph to solve it. Anyways, that has nothing to do with this blog entry. I only included it because right after we were done the test, Mr Cheng wrote about good qualities that make someone good at math. The top 3 qualities that I think make a good mathemagician (not a magician plus a smart math guy, just a smart math guy) are:
- knowing the fundamentals of math
- thinking outside of the box
- being patient and diligent
I think that having basic math skills are very important because without them, you would not be able to do any questions at all. eg. You need to know how to do subtraction and addition to do algebra.
Also, being able to think outside the box is really useful in solving math problems. Sometimes in math, questions are worded strangely or are just plain hard, but if you think outside the box and try unorthadox methods, then you can usually figure them out.
Lastly, you gotta be patient. Math takes a really long time sometimes, and it might get really annoying. Even though you might be frustrated, you have to be patient and work through the questions in order to succeed.
These are the three major things I think that you need in order to be good at math.
Friday, March 26, 2010
Saturday, March 20, 2010
solving radical expressions
Okay so a couple of weeks ago, we learned about radical expressions, which I think these are examples of. They are pretty hard, but after a while they get easier. These 3 examples we solved on Microsoft Word, and just figuring that out is harder than the actual math. On the other hand, I haven't done one of these in a few weeks, so I might not remember how to do them right. But anyways, these are radical expressions (I think).
The steps to solve this one are (stolen off Mr. Cheng):
1. resolve the negative exponent x^-2 by flipping it to the top
2. distribute the exponent 3/2 to all components
3. 3/2 means cube and then square root, so 16^(3/2) = 64, 25^(3/2) = 125
4. for the x term, power on power, so you multiply 3 with 3/2 to get 9/2
5. write x as a radical and simplify
Done! Yay!
To solve this second one, you:
1. resolve the negative exponents 81x^-2 and 49y^-4 by flipping them to the bottom and top accordingly
2. ditribute the exponent 5/2 to all components
3. 5/2 means find square root and then do to the power of 5, so 81^(5/2) = 59049, 49^(5/2) = 16807
4. for the x and y terms, power on power so you multiply 4 with 5/2 to get 20/2 and you multiply 2 by 5/2 to get 10/2
5. write x and y as radicals
6. simplify x and y
Yea! Done this one too!
The steps to solve this one are (stolen off Mr. Cheng):
1. resolve the negative exponent x^-2 by flipping it to the top
2. distribute the exponent 3/2 to all components
3. 3/2 means cube and then square root, so 16^(3/2) = 64, 25^(3/2) = 125
4. for the x term, power on power, so you multiply 3 with 3/2 to get 9/2
5. write x as a radical and simplify
Done! Yay!
To solve this second one, you:
1. resolve the negative exponents 81x^-2 and 49y^-4 by flipping them to the bottom and top accordingly
2. ditribute the exponent 5/2 to all components
3. 5/2 means find square root and then do to the power of 5, so 81^(5/2) = 59049, 49^(5/2) = 16807
4. for the x and y terms, power on power so you multiply 4 with 5/2 to get 20/2 and you multiply 2 by 5/2 to get 10/2
5. write x and y as radicals
6. simplify x and y
Yea! Done this one too!
Wednesday, March 17, 2010
cayley canadian math contest
Okay so a couple of weeks ago (Thursday, February 25, 2010 to be exact) our math class participated in an across Canada (I think) math contest. It wasn't the hardest thing I've ever done, but it was still pretty stressful. Most of questions were do-able, and I know I could I have gotten some of the harder ones if I had enough time. For one of the questions (number 16), I almost figured it out, but ran out of time at the last second. There was also one more question (number 14) that I thought I got right, but after we went over the contest in class, I learned the answer I put was wrong. :( In the end, I tried to leave exactly 5 questions that I did not know the answer to blank in order to get the max 10 free marks from unanswered questions, which was pretty good because there was in total 6 questions that I didn't know the answer to.
During the contest, I was quite focused. I think that the pressure helped me concentrate better, but from this contest I learned that being calm was the key to figuring out the answer. I also think that because near the end of the contest, I was running out of time and I was forced to work harder and faster - the pressure helped me do better.
My favourite question from this contest was number 23, but it would take way to long for me to explain what I did (and I don't quite remember everything I did anyways), and so I'm going to choose a different question. Another question that I liked was number 21, but it involves a drawing, and I don't really want to make it on paint, and so I'll choose an easier question. A fun and easy question was number 7. and so I'll do that one. I like it because it is fun and easy and simple. =D
The mean (average) of 5 consecutive integers is 9. What is the smallest of these 5 integers?
a) 4
b) 5
c) 6
d) 7
e) 8
If the Average of 5 consecutive integers is 9, then 9 must be the middle number. If there are 5 numbers, then the middle number must be 3. The number 3 is two less than 5. The number 7 is two less than 9. The answer is d) 7.
During the contest, I was quite focused. I think that the pressure helped me concentrate better, but from this contest I learned that being calm was the key to figuring out the answer. I also think that because near the end of the contest, I was running out of time and I was forced to work harder and faster - the pressure helped me do better.
My favourite question from this contest was number 23, but it would take way to long for me to explain what I did (and I don't quite remember everything I did anyways), and so I'm going to choose a different question. Another question that I liked was number 21, but it involves a drawing, and I don't really want to make it on paint, and so I'll choose an easier question. A fun and easy question was number 7. and so I'll do that one. I like it because it is fun and easy and simple. =D
The mean (average) of 5 consecutive integers is 9. What is the smallest of these 5 integers?
a) 4
b) 5
c) 6
d) 7
e) 8
If the Average of 5 consecutive integers is 9, then 9 must be the middle number. If there are 5 numbers, then the middle number must be 3. The number 3 is two less than 5. The number 7 is two less than 9. The answer is d) 7.
Friday, March 5, 2010
problem solving 3. question 19
Okay so yesterday in math class, we got the grade 10 Canadian Mathematics Contest from 2006. I came in late so I had to rush through all the questions, but there was one that stood out to me as both challenging and fun. I liked it because it was a little bit confusing and I almost got it wrong the first time because I read it wrong. From this I learned that I always have to read the questions very carefully and make sure I understand what they are asking. The question was number 19.
19. In a bin at the Cayley Convenience Store, there are 200 candies. Of these candies, 90 % are black and the rest are gold. After Yehudi eats some of the black candies, 80 % of the remaining candies in the bin are black. How many black candies did Yehudi eat?
a) 2
b) 20
c) 40
d) 100
e) 160
At first when I looked at this question, I automatically assumed that the answer was 20. Then I saw that in the question, it stated that 80% of the REMAINING candies were black. I then looked at the multiple choice answers given to you. I knew the answer wasn't a) 2, because the number is way too small. I also knew that b) 20 was wrong now, and so I tried out c) 40, to see if it worked out.
If 90 % of the 200 were black at the beginning, then that means that 180 of them were black.
200 x 0.9 = 180
If I take away c) 40, then I am left with 140 black candies and 20 gold candies,
200 - 180 = 20
and this is obviously not 80%.
140/160 = 0.875 = 87.5%
All of a sudden, the answer was really obvious. I didn't even have to look at the next multiple choice answer to know that the answer was d) 100. If I take away 100 from 180, then I am left with 80 black candies, and 20 gold candies, which works out to be the answer.
180 - 100 = 80
80/100 = 0.8 = 80%
The answer is d) 100
19. In a bin at the Cayley Convenience Store, there are 200 candies. Of these candies, 90 % are black and the rest are gold. After Yehudi eats some of the black candies, 80 % of the remaining candies in the bin are black. How many black candies did Yehudi eat?
a) 2
b) 20
c) 40
d) 100
e) 160
At first when I looked at this question, I automatically assumed that the answer was 20. Then I saw that in the question, it stated that 80% of the REMAINING candies were black. I then looked at the multiple choice answers given to you. I knew the answer wasn't a) 2, because the number is way too small. I also knew that b) 20 was wrong now, and so I tried out c) 40, to see if it worked out.
If 90 % of the 200 were black at the beginning, then that means that 180 of them were black.
200 x 0.9 = 180
If I take away c) 40, then I am left with 140 black candies and 20 gold candies,
200 - 180 = 20
and this is obviously not 80%.
140/160 = 0.875 = 87.5%
All of a sudden, the answer was really obvious. I didn't even have to look at the next multiple choice answer to know that the answer was d) 100. If I take away 100 from 180, then I am left with 80 black candies, and 20 gold candies, which works out to be the answer.
180 - 100 = 80
80/100 = 0.8 = 80%
The answer is d) 100
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