OK, for this radical equation, I'm going to just put the problem down because I have no clue how to show my mathematical progress through Microsoft word 0_0 Anyways let's get it started. This is how I solved this problem.
PROCESS:
1 Like I said before, I do not like to leave negatives in the equation so I'm going to solve them first. So, I move my x^-2 to the bottom and I move the y^-4 to the top and they become positive :)
2. I distribute my exponent 5/2 to every number/variables and it becomes like 81^5/2, y^10, 49^5/2 and x^5.
3. So now you might notice that we can simplify the numbers with exponents. When we do simplify them the result would be 59049 y^10/ 16807 x^5.
AHHH!! I totally forgot that I forgot to post a POST about how to solve a rational equation! Because of that I lost 10 mark! Although thanks to our most generous teacher Mr. Cheng, he told me he would change my mark if I post it before the first term. So enough of talking, let's et down to business.
Ok This expression seems quite confusing... Plus I haven't done this for WEEKS! Well let's solve it slowly:
Procedure of Solving: 1. For me, I really dislike negative numbers, so I usually solve them first. So I bring up the x^-2 up to make it positive and add it up with an x. So it would equal x^3.
2. Then I wold distribute the exponent 3/2 to the expression. So it would be 16^3/2 (x^3)^3/2 OVER 25^3/2.
3. We also learned from previous lessons, how to change exponents in fractions to radicals. So it would be Root 16^3 then we simplify the (x^3)^3/2 into x^9/2 OVER root 25^3. Now when you simply these two radicals, you wold get 64 for the numerator and 125 for the denominator.
4. Then we change the x^9/2 to Root x^9.
5. We are almost there! but not quite yet.. we still have to bring out the root x^9 into something MORE simpler, so it would become x^4 root x.
6. Finally you would put all of these together and would get the answer from the diagram above.
Math is something we go through from kindergarten, to end of high school and even life. However, we do not know what it is to become a good math student. To become really good in math, you need three things: Patience, perseverance and interrogation. These are the 3 main factors that I think would make someone into a good math solver.
Maybe many of you might have different opinions as I do. Why them? First, I thought patience was a vital factor because, solving math problem takes time. When solving a problem,DON'T RUSH! It's very common to make foolish mistakes trying to solve things as quick as possible. Take your time, nothing is going to get you. It doesn't hurt to be slow sometimes in your life.
Second, perseverance is the key. Most of us (including myself) always give up when questions seem so COMPLICATED. It's funny how when a question seems so hard, and when one of your math teacher solves it for you and you realize how easy it was. The problems just sound hard, we have to start thinking thinking THINKING! However, do not take too much time trying to solve one question in the test.
Third, INTERROGATE! Do not be afraid to ask lot of questions! I personally hated to ask questions. The reason was because it seemed like if I ask a question when no one is asking, it seemed like I was the only stupid one who didn't get it. So I end up not asking, go home and come back to school the next day and there it is, the TEST. Then I end up losing myself and go crazy for about half an hour and hand in my test, half of them all BLANK. Then I ask myself, "Why the heck didn't I ask?" This is for ALL OF YOU People's BENEFIT! Please don't act like you know EVERYTHING and end up screwing up yourself over. I will tell you once more, PLEASE ASK IF YOU DON'T KNOW! Who cares if people call you dumb for asking? You are learning for your own, who cares what other people say. As a matter of fact, Thomas Edison himself asked lot of questions although he was greatly despised by his teacher who thought all those questions seemed outrageously ridiculous.
I'm not saying you should become Thomas Edison or one of those genius kind of people. All I'm telling you is, as long as you try to practice 3 of these factors, I guarantee you, your grade average of your math would increase.
AHH!! We only have 1 week of Sprin BREAK LEFT!! This thought is making me so sad....although I do miss some of my friends at school =/ Well anyways during last few day, I had a great opportunity to sleep in and have some great time with some of those GREAT People :D However, I did get bored playing so decided to do some math over the break. I didn't do too much of them but did enough to keep my brain functioning. There was this one particular problem I really liked. This question was very challenging yet very fascinating and fun.
Question:A positive integer is to be placed in each box. The product of any four adjacent integers is alway 120. Wht is the value of x?
When I saw this questions at first I was like, "Whoa... how am I suppose to do this?" However, like I said in previous posts before, if you read and think carefully, you would be able to come up with your own way of solving this problem.
So this was how I solved it. First we look at numbers 2 and 4. When 2 and 4 multiplies, the product would be 8. Meaning, in between 2 and 4 must be two digits that would add up to 15. (120/8) So now let's solve the column beside number 4. We now know the two columns on the left of 4 is 15. So we multiply 15 with 4 and come up with 60. Next you would divide 120 with 60 and the product would equal to two. Next we also know that next two columns adjacent to number 2 equal to 15. Then we may be able to solve the colum right side of the variable x. 2x15= 30, 120/30= 4 The column on the right side of the variable is 4. Next let's figure out the column that's on the RIGHT side of the numer 4. 15x4= 60, 120/60= 2. so the number is 2.
Finally we have everything we need to solve for the "x". We know three numbers adjacent to the variable. 4, 2 and 3. So we muliply all those and divide 120 using the product.
Finally it's spring break! I'm very excited that I'm free from school and for two weeks I can sleep in and hang out with friends and do whatever. However, this doesn't mean I will be playing and be lazy all the time. I should some how function my brain so that it wouldn't shut down on the first day when I go back to school. -_-
So I decided to work on the P. Math 10 Practice Unit Tests package given by Mr. Cheng (how was your trip Mr.Cheng?). ^^
I went over few questions and most of things were very familiar to me and I thought I might challenge myself by trying to solve a question from the back. It was one of those annoying time questions that I really disliked. Although, I thought it wouldn't hurt to try, so I did. I carefully looked over the solution of that particular question and solved another question with similar logic.
Question:
Ann entered a race which required a run of 10 km north then a bike ride of 120km west. Determine her running speed if she cycled ten times faster than she ran and the total race took her 4 hours to complete. Answer to one decimal place.
Seems quite complicated doesn't it? It took a while for me to understand what this question was asking me and I needed to know how I could solve this problem and know how to set it up.
Let's think slowly and find solution little by little.
We all know that D x V = T (Distance x Velocity= Time) So let's make a chart based on the information we have. Now the chart above makes solving this problem easier.
Now we have to find her running speed and how are we going to do this? Well, let's try it this way. The second last sentence of the questions clarifies that it took about 4 hour to complete the race.
That was our first step of our equation. Now let's solve the rest of the equation.
This was how I solved this question. I really enjoyed solving this one (even though I don't really like these question and can't solve them very well) This won't be my only blog I'm going to post over the Spring break. (Notice my title! There is "1" at the end of the sentence meaning there are more to come!) >.<
So I will keep you guys posted and hope that you guys would have SPLENDID SPRING BREAK! WHOOT
Thursday March 4th, Mr. Cheng was gone to the Grad retreat with the grade 12's and our math class had Tuesday problem solving day moved to Thursday. We had a choice of choosing past math contests from gr. 9, gr. 10, gr 11 contests. However all of these contests required different numbers questions. For example, gr 9 contest you needed up to 22 questions done, gr 10 contest = 20 questions, gr 11 = 18 questions. So i chose the gr 10 math contest because I will have to take it next year again.
In the practice contest, I really enjoyed solving this one particular problem. It was number 15 in Part B. The question was:
In the multiplication shown, P and Q each represent a single digit, and the product is 32 951. What is the value of P+Q?
3 9 p x Q 3 ---------- ____________ 3 2 9 5 1
I honestly enjoyed solving this problem because it makes you think. So this is how I solved this problem. So first we have to realize if this question is asking for the sum or the product. Sum means addition and product means multiplication. Anyways, we have to go from right to left and use whatever we have to work with. So 3 x p = something 1. So we have to know what times 3 equals to digit that has the number 1 at the end. 3x1= 1 3x2= 6 3x3= 9 3x4= 12 3x5= 15 3x6= 18 3x7= 21 So we know that we have to multiply 7 to get number 21. Then 3x9 then next, 3x3.
Then the first answer will be 3 9 7 x q 3 --------- 1 1 9 1
___________ 3 2 9 5 1 Then now we work with the variable q. We know that q has to times 7 and we also know that the product must equal to 5. So we also know that q times 7 must equal to a digit with the last number 6. So this means that 7 x 8= 56. So we try replacing q with 8 .
This week Tuesday, it was the problem solving day where we can eat while we solve math problems during class. Sheets were handed out and everyone started to solve the problems given. There were three sections to the sheet, Part A, Part B, and Part C.
Part A was very easy and i could just do it myself, but Part B, it became somewhat challenging. Part C, I did try to solve one but it was really challenging and I did not have enough time to solve them D: Anyways, while I was going through the sheet, I really liked question number 15 and I really did enjoy solving it. Here is the question: Four points are on a line segment as shown. If AB:BC= 1:2 BC:CD= 8:5 and find the ratio of AB:BD.
This question seems quite confusing at first but when you look at it carefully, it's quite easy. So this is how I solved it. AB and BC is 1:2 meaning half of one another. BC and CD is 8:5 ratio. Both of these ratios include BC, except in different numbers. So one ratio says BC is 2 and other side says it's 8.
AB is half of BC meaning 8/2 = 4. So we can find out AB is 4. The question left to solve is, what is BD? To find BD, all you have to do is add BC and CD together: 8+5= 13. What is AB:BD? We found out what AB is 4 and BD is 13. So the ratio would be 4:13
It wasn't so hard was it? I mean I did find it confusing at first but when you read it several times, information that has been given to you would seem more clear and understandable.
