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Apr 30 2013 03:39pm
Quote (turtol @ 30 Apr 2013 15:18)
I have the practice exam out of 50 questions, and i don't understand anything on there. i'm not quite sure what to youtube, typing in like yx5 + x + 4 = yx2  i'm not quite sure what that's called, factoring? i just have to simplify it.


Get a textbook then since high school textbooks explain things in plain English without requiring any background in analysis or formal notation.
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May 1 2013 12:48pm
Quote (Azrad @ Apr 30 2013 01:49pm)
can you give a better example, because that can not be simplified...do you mean solve it explicitly for y?


Here's a few sample questions

Simplify a b2 c3 (2a2b)-1
c4 and eliminate any negative exponents.


all the numbers are exponents except for the 2a i have no clue how to start doing htis or anything...


(t+3)(2t-1)-3(t+2)

This post was edited by turtol on May 1 2013 12:49pm
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May 1 2013 01:19pm
You need to write that in a more explicit fashion (which is hard on a computer):
on way to write this on a computer is to write it as

a * b^2 * c^3 * (2a^2*b)^(-1)

that being said try to remember this:

x^(-1) = 1/x

y^(-2) = 1/(y^2)

1/[x^(-1)] = x/1 = x

1/[y^(-2)] = y^2/1 = y^2


so:
a * b^2 * c^3 * (2a^2*b)^(-1) = [a * b^2 * c^3]/[2a^2*b].... now you can do some canceling

[b*c^3]/[2a]



http://www.youtube.com/watch?v=JUGmviJC_pk

This post was edited by Azrad on May 1 2013 01:24pm
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May 1 2013 11:31pm
Quote (turtol @ Apr 29 2013 09:53pm)
I've got a math 11 exemption test on the 10th, i need to learn the entire curriculum before then =D what website would be best for this? or should i just hire a tutor? i can easily do this spending 4 hrs a day until then.

i only took up to math 10 i believe, up to the point where 4x + 4 = 5x

that's all that makes sense, simplifying the huge algebraic equations, and all this pre-calc stuff is nonesense to me =S



tl;dr, where to learn math 11? website? tutor? friend (pay them with beer!)?


LOL i thought this was funny, before the test it says "to get your percent simply divide correct answers by 50 then multipy by 100" AKA times your mark by 2 -.- if you can't figure out your percent on a test out of 50 you shouldn't even be attempting pre-calc...


id get the textbook that course is taught with look at what sections are taught read the sections then use khanacademy or mathisfun if you dont understand an entire section

if its just a problem or two try posting them on jsp its always helped me
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May 6 2013 03:10am
Quote (brigadier @ May 1 2013 10:31pm)
id get the textbook that course is taught with look at what sections are taught read the sections then use khanacademy or mathisfun if you dont understand an entire section

if its just a problem or two try posting them on jsp its always helped me


could you possibly tell me what these questions are? i don't know what exactly i'm supposed to be studying, i gotta learn this all by friday.

http://www.douglas.bc.ca/__shared/assets/DCMA_0100_Practice_Test_2012SEP81115.pdf


e/ if that link doesnt work, go to this one and it's the pre calc math placement practice test one

http://www.douglas.bc.ca/application-services/assessment-testing/information-sheets.html

This post was edited by turtol on May 6 2013 03:11am
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May 6 2013 06:18am
Quote (turtol @ 6 May 2013 09:10)
could you possibly tell me what these questions are? i don't know what exactly i'm supposed to be studying, i gotta learn this all by friday.
http://www.douglas.bc.ca/%5F%5Fshared/assets/DCMA%5F0100%5FPractice%5FTest%5F2012SEP81115.pdf
...


let me give you some start

1 d - easy to see if you turn the -1 term over, ie a(b^3)(c^2)*(c^4)/[2(a^2)b] and then consolidate

2 b - just multiply the formula out and then simplify [because i am lazy and it is multiple choice i just worked out the -9 :D ]

3 c - multiply the formula in c out and it matches the one above [actually i scanned over them and c is the only one giving 5y, then i checked for the other factors]

4 e - (x+1) can be eliminated [checking through the choices e was the only one having the (XXX-2)(YYY-3) so i started with that]

5 e - |t+1/2|=>2 for positive t=> 3/2 and for negative t<=-5/2
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May 6 2013 12:41pm
Quote (brmv @ May 6 2013 05:18am)
let me give you some start

1 d - easy to see if you turn the -1 term over, ie a(b^3)(c^2)*(c^4)/[2(a^2)b] and then consolidate

2 b - just multiply the formula out and then simplify [because i am lazy and it is multiple choice i just worked out the -9 :D ]

3 c - multiply the formula in c out and it matches the one above [actually i scanned over them and c is the only one giving 5y, then i checked for the other factors]

4 e - (x+1) can be eliminated [checking through the choices e was the only one having the (XXX-2)(YYY-3) so i started with that]

5 e - |t+1/2|=>2 for positive t=> 3/2 and for negative t<=-5/2


see, i don't understand why on earth you are doing those things =S I need to get 65% on this test =S do u know what these equations are called? so i could google how to do them from the beginning, i appreciate the effort put in though!
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May 6 2013 01:13pm
Quote (turtol @ May 6 2013 11:41am)
do u know what these equations are called?
simplifying negative exponents?

Quote (turtol @ May 6 2013 02:10am)

that just seems like a general algebra test. When do you need to take this exam?


This post was edited by Azrad on May 6 2013 01:20pm
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May 6 2013 02:44pm
I looked through that test and it looks like the first few questions deal with reducing fractions and simplifying terms. The point of it is to be able to write a long equation in as little space as possible. For example, I can write out a*a*a*a*a*a*a. However, I can also make it easier for myself by instead writing a^7 where ^ indicates that the next number is an exponent. Same goes for addition. That is, I can write a+a+a+a+a, but you should know that this is the same as 5*a, or 5a. Obviously, the 2nd way of writing stuff takes muuuch less space.

Another thing you need to know is negative exponents. Exponent is the small number on the upper right. Since d2jsp doesn't have included ways of writing it the way you will see it on the paper, I will use the ^ symbol. That means 5^2 = five squared, or five to the second power. You should know that if you have a term a^b that means that you are multiplying the term a by itself b amount of times. Sometimes you get negative exponents. That means that you are actually taking an exponent of the reciprocal of the term. Reciprocal means that if you have a fraction x/y, its reciprocal = y/x. So let's say you have (x/y)^(-5). This is the same as (y/x)^5.

So in question 1, you are faced with a*b^3*c^2*[(2a^2*b)/(c^4)]^(-1). First thing to do would be flip the fraction in order to get rid of the negative exponent. So
a*b^3*c^2*[(2a^2*b)/(c^4)]^(-1) = a*b^3*c^2*[(c^4)/(2a^2*b)]^(1), and since anything raised to the first power is just the term by itself, this equals a*b^3*c^2*[(c^4)/(2a^2*b)]. In multiplication, order doesn't matter. That's why a*b = b*a. Also, (a*b)*c = a*(b*c), that's why you can rewrite the term as [(a*b^3*c^2*c^4)/(2a^2*b)]. You notice that that you have two terms that have c in them right next to each other. What we're gonna do next is called combining like terms. c^2 and c^4 both have cs in them, so you can combine them together. They have different powers though, however, as I mentioned earlier, powers just mean multiplying the same term a certain amount of time. For the first c, we multiply it together twice, then we have the second term, in which we multiply the second c 4 times. So in total, you have (c*c)*(c*c*c*c) and this is the same as c^6. From this you should remember that if you ever have something that looks like a^x*a^y = a^(x+y). Going back to problem number 1, you get [(a*b^3*c^6)/(2a^2*b)]. Now let's look for more like terms. There's a at the top and bottom of the fraction, so you can cancel one of them out from both top and bottom (you should know that a/a = 1, and x*1 = x, which is why we are allowed to do that), so cancelling one of a's on both bottom and top gives us: [(b^3*c^6)/(2a*b)]. Notice that the 2 in front of a stays there, but instead we take the exponent. That's because that 2 is multiplied by a*a. We want to take out the actual a, not whatever a is being multiplied by. The only term left that repeats itself in here is b, so we do the same thing, that is, cancel it out. We get: [(b^2*c^6)/(2a)]. All the terms in here are now different, and so we can't reduce it anymore. Our answer looks just like d.
2. We start off with
(t+3)(2t-1) - 3(t+2)
They ask us to multiply it out, so let's follow directions. Now, i don't know if you heard this before, but there's a process for multiplying things that look like (a+b)(c+d) and it's called FOIL. The idea is that you start by multiplying First terms in both parentheses: a*b, then the Outside terms a*d, then the Inside terms b*d, then the Last terms, b*d. So let's do that in the question: t*2t+t*(-1)+3*2t+3*(-1) -3(t+2). Now we take care of those last terms, by multiplying the 3 out. So we get: t*2t+t*(-1)+3*2t+3*(-1) -3t + (-3)*2. Now let's simplify just a bit: 2t^2 - t+6t - 3 -3t -6. Now, we are going to once again combine like terms, but keep in mind that in addition, a and a^2 are not like terms. They have to have the same power. So we group them all together first: 2t^2 + (6t-3t-t) + (-3 - 6) and actually combine the like terms: 2t^2+2t -9. That looks just like b!

Let me know if I'm clear enough or if there's any parts you'd like me to explain in more depth or any details you don't seem to fully understand or if I'm writing too much details :P. I'll be happy to explain the rest of the questions once I get a reply from you on how you understand those two if you want. You can also pm me and maybe I could explain it better over skype or something.
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May 6 2013 02:46pm
Quote (Azrad @ May 6 2013 12:13pm)
simplifying negative exponents?


that just seems like a general algebra test. When do you need to take this exam?


on friday, i know basic math, something like this i can solve no problem

4x + x + 14 = 28

something like that i can do easily.
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