Mt.SAC MATH HELP

what school do you go to?

Dead Man
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Re: Mt.SAC MATH HELP

Post by Dead Man » Tue Jan 27, 2015 2:36 am

Also I just notice that at the bottom of that sheet are the answers.
3, 8, 9 says They are different answers Sparky

1. d
2. c
3. b (Says its not D)
4. a
5. c
6. a
7. c
8. d (Says its not A)
9. a (Says its not C)
10. b
11. c
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SS Sparky
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Re: Mt.SAC MATH HELP

Post by SS Sparky » Tue Jan 27, 2015 8:19 am

Crap sorry give me liek 10min to redo them

SS Sparky
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Re: Mt.SAC MATH HELP

Post by SS Sparky » Tue Jan 27, 2015 8:39 am

8. It was my stupid calculations error. I even wrote 9x=3 and x=3...which is obviously wrong xD So yea, that's obviously simple error so it is 1/3 as 9 x 1/3 = 3.

9. Just realized it can't be C because the base of a log can never be a negative. For example, that first part of the equation, log(x), would not be possible with a negative as the x because there is NO way 10 raised to a number can equal a negative number.

3. That was just me sucking at math haha. So, to get their correct answer, you gotta first make all the negative exponents into positive. I recommend you write out the numbers as I do it cuz it's easier to see it. The top becomes (1/x^2)-(1/y^2) and the bottoms is (1/x)+(1/y). You have to do these separately so each part has a common denominator. To make the common denominator, you have to mulitply so that they have the same denominators, which is the bottom of each fraction. Let's do the top first. (1/x^2)-(1/y^2) becomes (y^2/x^2y^2)-(x^2/y^2x^2). Notice that now they have the same denominators which is x^2y^2. That becomes ((y^2-x^2)/(x^2y^2)). The bottom we do the same thing. (1/x)+(1/y) becomes (y/xy)+(x/xy). Notice that if they are to be simplified, we can easily just cancel out the same variables. Back to the problem, we now have the bottom as (y+x)/(xy). Our equation is now ((y^2-x^2)/(x^2y^2))/((y+x)/(xy)). So when we have a fraction over a fraction, we do this. Here's an example. (a/b)/(c/d) will become (ad)/(bc). You see it? The bottom of the top becomes the bottom while the bottom of the bottom goes to the top. In our equation, it will be (y^2-x^2)(xy)/(x^2y^2)(y+x). Now we start cancelling. The top (xy) can cancel with the (x^2y^2) to make just (xy) on the bottom. That now gives us (y^2-x^2)/(xy)(y+x). Now's the stupid part. You need to be able to see that (y^2-x^2) can actually be FOILed, which would give (y+x)(y-x). So now the equation is (y+x)(y-x)/(xy)(y+x). You cancel the (y+x) and you're left with (y-x)/(xy) which is B. Stupid piece of crap long simplifying -.-

SS Sparky
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Re: Mt.SAC MATH HELP

Post by SS Sparky » Tue Jan 27, 2015 8:54 am

Since genius lame-o Okonkwo <3 got you on the 1st and 4th, I'll help with the middle 2.

2. So we look at this and ask, what can we do with it, and the only thing you can do is pray that by simplifying the top, you'll be able to start cancelling stuff. The top is x^2-xy. We can simplify this to x(x-y). We see that the bottom is (y-x), which is super close to what we have. It's only different by signs, so we can take out -1 from the top, too, because right now we technically have +x(+x-y). We take out a negative and we change it to -x(-x+y). We rearrange the parenthesis stuff and we get -x(y-x), which is the same as the bottom, so we can cancel. -x(y-x)/(y-x). We cancel the same things and get left with -x, which is A.

3. First you need to know that (x^A)/(x^B)=x^(A-B). ONLY exponents with the same bases can do this. (x^A)/(y^B) would not give you anything. Now that you know this, we can simplify the x^(1-a) to (x/x^a). We know that if there's a fraction on a fraction, we move the bottom of the top to the bottom. We will then get (x)/((x^a)(x^a). I think I've said this, but an exponent with the same base multiplying each other will give an addition part. For example, (x^2)(x^3)=x^(2+3)=x^5. So we have (x^a)(x^a) which will give x^(a+a)=x^2a. Now our equation is x/(x^2a). We now change it back to one of the answers provided since we know dividing exponents with same bases is the same as subtraction. We make it (x^(1-2a)) which is C.

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Okonkwo
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Re: Mt.SAC MATH HELP

Post by Okonkwo » Tue Jan 27, 2015 6:36 pm

lol deadman added two more problems (middle ones) to make me look like i didn't even know how to attempt them :cry:

Dead Man
Posts: 458
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Re: Mt.SAC MATH HELP

Post by Dead Man » Tue Jan 27, 2015 8:53 pm

Okonkwo wrote:lol deadman added two more problems (middle ones) to make me look like i didn't even know how to attempt them :cry:
LOL sorry. Its because I am doing a review test and those still hadn't came up. Really appreciate the help though that parabola thing was really helpful.
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