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Author Topic: Some Math help!  (Read 2585 times)

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Offline Lady Hitura CaprionTopic starter

Some Math help!
« on: October 10, 2009, 01:44:47 PM »
I suck at algebra so far, pure and simple. Our teacher doesn't teach it right. So! Can someone check over these two equations and tell me if I'm doing quadratic functions right please?
 Yes i did them in paint, they suck I know.

Offline Oniya

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Re: Some Math help!
« Reply #1 on: October 10, 2009, 01:51:39 PM »
That is the correct application, and your simplification (i.e., the actual math parts of it) looks correct as well.

(And for being done in Paint, you did a pretty clean job as well.  :D)

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #2 on: October 10, 2009, 07:33:31 PM »

Offline Oniya

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Re: Some Math help!
« Reply #3 on: October 10, 2009, 07:54:40 PM »
Looks right to me.  Another way to do it is to let Y=sqrt(x), which gives you y^2-5y-24 as your starting equation.  You just have to substitute back after you're done.
« Last Edit: October 10, 2009, 07:56:34 PM by Oniya »

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #4 on: October 11, 2009, 11:24:04 AM »
Need help figuring out how to set up a problem since my teacher NEVER went over this with us and the book has no examples of it.

A boat travels 10 miles upstream and 10 miles downstream. The total time for both parts of the trip is 5 1/3 hours. The speed of the stream is 1 mph. What is the speed of the boat in still water?

Offline Oniya

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Re: Some Math help!
« Reply #5 on: October 11, 2009, 11:41:19 AM »
Have you covered systems of equations yet? (solving for two variables using two equations)

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #6 on: October 11, 2009, 11:45:31 AM »
Yes I believe we have.

Offline Oniya

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Re: Some Math help!
« Reply #7 on: October 11, 2009, 12:40:27 PM »
Actually, I think I was thinking too hard.  Distance = rate * time.  Rate = distance/time.  Time = Distance/rate.

16/3 (or 5 1/3) = 10/rate upstream + 10/rate downstream.

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #8 on: October 11, 2009, 10:01:39 PM »
Okay new one

How to do

(x +2)^3

I know it goes to

(x+2)(x+2)(x+2) but i dont know if i do two then the next or all three at once or what @.@

Offline Oniya

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Re: Some Math help!
« Reply #9 on: October 11, 2009, 10:20:28 PM »
Multiplication is associative.  (AB)C = A(BC) = (AC)B

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #10 on: October 24, 2009, 03:50:01 PM »
Factoring

x^5 - 3x^3 -8x^2 + 24 > 0

i know its factor by grouping buuut... I get as far as

x^3(x^2-3) -8(x^2+3) before I get lost...

Offline Oniya

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Re: Some Math help!
« Reply #11 on: October 24, 2009, 05:47:55 PM »
Remember that it takes two negatives to get a positive.

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #12 on: October 24, 2009, 05:49:58 PM »
xD no at that point i dont know what to do at ALL to further factor it!

Offline Oniya

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Re: Some Math help!
« Reply #13 on: October 24, 2009, 06:01:43 PM »
You have a slight mistake in your second grouping.  -8(3) = -24, not 24.

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #14 on: October 24, 2009, 06:03:39 PM »
ooh... okay thanks :D

Offline Voichin

Re: Some Math help!
« Reply #15 on: October 30, 2009, 10:40:40 AM »


Oh my God, where does your maths teacher, no all your maths teachers since junior high, plus those in Chemistry, Physics and Computer science come from? Adeptus Inumeratus?

(x-5\sqrt(x)-24)^2=x^2-25x-24^2-2x5\sqrt(x)-2x24+2*5\sqrt(x)24. The rest you can do yourself, I hope.

You can't simply put square on each term of the sum - (1+1+1)^2=9 but 1^2+1^2+1^2=3! The general formula is as follows:

(a_1+a_2+....a_n)^2= a_1^2+a_2^2+....a_n^2+2a_1a_2+2a_1a_3+...2a_{n-1}a_n

You take all squares and add to them twice the product of every two terms!

« Last Edit: October 30, 2009, 04:07:41 PM by Voichin »

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #16 on: October 30, 2009, 12:22:26 PM »
typed out that equation is too confusing and hard to follow. I was following my book not my teacher in that anyway, I was absent that day o-o

Offline Voichin

Re: Some Math help!
« Reply #17 on: October 30, 2009, 01:43:28 PM »
Hmmm, now tell me what is confusing you?

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #18 on: October 30, 2009, 01:48:19 PM »
Its just hard to read in text assembled like that @.@

Offline Voichin

Re: Some Math help!
« Reply #19 on: October 30, 2009, 03:44:21 PM »
Put it in a tex editor and enjoy the magic :)

http://thornahawk.unitedti.org/equationeditor/equationeditor.php
« Last Edit: October 30, 2009, 04:08:24 PM by Voichin »

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #20 on: October 31, 2009, 10:06:24 PM »
1000 < 25t / 3t^2 + 15

"The population of a certain organism, in thousands, is given by the function P(t) = 25t / 3t^2 + 15, where t is time, in days. Find the interval on which the population is greater than 1000.

I know to set it up as:

1000 < 25t / 3t^2 + 15

then

0 < 25t / 3t^2 + 15 - 1000 aaaand then i blank

Offline Oniya

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Re: Some Math help!
« Reply #21 on: October 31, 2009, 11:08:40 PM »
Is the 15 in the denominator, or only the 3t^2?

Offline Lady Hitura CaprionTopic starter

Re: Some Math help!
« Reply #22 on: October 31, 2009, 11:33:23 PM »
1000 < (25t/3t^2 + 15)

Offline Vekseid

Re: Some Math help!
« Reply #23 on: November 01, 2009, 12:20:08 AM »
The equation gives the population in thousands - not individuals. Trying to solve it for 1000 is nonsense because

25t/3t^2 = 8.33333/t - the population is steadily reducing from an infinity at zero. It never rises.

What you are looking for is to solve t in 0 = 25t/3t^2 + 15 - 1

Offline Oniya

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Re: Some Math help!
« Reply #24 on: November 01, 2009, 12:24:08 AM »
It still makes a difference whether the initial equation is 25t / (3t^2 + 15)
or 25t / (3t^2) + 15