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Old 09-10-2008, 11:36 PM   #1
The Fonz
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It's been a while since I was in school and it's showing, the simplest problems are throwing me off. If anyone could go thru these questions and show the step by step process I'd appreciate it:

1. (.01)^-3/2

2. (9/49)^3/2

3. (x^2 - 5x +12)^1/5 = 2^3/5

4. (x-6)^1/3

5. 16^-1/4
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Old 09-10-2008, 11:41 PM   #2
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Cheater.

If that's basic, I'm even dumber than I thought.
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Old 09-10-2008, 11:43 PM   #3
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I can't even do long division anymore, let alone this hippy "new math" stuff.
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Old 09-10-2008, 11:44 PM   #4
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bedmas

brackets, exponents, divison, multiplication, addition, subtraction.. Just follow order of operations.
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Old 09-10-2008, 11:44 PM   #5
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Put into a calculator? Solve for x in 3 & 4 and calculate value for the others?
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Old 09-10-2008, 11:45 PM   #6
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Yeah I think I could figure it out if I knew what ^ meant

Like burn this city said Bedmas
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Old 09-10-2008, 11:46 PM   #7
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Teacher says a calculator is not necessary
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Old 09-10-2008, 11:46 PM   #8
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Quote:
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Yeah I think I could figure it out if I knew what ^ meant

Like burn this city said Bedmas
Frequently known as "to the power of" 6 squared would read as "6^2=36"

A fraction such as 1/2 would actually mean the square root.
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Old 09-10-2008, 11:46 PM   #9
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Quote:
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Yeah I think I could figure it out if I knew what ^ meant
exponent.

ex)16^-1/4
= 16 to the power of -1/4
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Old 09-11-2008, 12:09 AM   #10
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Quote:
Originally Posted by The Fonz View Post
It's been a while since I was in school and it's showing, the simplest problems are throwing me off. If anyone could go thru these questions and show the step by step process I'd appreciate it:

1. (.01)^-3/2 = 1/((0.01)^3/2) = 1/((sqrt(0.01))^3) = 1/(0.1^3) = 1/0.001 = 1,000

2. (9/49)^3/2 = (sqrt(9/49))^3 = (3/7)^3 = (3^3)/(7^3) = 27/343

3. (x^2 - 5x +12)^1/5 = 2^3/5 so x^2-5x+12 = 2^3 = 8 so x^2-5x+4 = 0 = (x-4)(x-1) therefore x = 4 or 1, either answer should work.

4. (x-6)^1/3 = I don't understand what this is asking...

5. 16^-1/4 = 1/(16^1/4)= 1/2 = 0.5
Here's a start for you...

Last edited by V; 09-11-2008 at 12:16 AM.
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Old 09-11-2008, 12:19 AM   #11
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Here's a start for you...
You're the man!
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Old 09-11-2008, 12:21 AM   #12
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4x4 would disagree...
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Old 09-13-2008, 04:38 PM   #13
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Ok, I've got another problem.

How do you factor (m-n)^2 - 36
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Old 09-13-2008, 04:44 PM   #14
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Ok, I've got another problem.

How do you factor (m-n)^2 - 36
Say (m-n) = x

x^2 - 36 = (x-6)(x+6)

therefore you get ((m-n)-6)((m-n)+6) = (m-n-6)(m-n+6)
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Old 09-13-2008, 05:18 PM   #15
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V to the rescue again!

I feel like an idiot after you answer my questions, they seem so simple once you explain it.
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Old 09-13-2008, 06:31 PM   #16
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It always seems easier when someone else does it for you. This doesn't just apply to mathematical equations.
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Old 09-13-2008, 07:09 PM   #17
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Fonz, whenever you see the numbers 0.01, 4, 9, 16, 25, 36, 49, 64, 81, 100, 121, 144 etc (for ^2 for 0.1, 2, 3, 4, 5, 6, 7, 8, 9) for this kind of this stuff, your first instinct should be find some way you can take the sq root, be it simplifying or difference of squares.
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Old 09-13-2008, 11:24 PM   #18
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Quote:
Originally Posted by The Fonz View Post
V to the rescue again!

I feel like an idiot after you answer my questions, they seem so simple once you explain it.
How do you think I feel?? Not only am I posting on a messageboard, I'm doing math problems at the same time!!

What a nerd...
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Old 09-13-2008, 11:33 PM   #19
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Quote:
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Ok, I've got another problem.

How do you factor (m-n)^2 - 36
That question is a classic Math Jerk question. All it is a simple factor of x^2 -y^2 but they try to trick you by throwing the m-n that they try to get you to expand and if you do that you get lost. Its questions like that why I did engineering instead of math
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Old 09-14-2008, 02:01 PM   #20
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I'm so screwed for university math...
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