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Easymathhelp Easy Math Help Easy Math Help It 1 Easy Math Help Simple Math-- Need help =/

Easymathhelp Easy Math Help Easy Math Help It 1 Easy Math Help

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May6-08, 03:34 PM                  #1

Posts: 39
Simple Math-- Need help =/
[b]1. The problem statement, all variables and given/known data[/b
I am confused on how this equation:
2h + (16/h-8 +2) (-8/(h-8)^2
is simplified to this equation:
2h - 128 (h-8)^3 - 16/(h-8)^2
2. Relevant equations

Simplifying skills.

3. The attempt at a solution

Well, I know that they got the number 128 from multiplying 16 and -8 together and how they got -16 by multiplying 2 and -8 together. Now, what I don't get is getting (h-8)^3 from (h-8)... SOMEONE PLEASE HELP?
 
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May6-08, 04:15 PM                  #2

Posts: 751
If you have (A + B) * -C

and you want to expand it out you do -A*C - B*C ....do you see?

in your question A = 16/(h-8)

also if you have (1/y)(2/y^2)

you'll get 2/y^3

the top multiplied together and the bottom multiplied together... do you see?
 
May6-08, 04:18 PM                  #3

Posts: 39
do you mean that 2h - 128/(h-8)^3 - 16/(h-8)^2? if you meant that, then the (h-8)^3 comes from multiplying (h-8) times ( h-8)^2, what you do is you add the exponents.

It is like x times x, you add the exponents if the bases are the same, it becomes x^2, 5 times 5 is 5^2.
 
May6-08, 04:24 PM                  #4

Posts: 39
I'll be right back. I'm at school right now and bell is going to ring soon. Thanks for the replies, I'll look into it once I get home.
 
May11-08, 01:03 AM                  #5

Posts: 19
[tex]2h+( \frac{}{} + 2 ) ( \frac{}{} ) [/tex]

let [tex]x = \frac{}{} [/tex], then:

[tex]2h+( \frac{}{} + 2 ) x [/tex]

[tex]2h+ \frac{}{} \times x + 2 \times x [/tex]

sub [tex]x = \frac{}{} [/tex] back in:

[tex]2h+ \frac{}{} \times \frac{}{} + 2 \times \frac{}{} [/tex]

[tex]2h+ \frac{}{} + \frac{}{} [/tex]

[tex]2h+ \frac{}{} + \frac{}{} [/tex]

Does that clear it up?
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