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Simplify this :

Mathematics
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\[\frac{ [1-(1+3h+3h^2+3h^3)]-(2) }{ h }\]
precalculus? or no?
yes it is

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Other answers:

you should do 1 - (1 + 3h + 3h^2 + 3h^3) first
ohh okay
do i just get rid of the first round bracket?
correct. Distribute the - through [1 -1*( everything inside parenthesis) ]
oh so you mutiply everything by -1 ?:S
sorry had to step away..
it's alright :P
correct: \[\frac{ [1-(1+3h+3h^2+3h^3)]-(2) }{ h } = \frac{ [1-1-3h-3h^2-3h^3]-2 }{ h }\]
but I'm curious: did you originally have to do 1-(1+h)^3 in the top?
well i have this equation: \[y=1-x^3\] and i need to find the average slope of the line passig through A (-1,2) and Δx=h
so that was what i came up with
I think before you go any further, there is something that need to be corrected.
ohh what ?
Here x = -1, so you want \[f(-1+h) = 1-(-1+h)^3 = 1-[(-1)^3 + 3(-1)^2h + 3(-1)h^2 + h^3\] \[ = 1-(-1+3h-3h^2+h^3)\] So the top would be \[\frac{ [1-(-1+3h-3h^2+h^3)]-(2) }{ h } = \frac{ [1+1-3h+3h^2-h^3]-2 }{ h }\]
then the -2 would cancel out with 2
and we would factor out the h ?
you got it.
so m= \[3+3h-h^2\]
yep.
oh wrong signs
hah. right. those pesky negatives...
but my text book says the final answer is: \[-h2+3h-3\] :S :S
just rearrange the terms in decending order in h
ohh it's the same thing .. :$ LOL i thought it was different haha :$
thank-you so much :D !
np!
if part b says to calculate slope for h= 10,2,1,1/2 do i find the limits for his @cruffo ?
Sorry for the delay.
Should just be able to plug in the numbers for h in you previous answer, \[\large -h^2 + 3h - 3\]
thanks :D

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