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h(3) h(x)= 5|x|/x

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i plug this one in calculator and got 5 as final answer
That looks right.

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Other answers:

alright cool :D
for the next one it's asking for h(-2/3) but you can't use a negative for absolute value right? or is it ?
umm why over 5 ? isn't it over -2/3 ?
@Shido88 Sorry, I got distracted. Back now. Absolute value as a function either leaves the thing inside of it alone if it is 0 or positive, or multiplies it b negative 1 if it is negative. So |x|=x IF x is greater than or equal to 0, and |x| = -x IF x is less than 0. So for h(-2/3) you would have\[h(\frac{-2}{3})=5\frac{|\frac{ - 2}{3}|}{\frac{-2}{3}}=5\frac{\frac{2}{3}}{\frac{-2}{3}}\]
Hmm, looks like the negative signs didn't display properly... Well, the 2s that are shifted way over to the right are supposed to have negative signs in front of them.
so basictly anything negative inside the absolute value will turn into positive ?
and once u turn the negative into positive then the absolute value sign goes away ?
In this case, yes.
what about h(3a) ?
Ahh, well this depends on if a is positive or not. :P Find the answer for if a is positive, and then the answer for if a is negative. Then you say which one it is based on a.
hmm ? what ? lol
If \[a \ge 0\]then\[h(3a)=5\]and if\[a < 0\]then\[h(3a)=-5\]
i see
what do i have to do to know if a>0 ?
how did you get h(3a)= 5 ? do you just cancel out the 3a ?
oh i see
so if it was h(a-2) would the final answer be 5 anyway ?
Either 5 or -5, depending on a.
gtg now, thanks again for helping me Aylin :)

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