Shido88
h(3)
h(x)= 5|x|/x
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Shido88
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@Aylin
Shido88
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i plug this one in calculator and got 5 as final answer
Aylin
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That looks right.
Shido88
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alright cool :D
Shido88
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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 ?
Shido88
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umm why over 5 ?
isn't it over -2/3 ?
Aylin
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@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\[h(\frac{-2}{3})=5\frac{|\frac{ - 2}{3}|}{\frac{-2}{3}}=5\frac{\frac{2}{3}}{\frac{-2}{3}}\]
Aylin
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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.
Shido88
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so basictly anything negative inside the absolute value will turn into positive ?
Aylin
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Yep!
Shido88
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and once u turn the negative into positive then the absolute value sign goes away ?
Aylin
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In this case, yes.
Shido88
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what about h(3a) ?
Aylin
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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.
Shido88
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hmm ? what ? lol
Aylin
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If \[a \ge 0\]then\[h(3a)=5\]and if\[a < 0\]then\[h(3a)=-5\]
Shido88
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i see
Shido88
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what do i have to do to know if a>0 ?
Shido88
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how did you get h(3a)= 5 ?
do you just cancel out the 3a ?
Aylin
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Yeah.\[5\frac{3a}{3a}=5\]
Shido88
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oh i see
Shido88
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so if it was h(a-2)
would the final answer be 5 anyway ?
Aylin
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Either 5 or -5, depending on a.
Shido88
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ok!
Shido88
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gtg now, thanks again for helping me Aylin :)