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OpenStudy (anonymous):
Help. Help. Help
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OpenStudy (anonymous):
:O
OpenStudy (anonymous):
Can you be obsessed with my question? lmao
OpenStudy (anonymous):
Mhm
OpenStudy (anonymous):
Find the zeros of each function. State the multiplicity of multiple zeros. 53) (x+2)^2(x-5)^4
OpenStudy (anonymous):
what do you need to plug into x to get zero?
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OpenStudy (anonymous):
it's given away in the parenthesis
OpenStudy (anonymous):
-2?
OpenStudy (anonymous):
that's one of them :)
OpenStudy (anonymous):
5?
OpenStudy (anonymous):
so what does multiplicity mean?
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OpenStudy (anonymous):
sorry, I had to look it up :) it's how many zero's there are
OpenStudy (anonymous):
Ok so (-2+2)^2(5-5)^4
OpenStudy (anonymous):
no, sorry :)
OpenStudy (anonymous):
when it's (x+2)^2 it's really (x+2)*(x+2) and (x-5)^4 is really (x-5)*(x-5)*(x-5)*(x-5)
OpenStudy (anonymous):
so there are 6 places that there can be a zero :)
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OpenStudy (anonymous):
does that help?
OpenStudy (anonymous):
omg
yes
OpenStudy (anonymous):
Good :) take care
OpenStudy (anonymous):
waaaaait
OpenStudy (anonymous):
Let me verify
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OpenStudy (anonymous):
ok go
OpenStudy (anonymous):
51) f(t)=4t+3t^3+2t-7
4t + 3t(3t)(3t) +2t-7 5 possible zeros?
OpenStudy (anonymous):
I'm sorry friend :? I gotta run.. Just do the same thing you did in the last one!! look for what will make it zero!! Good luck :D
OpenStudy (anonymous):
Okay ty take care
OpenStudy (anonymous):
you too
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