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OpenStudy (johnweldon1993):
We can factor a 'z' out of this since both terms have at least 1 'z'
\[\large z(z^2 - 4) = 0\]
Now we can see there are 3 different solutions to this equation
OpenStudy (candy13106):
ok
OpenStudy (johnweldon1993):
Since we have 'z' being multiplied to whatever is in those parenthesis, we know that 'z' = 0 is actually a solution to this...since 0 times anything = 0
Now we just focus on the parenthesis
\[\large z^2 - 4 = 0\]
What 2 values of 'z' make this equation = 0?
OpenStudy (candy13106):
ok
OpenStudy (candy13106):
@Holly00d1248
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OpenStudy (holly00d1248):
is it z*3 or z to the 3rd power
OpenStudy (candy13106):
z to the 3rd power
OpenStudy (candy13106):
can u hurry up soory i just have to go skate in 5 min
OpenStudy (holly00d1248):
Z=4
OpenStudy (candy13106):
ok thanks
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OpenStudy (candy13106):
my teacher doesnt even care if i do them wrong
OpenStudy (holly00d1248):
why not
OpenStudy (candy13106):
but she just checks if we do them
OpenStudy (candy13106):
i think its becuase we r the honors class
OpenStudy (candy13106):
ok i gtg ttyl bye
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OpenStudy (candy13106):
thanks
OpenStudy (holly00d1248):
Its not that she don't care it that you tried is what she cares about