Mathematics
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OpenStudy (anonymous):
sqrt(x) (x-17)
multiply and find the domain
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OpenStudy (zzr0ck3r):
domain is all numbers that it can be, the only things it cant be is a sqrt of a negative number so domain is all x such that x>=0
OpenStudy (zzr0ck3r):
sqrt(x)*x-17sqrt(X)
OpenStudy (anonymous):
is it: \[\sqrt{x}(x-17)\]
or:\[\sqrt{x(x-17)}\]
OpenStudy (anonymous):
the first one
OpenStudy (anonymous):
\[\sqrt{x}(x-17)\]
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OpenStudy (anonymous):
so then what @zzr0ck3r said
OpenStudy (anonymous):
i dont understand the form of that answer can you put it in the equation format
OpenStudy (zzr0ck3r):
A(B-C) = AB-AC
OpenStudy (anonymous):
x>0
OpenStudy (anonymous):
no, the answer to the multiplication problem
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OpenStudy (anonymous):
\[x \sqrt{x}-17\sqrt{x}\]
OpenStudy (zzr0ck3r):
\[\sqrt{x}*x - \sqrt{x}*17\]
OpenStudy (anonymous):
k thanks
OpenStudy (anonymous):
and also your answer about x>0 is wrong it has to be \[x \ge0\]
OpenStudy (anonymous):
ye, my bad
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OpenStudy (anonymous):
what about \[\sqrt{x}\div x-17\]
OpenStudy (anonymous):
can that be simplified or do you think i should just write it out like that
OpenStudy (anonymous):
is it::
\[\sqrt{x} \over x-17\]
OpenStudy (anonymous):
yes
OpenStudy (anonymous):
so denominator can't be 0. That's exclude x=-17. And sqrt excludes the x<0.
so x>=0 and x not equal -17
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OpenStudy (anonymous):
sry, 17
OpenStudy (anonymous):
not -17
OpenStudy (anonymous):
right