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Mathematics 17 Online
OpenStudy (anonymous):

sqrt(x) (x-17) multiply and find the domain

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):

\[\sqrt{x}(x-17)\]

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

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

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

OpenStudy (anonymous):

sry, 17

OpenStudy (anonymous):

not -17

OpenStudy (anonymous):

right

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