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Algebra 19 Online
OpenStudy (anonymous):

x^2>36

OpenStudy (anonymous):

check all that apply A.0 B. 6 C. 7 D. -15 F. -6 D. -3

OpenStudy (usukidoll):

\[x^2 > 36\] well I'm not giving direct answers because that's against code of conduct... but all you have to do is to square root both sides \[\sqrt{x^2}>\sqrt{36}\] then take the square root of x^2 and 36

OpenStudy (anonymous):

okay

OpenStudy (anonymous):

okay

OpenStudy (usukidoll):

we need an x... we can't leave it like that... what's the square root of x^2?

OpenStudy (usukidoll):

|dw:1437632210874:dw| I just re-wrote it to exponential form to make it a bit easier to see what is 2 divided by 2 ?

OpenStudy (anonymous):

The definition of \[x^2= \pm \sqrt{x}\]

OpenStudy (anonymous):

so if \[X^2 ={25}\] then \[x=\pm \sqrt{25} = x=\pm5\]

OpenStudy (anonymous):

How wan you solve this now? x^2>36

OpenStudy (anonymous):

so far your work is corret

OpenStudy (anonymous):

keep it going and you will get your answer

OpenStudy (usukidoll):

:)

OpenStudy (anonymous):

lol

OpenStudy (anonymous):

did she or he find the answer

OpenStudy (usukidoll):

no not yet. the user is offline at the moment.

OpenStudy (anonymous):

oh

OpenStudy (anonymous):

i'll wait

OpenStudy (anonymous):

isn't the answer c &d ..... since the absolute value has to be bigger than 6

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