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OpenStudy (sbuck98):
OpenStudy (anonymous):
30-60-90 triangles are always in the format|dw:1456441777453:dw|
OpenStudy (anonymous):
So based on this what do you think x equals?
OpenStudy (sbuck98):
15
OpenStudy (sbuck98):
@Brill
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OpenStudy (anonymous):
No
The side it gives you in your problem is the one across from the 60 degree angle
In the diagram I gave you the side across from the 60 degrees is sqrt3 times x
Or in other words\[x \sqrt{3}=15\]
OpenStudy (anonymous):
How could you solve for x in the situation?
OpenStudy (sbuck98):
|dw:1456449861328:dw| @Brill
OpenStudy (anonymous):
Are you sure?
Wouldn't you isolate x by DIVIDING sqrt3?
OpenStudy (sbuck98):
I didn't think you did.
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OpenStudy (sbuck98):
@Brill
OpenStudy (sbuck98):
Help me with 3 more?
OpenStudy (anonymous):
Wait
I'd be glad to but what was your answer for the previous one because the drawing is incorrect
OpenStudy (sbuck98):
D. @Brill
OpenStudy (anonymous):
Uhh D is incorrect
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OpenStudy (sbuck98):
So B is my only other option....... @Brill
OpenStudy (anonymous):
No it isn't
OpenStudy (anonymous):
Remember when you divide 15 by sqrt3 you have to rationalize
OpenStudy (sbuck98):
I told you 5 sqrt 3 and you said no.....
OpenStudy (sbuck98):
@Brill
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OpenStudy (anonymous):
OH MY BAD I AM SO VERY SORRY
I FORGOT TO SIMPLIFY 15sqrt3/3
I AM SO SO VERY SORRY