Mathematics
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OpenStudy (anonymous):
I am in need of help, please......
2/7 = z/-3 (solve using the addition principle) ??
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OpenStudy (saifoo.khan):
it can be solved by another method as well.
OpenStudy (saifoo.khan):
\[\frac27 = \frac{z}{-3}\]
OpenStudy (anonymous):
hello
OpenStudy (saifoo.khan):
Cross multiply,
\[z \times 7 = 2 \times -3\]
\[7z = -6\]
z = -6/7
OpenStudy (saifoo.khan):
hi
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OpenStudy (anonymous):
okay that will work as it wants it to be done in the addition principal?
OpenStudy (anonymous):
okay, I can understand that but why in the heck do they confuse you when asking that it be done solving using the addition principle?
OpenStudy (anonymous):
@saifo.khan
OpenStudy (saifoo.khan):
umm, let me see.
OpenStudy (anonymous):
hey I have two more problems can you help me with them?
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OpenStudy (anonymous):
5x - x = x + 3x
OpenStudy (saifoo.khan):
sure.
OpenStudy (saifoo.khan):
5x - x = x + 3x
4x = 4x
OpenStudy (anonymous):
does it go any more to reduce? or just stays 4x=4x
OpenStudy (saifoo.khan):
idk in which form u need the answer.
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OpenStudy (anonymous):
last one is,
Q = p-q/2, for p
OpenStudy (anonymous):
sorry long lag on my end on the computer
OpenStudy (saifoo.khan):
2Q = p -q
p = 2Q +q
OpenStudy (saifoo.khan):
no problem, this website has old tradition of lagging.
OpenStudy (anonymous):
this is a relativity formula from physics?
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OpenStudy (saifoo.khan):
ummm, i dont remember. LOL
OpenStudy (anonymous):
oh okay, thanks
OpenStudy (saifoo.khan):
uw/
OpenStudy (anonymous):
lol, this is how it is written in the book, as follows
OpenStudy (anonymous):
Q = p-q over 2, for p
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OpenStudy (saifoo.khan):
\[Q=p-\frac q2\]
OpenStudy (saifoo.khan):
\[p = Q + \frac q2\]
OpenStudy (anonymous):
so, is that how I would re-write it?
OpenStudy (saifoo.khan):
\[p = Q + \frac q2\]
OpenStudy (anonymous):
thanks so much for your help, hope to see you around again in here....have a good nite!!
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OpenStudy (saifoo.khan):
you welcome!!
Yeah, sure! ^_^
OpenStudy (saifoo.khan):
well it's morning here. LOL
OpenStudy (anonymous):
lol, hey are you still there???