Mathematics
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OpenStudy (anonymous):
algebera help @ganeshie8
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OpenStudy (anonymous):
OpenStudy (anonymous):
@CGGURUMANJUNATH
OpenStudy (anonymous):
plss
OpenStudy (cggurumanjunath):
x^3/x^(1/12)
OpenStudy (cggurumanjunath):
x^a/x^b=x^(a-b)
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OpenStudy (cggurumanjunath):
is it clear @plohrr ?
OpenStudy (cggurumanjunath):
x^3/x^(1/12)=x^(3-1/12)
OpenStudy (anonymous):
so x^3/x^(1/12) is the form it was in originally right
OpenStudy (cggurumanjunath):
yes
OpenStudy (anonymous):
ok and the second reply you posted is a formula?
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OpenStudy (cggurumanjunath):
x^(3-1/12)=x^(36/12-1/12)
OpenStudy (cggurumanjunath):
yes
OpenStudy (cggurumanjunath):
3=36/12
OpenStudy (cggurumanjunath):
we are taking lcm .
OpenStudy (cggurumanjunath):
x^(3-1/12)=x^(36/12-1/12)=x^(35/12)
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OpenStudy (anonymous):
ok wait too fast cx
OpenStudy (cggurumanjunath):
x^(35/12) is the final answer .
OpenStudy (cggurumanjunath):
ok.
OpenStudy (anonymous):
so the first step in simplifying is the sub it it in formula
OpenStudy (cggurumanjunath):
yes
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OpenStudy (anonymous):
and what formula is that called.
OpenStudy (cggurumanjunath):
go through the above link.
OpenStudy (cggurumanjunath):
it is the law of exponents.
OpenStudy (anonymous):
mmm i see
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OpenStudy (anonymous):
and the next step is ?
OpenStudy (cggurumanjunath):
x^3/x^(1/12)=x^(3-1/12)
OpenStudy (anonymous):
what did you do there
OpenStudy (anonymous):
simplify ?
OpenStudy (cggurumanjunath):
yes
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OpenStudy (cggurumanjunath):
i have used 3rd basic law of exponent .
OpenStudy (cggurumanjunath):
subtract exponents
OpenStudy (anonymous):
ohh ok
OpenStudy (anonymous):
x^(3-1/12)=x^(36/12-1/12) simplified is 3=36/12?
OpenStudy (cggurumanjunath):
x^3/x^(1/12)=x^(3-1/12)
3 and 1/12 u have to take lcm for subtracting these , u know how to take lcm right ? @plohrr
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OpenStudy (cggurumanjunath):
3-1/12
OpenStudy (cggurumanjunath):
u know how to take lcm right ?
OpenStudy (anonymous):
yep
OpenStudy (cggurumanjunath):
very good.
OpenStudy (cggurumanjunath):
x^(3-1/12)=x^(36/12-1/12)=x^(35/12)
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OpenStudy (cggurumanjunath):
x^(35/12) is the final answer .
OpenStudy (cggurumanjunath):
is it clear @plohrr ?
OpenStudy (anonymous):
yes ! thank you so much
!!
OpenStudy (cggurumanjunath):
:)
OpenStudy (cggurumanjunath):
can u do the second part of the q .
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OpenStudy (cggurumanjunath):
32 nd sqrt (x*x*x^5)
OpenStudy (cggurumanjunath):
use the first basic law of exponent here.
OpenStudy (cggurumanjunath):
(x*x2*x^5) = (x^8)
OpenStudy (anonymous):
not really sure of that one
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OpenStudy (cggurumanjunath):
32 nd sqrt (x*x2*x^5) =32 nd sqrt (x^8)
OpenStudy (cggurumanjunath):
is it clear or not @plohrr ?
OpenStudy (cggurumanjunath):
am i fast ? @plohrr
OpenStudy (anonymous):
yes its clear now
OpenStudy (cggurumanjunath):
32 nd sqrt (x^8) =(x^8)^(1/32)
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OpenStudy (cggurumanjunath):
is it clear @plohrr ?
OpenStudy (cggurumanjunath):
32 nd sqrt (x^8) =(x^8)^(1/32)=(x^(8/32))
OpenStudy (cggurumanjunath):
32 nd sqrt (x^8) =(x^8)^(1/32)=(x^(8/32))=(x^(1/4))
OpenStudy (anonymous):
… so the simplified is x^(1/4))
OpenStudy (cggurumanjunath):
yes .
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OpenStudy (cggurumanjunath):
they ( Megan and Julie ) started with expressions that are not equal
OpenStudy (anonymous):
so for the first: x^(35/12
and the second : (x^(1/4))
OpenStudy (cggurumanjunath):
that is the conclusion we draw from the given equation.
OpenStudy (cggurumanjunath):
yes
OpenStudy (cggurumanjunath):
both are not same right ? @plohrr
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OpenStudy (anonymous):
yes thats true . and for the first we used the 3rd basic law of exponents. and for the second, we used the first?
OpenStudy (cggurumanjunath):
yes .
OpenStudy (anonymous):
:D can you help with some more
OpenStudy (cggurumanjunath):
sure . Always.
OpenStudy (anonymous):
ok hold on a sec
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OpenStudy (cggurumanjunath):
ok.
OpenStudy (cggurumanjunath):
go ahead .