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Solve for d. sqrt--d^2+1 =1-d
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\[\sqrt{d^2-1}=1-d\]
\[is this: \sqrt{d ^{2}+1}=1-d ?\]
yes angela thats the exact problem (:
if it is : \[d ^{2}+1=(1-d)^{2}\] solve it for d
angela has the correct format of the problem
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\[d ^{2}+1=1-2d+d ^{2}\]
d=1
d^2-1=(1-d)^2 d^2-1=1-2d-d^2 d^2-1-1+2d+d^2=0 2d^2+2d-2=0 d^2+d-1=0 D=b^2-4ac=1-4*1*(-1)=5 d1=(-1-sqr5)/2 d2=(sqr5-1)2
1d2=(-b+-sqrD)/2a
\[d ^{2} - canceled out as well as 1\] d=0
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inik is right
angela21073.. per original: should be d^2 +1 (not -)...?
it is 0
good :)
d=0
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sorry didn't get it...where? can u write it?
I got it
:(:(:( All that for nothing :(:(:(:( Sorry Jazmine :(
its okay angela u have helped me alot before too (: its just one mistake
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