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Mathematics 7 Online
OpenStudy (anonymous):

What would be the answer to this equation? Im going to write it as an answer to the question so I can use the equation form.

OpenStudy (anonymous):

\[(i \sqrt{-363})(\sqrt{-243})\]

OpenStudy (phi):

find the prime factors of 363 and 243, and look for perfect squares. Also, factor i from each root, to get i^3 out front. i^3= -i

OpenStudy (anonymous):

i got \[-99i \sqrt{3}\]

OpenStudy (amistre64):

yeah, remember that you need to factor the "i" out first :) i remember half the time lol

OpenStudy (anonymous):

from factoring 11 and 9 out of the two square roots and leaving 3 under the sign.

OpenStudy (phi):

I think you should get two sqrt(3) factors

OpenStudy (anonymous):

after the first factor i get: \[(11i ^{2}\sqrt{3})(9i \sqrt{3})\]

OpenStudy (phi):

Yep. and sqrt(3)*sqrt(3)= 3

OpenStudy (anonymous):

ohh...duhh me! omg, thanks! :)

OpenStudy (amistre64):

363 243 ------- 1089 1452 726 --------- 88209 ------------------------- 2 9 7 --------- / 8 82 09. 4 ---- 4 82 ; 4n * n < 482, n=9 : 49*9 = 441 4 41 ------ 41 09 ; 58n*n < 3109 , n=7 : 587*7 = 4109 41 09 -------- 0 ................................ i^3 * 297

OpenStudy (amistre64):

its funner on an abacus :)

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