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mindreader67
Group Title
(6) with exponent of 12 times (6) with exponent of 5 times (6) with exponent of 2?
 one year ago
 one year ago
mindreader67 Group Title
(6) with exponent of 12 times (6) with exponent of 5 times (6) with exponent of 2?
 one year ago
 one year ago

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jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
Hi, welcome to openstudy:) When you multiply a power with the same base, you just add the exponenets together. In this case: \[6^{12} \times 6^5 \times 6^2\] you would add 12 by 5 by 2: \[(6)^{12+5+2}\] Can you do the rest from here?
 one year ago

jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
Can you add the exponents?
 one year ago

mindreader67 Group TitleBest ResponseYou've already chosen the best response.0
yes but the 6's are all in perentheses does that mean anything ?
 one year ago

jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
Not really. You would get \[(6)^{19}\] You see, it's okay if it's a negative, because all three of them are negative. They have to be exactly the same for you to be able to add the exponents like that :) Does that help?
 one year ago

mindreader67 Group TitleBest ResponseYou've already chosen the best response.0
okay and if they weren't exactly the same what would u do? like say one of the 6's is a 4
 one year ago

jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
The parentheses just indicate that the negative is part of the 6, so that it isn't confused with a subtraction sign. If one of the 6's was a 4, lets say \[(−6)^{12} × (−6)^{5} × (4)^2\] Then it would be \[(−6)^{12+5} × (4)^2\] or just \[(−6)^{17} × (4)^2\] To simplify further, you add the powers together and multiply the bases, so \[(−6)^{17} × (4)^2 = 24^{19} \]
 one year ago

jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
Does that make sense?
 one year ago

jollysailorbold Group TitleBest ResponseYou've already chosen the best response.0
Yes? No? I hope this helped :)
 one year ago

mindreader67 Group TitleBest ResponseYou've already chosen the best response.0
ohhhh okay
 one year ago
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