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sasogeek
 2 years ago
Best ResponseYou've already chosen the best response.3ok umm if you have \( (ab)^2\) it is the same as \( (a^22ab+b^2) \) so we simply have to apply this here and then simplify... \[ (2t1)^22(2t1)1 \]\[=( \ (2t)^22(2t)+1\ \ )(4t2)1\]\[ =4t^24t+14t+21\]\[=4t^28t+2\] hence \(f(2t1)=4t^28t+2 \)

nanda082
 2 years ago
Best ResponseYou've already chosen the best response.0the answer is simple based on (a+b)^2=a^2+b^2+2ab; so the answer would be (2t11I^2=4t^2+14t and then 2(2t1)=4t2 so over all goes like this 4t^2+14t(4t2)1=4t^2+14t4t+21=4t^28t+2
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