Oliver

10-26-2013, 04:15 AM

The operation @ is defined as a@b=(a+b)^2-a. Which one of the following is the result of a@a@a?

(A) 4a^2-a

(B) 16a^2-a

(C) 4a^4-a

(D) 16a^4-a

Thanks!:)

(A) 4a^2-a

(B) 16a^2-a

(C) 4a^4-a

(D) 16a^4-a

Thanks!:)

View Full Version : Newly defined symbol question

Oliver

10-26-2013, 04:15 AM

The operation @ is defined as a@b=(a+b)^2-a. Which one of the following is the result of a@a@a?

(A) 4a^2-a

(B) 16a^2-a

(C) 4a^4-a

(D) 16a^4-a

Thanks!:)

(A) 4a^2-a

(B) 16a^2-a

(C) 4a^4-a

(D) 16a^4-a

Thanks!:)

nicole

10-26-2013, 04:16 AM

a@a@a= a@(a@a)= a@ ((a+a)^2-a) = a@ (4a^2-a)= (a+4a^2-a)^2-a

= (4a^2)^2-a=16a^4-a

Therefore the correct answer is (D).

Hope these help!!:)

= (4a^2)^2-a=16a^4-a

Therefore the correct answer is (D).

Hope these help!!:)

Oliver

10-26-2013, 04:16 AM

Thank you very much for you quickly answer!!:)

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