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27 Nov 2019, 04:31
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A
Be sure to select an answer first to save it in the Error Log before revealing the correct answer (OA)!
Difficulty:
55%
(hard)
Question Stats:
66% (02:37) correct
34%
(01:58)
wrong
based on 198
sessions
History
Date
Time
Result
Not Attempted Yet
The expansion of \((203x^2 - 199x - 3)(72y^2 + 27y + 1)\) is a1x^2y^2 + a2x^2y + a3x^2 +…..+ a7y^2 + a8y + a9. What is a1 + a2 + a3 +…..+ a7 + a8 + a9?
A. \(100\)
B. \(122\)
C. \(125\)
D. \(200\)
E. \(228\)
A. \(100\)
B. \(122\)
C. \(125\)
D. \(200\)
E. \(228\)
27 Nov 2019, 13:09
Kudos
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\((203x^2 - 199x - 3)(72y^2 + 27y + 1)\) = a1x^2y^2 + a2x^2y + a3x^2 +…..+ a7y^2 + a8y + a9
Put x=1 and y=1
(203-199-3)(72+27+1)=\( a1 + a2 + a3 +…..+ a7 + a8 + a9\)
1*100= \(a1 + a2 + a3 +…..+ a7 + a8 + a9\)
100= \(a1 + a2 + a3 +…..+ a7 + a8 + a9\)
Put x=1 and y=1
(203-199-3)(72+27+1)=\( a1 + a2 + a3 +…..+ a7 + a8 + a9\)
1*100= \(a1 + a2 + a3 +…..+ a7 + a8 + a9\)
100= \(a1 + a2 + a3 +…..+ a7 + a8 + a9\)
MathRevolution
29 Nov 2019, 01:40
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=>
When we substitute x and y with \(1’s\),
we have a1 + a2 + a3 +…..+ a7 + a8 + a9
= a1\((1)^2(1)^2\) + a2\((1)^2(1)\) + a3\((1)^2\) +…..+ a7\((1)^2\) + a8\((1)\) + a9.
\(= (203(1)^2 - 199(1) - 3)(72(1)^2 + 27(1) + 1)\)
\(= 1*100\)
\(= 100\)
Therefore, A is the answer.
Answer: A
When we substitute x and y with \(1’s\),
we have a1 + a2 + a3 +…..+ a7 + a8 + a9
= a1\((1)^2(1)^2\) + a2\((1)^2(1)\) + a3\((1)^2\) +…..+ a7\((1)^2\) + a8\((1)\) + a9.
\(= (203(1)^2 - 199(1) - 3)(72(1)^2 + 27(1) + 1)\)
\(= 1*100\)
\(= 100\)
Therefore, A is the answer.
Answer: A
