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avanturin [10]
4 years ago
15

Expand and evaluate the series n=5 ai=2×i^2+3

Mathematics
1 answer:
Ksju [112]4 years ago
5 0
Hello,

a_{1}=2*1^2+3=2*1+3=2+3=5\\\\
 a_{2}=2*2^2+3=2*4+3=8+3=11\\\\
 a_{3}=2*3^2+3=2*9+3=18+3=21\\\\




see file jointed
==========
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5(-3x - 2) - (x - 3) = -4 (4x +5) + 13
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Start off by distributing the numbers into the parentheses:

5(-3x - 2) - (x - 3) = -4(4x + 5) + 13
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(Note: It's super important to be careful when opening up negative parentheses! -(x-3) is not just - x - 3, it is actually -x + 3 since the negative is distributed in every number!)

-15x - 10 - x + 3 = -16x - 20 + 13
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3 years ago
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Step-by-step explanation:

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