Question
Add the following:
(i) 5a2b2+10ab−12a2+15b−20 and 34ab+24a2+22b+24
(ii) abc+6bcd+abd−9ab+4bc−9a−43 and 4acd+10ac−54b−32c
(iii) 5bc+9ad−3c−12 and 24−5c−5b+2ab [4 MARKS]
Answer: Option A
:
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Concept: 1 Mark
Application: 1 Mark each
(i) 5a2b2+10ab−12a2+15b−20+34ab+24a2+22b+24
=5a2b2+44ab+12a2+37b+4
(ii) abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c
=abc+6bcd+abd−9ab+4bc−9a−43+4acd+10ac−54b−32c
(iii) 5bc+9ad−3c−12+24−5c−5b+2ab
=5bc+9ad−8c+12−5b+2ab
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