let's see who answers first and is brainliest!!!
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4a‐¹b‐²c³ × 6a³b²c
12a²c⁴b
= c³× 2× a³b²c
a²c⁴b × a¹b²
= 2a³b²c³+¹
a²+¹ c⁴ b¹+²
= 2a³b²c⁴
a³c⁴b³
= 2a³-³ b²-³c⁴-⁴
= 2a⁰b-¹c⁰
= 2× 1 × b-1¹ × 1 (a⁰ = 1)
= 2 × b-¹
= 2b-¹
Option (d) is the answer
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