72a^3b^4÷(-8a^2b^2)
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3
Answer:
.
Find the sum of first 16 terms of an Arithmetic Progression whose
4th and 9th terms are - 15 and - 30 respectively.
Answered by
2
Step-by-step explanation:
Simplify ———————————
23a2b4
Dividing exponential expressions :
3.1 a3 divided by a2 = a(3 - 2) = a1 = a
Dividing exponential expressions :
3.2 b3 divided by b4 = b(3 - 4) = b(-1) = 1/b1 = 1/b
Canceling Out:
3.3 Canceling out 23 as it appears on both sides of the fraction line
Final result :
9a
——
b
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