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14-10-2021
Mathematics

contestada

factor by grouping jk^2+3jk+9k+27

Respuesta :

hf0028321 hf0028321

16-10-2021

Answer:

(jk+9)(k+3)

Step-by-step explanation:

jk(k+3)+9(k+3)=(jk+9)(k+3)

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