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stealth61 [152]
3 years ago
6

What is the product of the expressions? Assume y ≠ 0. (2y-1/y2)(3y2/7)

Mathematics
1 answer:
d1i1m1o1n [39]3 years ago
6 0

Answer:

(2y-1/y_2)(3y_2/7)

= 2y*(3y_2/7) - (1/y_2)(3y_2/7)

= 6y*y_2/7 - 3/7 = (6y*y_2 - 3)/7

Note: I'm not sure if the 2y-1 was written correctly. If it were intended as the entire thing being a numerator or as 2y_1, this answer is inaccurate.

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