Question

Find the sum of the following horizontally. a) 6m^2n + 3mn – 2n^2, n^2 – mn – nm^2, mn – 3n^2 – 4m^2n​

Answer 1

Answer:

this is the correct answer of this question

Step-by-step explanation:

The three terms are

(6m²n + 4mn - 2n² + 5),

(n² - nm² + 3) and

(mn - 3n² - 2m²n - 4)

So, the required sum be

= (6m²n + 4mn - 2n² + 5)

+ (n² - nm² + 3)

+ (mn - 3n² - 2m²n - 4)

= (6 - 2)m²n + (4 + 1)mn

+ (- 2 + 1 - 3)n² - nm²

+ (5 + 3 - 4)

= 4m²n + 5mn - 4n² - nm² + 4