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6m^2n + 4mn – 2n^2 + 5, n^2 – nm^2 + 3, mn – 3n^2 – 2m^2n – 4
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Answered by
4
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
#
Answered by
3
Hiii. .....friend
The answer is here,
Addition :-
:-)Hope it helps u.
The answer is here,
Addition :-
:-)Hope it helps u.
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