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NISA [10]
3 years ago
12

(3a^2–7a^2-3a^4)+(4a^2-a^4-4a)

Mathematics
2 answers:
Blababa [14]3 years ago
6 0
Rearrange the equation

(3a^2–7a^2-3a^4)+(4a^2-a^4-4a)

(-3a^4 +3a^2 - 7a^2) + (-a^4 + 4a^2 - 4a)

-4a^4 - 4a

-4a ( a^3 + 1)
vazorg [7]3 years ago
5 0
There is no need for parenthesis, so you can just combine like terms.
3 a^{2} -7 a^{2} - 3a^{4}  + 4 a^{2} - a^{4} -4a   \\  \\ = -4 a^{4} - 4a  \\  \\ = -4a( a^{3} + 1)
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the answer is

<span>the values ​​of the range are different because the domain in which the inverse function exists are different</span>  
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