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dcjd
@dcjd
January 2023
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Знайдіть область визначення
√16-х^2
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robertoborness
Ответ:
16-x²≥0
(4-х)(4+х)≥0
х∈[-4;4]
1 votes
Thanks 1
dcjd
спасибо)
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Answers & Comments
Ответ:
16-x²≥0
(4-х)(4+х)≥0
х∈[-4;4]