Предмет: Алгебра, автор: vik15061979

Найдите область определения функции y=x-3÷3x+12

Ответы

Автор ответа: arina8024

4

Ответ:

y=x-3/3x+12

x-3/3x+12

x-3

3x+12

Область определения рациональной функции - все значения х, для которых знаменатель не равен 0.

Автор ответа: oleksandra1573

3

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