ग्राफ \(y=x^3+27\) (x)-अक्ष को किस बिंदु पर काटता है?
At which point does the graph \(y=x^3+27\) cut the (x)-axis?
Explanation opens after your attempt
A. ((-3,0))
Concept
For the (x)-axis, \(x^3+27=0\), so (x=-3). In exams, find the real root of a cubic carefully.
Why this answer is correct
The correct answer is A. ((-3,0)). For the (x)-axis, \(x^3+27=0\), so (x=-3). In exams, find the real root of a cubic carefully.
Exam Tip
(x)-अक्ष के लिए \(x^3+27=0\) इसलिए (x=-3)। परीक्षा में घन समीकरण का वास्तविक मूल ध्यान से निकालें।
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