कितने पूर्णांक (x) असमानता \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\) को संतुष्ट करते हैं?
How many integers (x) satisfy \(\frac{2x-3}{5}<x-1\le\frac{x+8}{2}\)?
Explanation opens after your attempt
D. अनंतinfinitely many
Concept
The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).
Why this answer is correct
The correct answer is D. अनंत / infinitely many. The first part gives \(x>-\frac{2}{3}\), and the second gives \(x\le10\). Integer solutions are (0) through (10), totaling (11).
Exam Tip
पहले भाग से \(x>-\frac{2}{3}\) और दूसरे से \(x\le10\) मिलता है। पूर्णांक हल (0) से (10) तक हैं, कुल (11)।
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