असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+y\ge 9\) का हल क्षेत्र कैसा है?

What type of solution region is given by \(x\ge 0\), \(y\ge 0\), and \(x+y\ge 9\)?

Explanation opens after your attempt
Correct Answer

D. असीमित क्षेत्रunbounded region

Step 1

Concept

In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

Step 2

Why this answer is correct

The correct answer is D. असीमित क्षेत्र / unbounded region. In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

Step 3

Exam Tip

प्रथम चतुर्थांश में (x+y=9) के ऊपर का क्षेत्र अनंत तक फैलता है। इसलिए हल क्षेत्र असीमित है।

Question me issue ya doubt hai?

Answer, explanation, typing mistake ya suggestion directly hamari team ko bhejein. 📱Helpline (Call / WhatsApp): +91 7272824365

Related Mathematics Questions

FAQs

Mathematics Answer, Explanation and Revision Hints

असमानताओं \(x\ge 0\), \(y\ge 0\), \(x+y\ge 9\) का हल क्षेत्र कैसा है? / What type of solution region is given by \(x\ge 0\), \(y\ge 0\), and \(x+y\ge 9\)?

Correct Answer: D. असीमित क्षेत्र / unbounded region. Explanation: प्रथम चतुर्थांश में (x+y=9) के ऊपर का क्षेत्र अनंत तक फैलता है। इसलिए हल क्षेत्र असीमित है। / In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

Which concept should I revise for this Mathematics MCQ?

In the first quadrant, the region above (x+y=9) extends infinitely. Hence the solution region is unbounded.

What exam hint can help solve this Mathematics question?

प्रथम चतुर्थांश में (x+y=9) के ऊपर का क्षेत्र अनंत तक फैलता है। इसलिए हल क्षेत्र असीमित है।