0 reads0 ratings0 helpful
In the first quadrant, what is the \(x\)-intercept of the region formed by \(4x+y\le 20\)?प्रथम चतुर्थांश में \(4x+y\le 20\) से बने क्षेत्र का \(x\)-अवरोध कौन-सा है?
Answer and explanationउत्तर और व्याख्या
Correct answer:सही उत्तर: \((5,0)\)\((5,0)\)
To find the \(x\)-intercept, use the boundary line \(4x+y=20\) and put \(y=0\). This gives \(4x=20\), so \(x=5\). Hence, the \(x\)-intercept is \((5,0)\). The point \((0,5)\) lies on the \(y\)-axis, so it is not an \(x\)-intercept. Exam tip: for an \(x\)-intercept, always set \(y=0\).\(x\)-अवरोध ज्ञात करने के लिए सीमा रेखा \(4x+y=20\) में \(y=0\) रखते हैं। तब \(4x=20\), अतः \(x=5\)। इसलिए \(x\)-अवरोध \((5,0)\) है। \((0,5)\) \(y\)-अक्ष पर है, इसलिए वह \(x\)-अवरोध नहीं है। परीक्षा टिप: \(x\)-अवरोध के लिए हमेशा \(y=0\) रखें।
Student feedback
Was this question useful?
Was this question useful?
👍 0 Helpful 👎 0 Not helpful
Difficulty
Easy0%
Medium0%
Hard0%
Was the explanation clear?
Yes 0% No 0%
0 responsesStudent Reviews
No published reviews yet.
