\( -\frac{11}{3}\approx-3.667\) and \( -\sqrt{13}\approx-3.606\), so \( -\frac{11}{3}\) is smaller. On a number line, the smaller number lies farther left.
Step 2
Why this answer is correct
The correct answer is A. \( -\frac{11}{3}\). \( -\frac{11}{3}\approx-3.667\) and \( -\sqrt{13}\approx-3.606\), so \( -\frac{11}{3}\) is smaller. On a number line, the smaller number lies farther left.
Step 3
Exam Tip
\( -\frac{11}{3}\approx-3.667\) और \( -\sqrt{13}\approx-3.606\), इसलिए \( -\frac{11}{3}\) अधिक छोटा है। संख्या रेखा पर छोटी संख्या अधिक बाईं ओर होती है।
Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Perfect squares quickly give the interval.
Step 2
Why this answer is correct
The correct answer is A. (2) और (3) / (2) and (3). Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Perfect squares quickly give the interval.
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\), इसलिए \(\sqrt{5}\), (2) और (3) के बीच है। पूर्ण वर्गों से अंतराल जल्दी मिल जाता है।
\(\sqrt{17}\) is irrational, so its decimal expansion is non-terminating non-recurring. In exams distinguish irrational decimals from recurring decimals.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{17}\). \(\sqrt{17}\) is irrational, so its decimal expansion is non-terminating non-recurring. In exams distinguish irrational decimals from recurring decimals.
Step 3
Exam Tip
\(\sqrt{17}\) अपरिमेय है, इसलिए इसका दशमलव अनवसानी अनावर्ती होगा। परीक्षा में अपरिमेय और आवर्ती दशमलव में अंतर रखें।
\(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), which is irrational. In exams do not treat addition like multiplication.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{20}+\sqrt{45}\). \(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), which is irrational. In exams do not treat addition like multiplication.
Step 3
Exam Tip
\(\sqrt{20}+\sqrt{45}=2\sqrt{5}+3\sqrt{5}=5\sqrt{5}\), जो अपरिमेय है। परीक्षा में योग को गुणन जैसा न मानें।
Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 2
Why this answer is correct
The correct answer is A. यह (p(x)) का शून्यक है / It is a zero of (p(x)). Since (p\(1+\sqrt{3}\)=0), \(1+\sqrt{3}\) is a zero. To prove a number is a zero, show that the polynomial value is (0).
Step 3
Exam Tip
(p\(1+\sqrt{3}\)=0), इसलिए \(1+\sqrt{3}\) शून्यक है। किसी संख्या को शून्यक सिद्ध करने के लिए बहुपद का मान (0) दिखाएँ।
B. एक गुणांक अपरिमेय है/One coefficient is irrational
Step 1
Concept
The constant term \(-\sqrt{2}\) is irrational, while the other coefficients are rational. Check coefficient type before applying root rules.
Step 2
Why this answer is correct
The correct answer is B. एक गुणांक अपरिमेय है / One coefficient is irrational. The constant term \(-\sqrt{2}\) is irrational, while the other coefficients are rational. Check coefficient type before applying root rules.
Step 3
Exam Tip
स्थिर पद \(-\sqrt{2}\) अपरिमेय है, जबकि बाकी गुणांक परिमेय हैं। शून्यक नियम लागू करने से पहले गुणांकों का प्रकार देखें।
If it were rational, its square \(3+\sqrt{5}\) would be rational, which it is not. Hence it is real irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय वास्तविक संख्या / Irrational real number. If it were rational, its square \(3+\sqrt{5}\) would be rational, which it is not. Hence it is real irrational.
Step 3
Exam Tip
यदि यह परिमेय होता तो उसका वर्ग \(3+\sqrt{5}\) परिमेय होता, जो नहीं है। इसलिए यह वास्तविक अपरिमेय है।
The first product is (25-6=19) and \(\sqrt{24}=2\sqrt{6}\) is irrational. A rational plus an irrational is irrational.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. The first product is (25-6=19) and \(\sqrt{24}=2\sqrt{6}\) is irrational. A rational plus an irrational is irrational.
Step 3
Exam Tip
पहला गुणनफल (25-6=19) है और \(\sqrt{24}=2\sqrt{6}\) अपरिमेय है। परिमेय और अपरिमेय का योग अपरिमेय होता है।
A. यह अपरिमेय संख्या है/It is an irrational number
Step 1
Concept
The decimal is infinite and has no fixed repeating block. So it is an irrational number.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय संख्या है / It is an irrational number. The decimal is infinite and has no fixed repeating block. So it is an irrational number.
Step 3
Exam Tip
दशमलव अनंत है और इसमें कोई निश्चित आवर्ती समूह नहीं है। इसलिए यह अपरिमेय संख्या है।
(20) is not a perfect cube so \(\sqrt[3]{20}\) is irrational. Identify perfect cubes in cube roots.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. (20) is not a perfect cube so \(\sqrt[3]{20}\) is irrational. Identify perfect cubes in cube roots.
Step 3
Exam Tip
(20) पूर्ण घन नहीं है इसलिए \(\sqrt[3]{20}\) अपरिमेय है। घनमूल में पूर्ण घन पहचानें।
A. यह अपरिमेय संख्या है/It is an irrational number
Step 1
Concept
This decimal has no fixed repeating block. So it is non terminating and non repeating.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय संख्या है / It is an irrational number. This decimal has no fixed repeating block. So it is non terminating and non repeating.
Step 3
Exam Tip
इस दशमलव में निश्चित आवर्ती समूह नहीं है। इसलिए यह अनंत और अनावर्ती है।
A non terminating and non repeating decimal identifies an irrational number. Check carefully if no repeating block appears.
Step 2
Why this answer is correct
The correct answer is A. अपरिमेय संख्या / Irrational number. A non terminating and non repeating decimal identifies an irrational number. Check carefully if no repeating block appears.
Step 3
Exam Tip
अनंत और अनावर्ती दशमलव अपरिमेय संख्या की पहचान है। आवर्ती भाग न दिखे तो सावधानी से जाँचें।
(12) is not a perfect cube so \(\sqrt[3]{12}\) is irrational. Check perfect cubes in cube roots.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. (12) is not a perfect cube so \(\sqrt[3]{12}\) is irrational. Check perfect cubes in cube roots.
Step 3
Exam Tip
(12) पूर्ण घन नहीं है इसलिए \(\sqrt[3]{12}\) अपरिमेय है। घनमूल में पूर्ण घन देखें।
A. यह अपरिमेय संख्या है/It is an irrational number
Step 1
Concept
The decimal is infinite and has no fixed repeating block. So it is irrational.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय संख्या है / It is an irrational number. The decimal is infinite and has no fixed repeating block. So it is irrational.
Step 3
Exam Tip
दशमलव अनंत है और कोई निश्चित आवर्ती समूह नहीं है। इसलिए यह अपरिमेय है।
\(\frac{3}{\sqrt{3}}=\sqrt{3}\) because \(3=\sqrt{3}\times\sqrt{3}\). Simplify when a root is in the denominator.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\frac{3}{\sqrt{3}}=\sqrt{3}\) because \(3=\sqrt{3}\times\sqrt{3}\). Simplify when a root is in the denominator.
Step 3
Exam Tip
\(\frac{3}{\sqrt{3}}=\sqrt{3}\) क्योंकि \(3=\sqrt{3}\times\sqrt{3}\)। हर में जड़ हो तो सरल करें।
A. यह अपरिमेय संख्या है/It is an irrational number
Step 1
Concept
It has no fixed repeating block and the decimal is infinite. So it is irrational.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय संख्या है / It is an irrational number. It has no fixed repeating block and the decimal is infinite. So it is irrational.
Step 3
Exam Tip
इसमें कोई निश्चित आवर्ती समूह नहीं है और दशमलव अनंत है। इसलिए यह अपरिमेय है।
(16) is not a perfect cube so its cube root is irrational. Identify perfect cubes in cube roots.
Step 2
Why this answer is correct
The correct answer is A. यह अपरिमेय है / It is irrational. (16) is not a perfect cube so its cube root is irrational. Identify perfect cubes in cube roots.
Step 3
Exam Tip
(16) पूर्ण घन नहीं है इसलिए इसकी घनमूल अपरिमेय है। घनमूल में पूर्ण घन पहचानें।
\(\frac{\sqrt{3}}{3}\) is irrational and lies between (0) and (1). Dividing by a non zero rational keeps irrationality.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{3}}{3}\). \(\frac{\sqrt{3}}{3}\) is irrational and lies between (0) and (1). Dividing by a non zero rational keeps irrationality.
Step 3
Exam Tip
\(\frac{\sqrt{3}}{3}\) अपरिमेय है और इसका मान (0) और (1) के बीच है। गैर शून्य परिमेय से भाग देने पर अपरिमेयता रहती है।
