\(\pi\approx3.14\) and \(\sqrt{10}\approx3.16\), so \(\pi<\sqrt{10}\). Good approximations help compare close irrationals.
Step 2
Why this answer is correct
The correct answer is A. \(\pi<\sqrt{10}\). \(\pi\approx3.14\) and \(\sqrt{10}\approx3.16\), so \(\pi<\sqrt{10}\). Good approximations help compare close irrationals.
Step 3
Exam Tip
\(\pi\approx3.14\) और \(\sqrt{10}\approx3.16\), इसलिए \(\pi<\sqrt{10}\)। निकट अपरिमेयों की तुलना में अच्छे अनुमान उपयोगी होते हैं।
Since \(2^2=4\) and \(3^2=9\), \(\sqrt{7}\) lies between (2) and (3). In exams, compare squares first.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि \(2^2<7<3^2\) / Because \(2^2<7<3^2\). Since \(2^2=4\) and \(3^2=9\), \(\sqrt{7}\) lies between (2) and (3). In exams, compare squares first.
Step 3
Exam Tip
\(2^2=4\) और \(3^2=9\), इसलिए \(\sqrt{7}\) संख्या रेखा पर (2) और (3) के बीच होगा। परीक्षा में पहले वर्गों की तुलना करें।
\(\sqrt{2}\approx1.41\), so \(-1.5<-\sqrt{2}<-1.3\). For negative numbers, the farther left number is smaller.
Step 2
Why this answer is correct
The correct answer is A. \(-1.5,-\sqrt{2},-1.3\). \(\sqrt{2}\approx1.41\), so \(-1.5<-\sqrt{2}<-1.3\). For negative numbers, the farther left number is smaller.
Step 3
Exam Tip
\(\sqrt{2}\approx1.41\), इसलिए \(-1.5<-\sqrt{2}<-1.3\) है। ऋणात्मक संख्याओं में अधिक बाईं संख्या छोटी होती है।
\(\sqrt{11}\) lies between (3) and (4), so \(-\sqrt{11}\) lies between (-4) and (-3). The negative sign changes the side.
Step 2
Why this answer is correct
The correct answer is C. (-4) और (-3) / (-4) and (-3). \(\sqrt{11}\) lies between (3) and (4), so \(-\sqrt{11}\) lies between (-4) and (-3). The negative sign changes the side.
Step 3
Exam Tip
\(\sqrt{11}\) (3) और (4) के बीच है इसलिए \(-\sqrt{11}\) (-4) और (-3) के बीच होगा। ऋणात्मक चिन्ह दिशा बदल देता है।
\(\frac{4}{3}\approx1.33\) and \(\sqrt{2}\approx1.41\), so the order is \(\frac{4}{3}<\sqrt{2}<1.5\). Estimate values for comparison.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4}{3},\sqrt{2},1.5\). \(\frac{4}{3}\approx1.33\) and \(\sqrt{2}\approx1.41\), so the order is \(\frac{4}{3}<\sqrt{2}<1.5\). Estimate values for comparison.
Step 3
Exam Tip
\(\frac{4}{3}\approx1.33\), \(\sqrt{2}\approx1.41\), इसलिए क्रम \(\frac{4}{3}<\sqrt{2}<1.5\) है। तुलना के लिए अनुमान लगाएं।
B. यह (2) और (3) के बीच है/It is between (2) and (3)
Step 1
Concept
Since \(2^2<7<3^2\), \(\sqrt{7}\) lies between (2) and (3). Taking \(\sqrt{7}\) as (7) is a common mistake.
Step 2
Why this answer is correct
The correct answer is B. यह (2) और (3) के बीच है / It is between (2) and (3). Since \(2^2<7<3^2\), \(\sqrt{7}\) lies between (2) and (3). Taking \(\sqrt{7}\) as (7) is a common mistake.
Step 3
Exam Tip
क्योंकि \(2^2<7<3^2\), इसलिए \(\sqrt{7}\), (2) और (3) के बीच होगा। \(\sqrt{7}\) को (7) समझना सामान्य गलती है।
Since \(3^2<10<4^2\), \(\sqrt{10}\) lies between (3) and (4). Use nearby perfect squares to locate roots.
Step 2
Why this answer is correct
The correct answer is B. (3) और (4) / (3) and (4). Since \(3^2<10<4^2\), \(\sqrt{10}\) lies between (3) and (4). Use nearby perfect squares to locate roots.
Step 3
Exam Tip
क्योंकि \(3^2<10<4^2\), इसलिए \(\sqrt{10}\), (3) और (4) के बीच होगा। वर्गमूल की स्थिति के लिए पास के पूर्ण वर्ग देखें।
Since \(1^2=1\) and \(2^2=4\), \(\sqrt{3}\) lies between (1) and (2). In exams, bracket square roots using perfect squares.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). Since \(1^2=1\) and \(2^2=4\), \(\sqrt{3}\) lies between (1) and (2). In exams, bracket square roots using perfect squares.
Step 3
Exam Tip
क्योंकि \(1^2=1\) और \(2^2=4\), इसलिए \(\sqrt{3}\) (1) और (2) के बीच है। परीक्षा में वर्गमूल को पूर्ण वर्गों से घेरें।
Since \(4^2=16\) and \(5^2=25\), \(\sqrt{20}\) lies between (4) and (5). In exams, find nearby perfect squares.
Step 2
Why this answer is correct
The correct answer is C. (4) और (5) / (4) and (5). Since \(4^2=16\) and \(5^2=25\), \(\sqrt{20}\) lies between (4) and (5). In exams, find nearby perfect squares.
Step 3
Exam Tip
क्योंकि \(4^2=16\) और \(5^2=25\), इसलिए \(\sqrt{20}\) (4) और (5) के बीच है। परीक्षा में नजदीकी पूर्ण वर्ग खोजें।
Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Use squares to locate square roots quickly.
Step 2
Why this answer is correct
The correct answer is B. (2) और (3) / (2) and (3). Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3). Use squares to locate square roots quickly.
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\), इसलिए \(\sqrt{5}\), (2) और (3) के बीच होगा। वर्गों से वर्गमूल की स्थिति जल्दी मिलती है।
\(\sqrt{8}\) lies between (2) and (3), so \(-\sqrt{8}\) lies between (-3) and (-2). The negative sign changes the direction.
Step 2
Why this answer is correct
The correct answer is A. (-3) और (-2) / (-3) and (-2). \(\sqrt{8}\) lies between (2) and (3), so \(-\sqrt{8}\) lies between (-3) and (-2). The negative sign changes the direction.
Step 3
Exam Tip
\(\sqrt{8}\) (2) और (3) के बीच है, इसलिए \(-\sqrt{8}\) (-3) और (-2) के बीच होगा। ऋण चिह्न दिशा बदल देता है।
Since \(2^2=4\) and \(3^2=9\), \(\sqrt{5}\) lies between (2) and (3). Look at the nearest perfect squares.
Step 2
Why this answer is correct
The correct answer is A. (2) और (3) / (2) and (3). Since \(2^2=4\) and \(3^2=9\), \(\sqrt{5}\) lies between (2) and (3). Look at the nearest perfect squares.
Step 3
Exam Tip
क्योंकि \(2^2=4\) और \(3^2=9\), इसलिए \(\sqrt{5}\) (2) और (3) के बीच है। निकटतम पूर्ण वर्ग देखें।
\(\sqrt{27}=3\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Hence the value is \(6\sqrt{3}\).
Step 2
Why this answer is correct
The correct answer is A. \(6\sqrt{3}\). \(\sqrt{27}=3\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) and \(\sqrt{12}=2\sqrt{3}\). Hence the value is \(6\sqrt{3}\).
Step 3
Exam Tip
\(\sqrt{27}=3\sqrt{3}\), \(\sqrt{75}=5\sqrt{3}\) और \(\sqrt{12}=2\sqrt{3}\) है। इसलिए मान \(6\sqrt{3}\) है।
Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{13}-\sqrt{12}\). Since (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), the reciprocal is \(\sqrt{13}-\sqrt{12}\). In exams quickly identify conjugates where (a-b=1).
Step 3
Exam Tip
क्योंकि (\(\sqrt{13}+\sqrt{12}\)\(\sqrt{13}-\sqrt{12}\)=1), इसलिए व्युत्क्रम \(\sqrt{13}-\sqrt{12}\) है। परीक्षा में (a-b=1) वाले संयुग्मी जल्दी पहचानें।
The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 2
Why this answer is correct
The correct answer is A. \(x^2-10x+1\). The companion zero is \(5-2\sqrt{6}\), with sum (10) and product (25-24=1). In exams form the polynomial using the conjugate.
Step 3
Exam Tip
साथी शून्यक \(5-2\sqrt{6}\) होगा, योग (10) और गुणनफल (25-24=1) है। परीक्षा में संयुग्मी लेकर बहुपद बनाएं।
