If (0<a<1), then \(0<\sqrt{a}<1\); here (a=0.49). For decimal square roots, identify the bounds first.
Step 2
Why this answer is correct
The correct answer is A. क्योंकि (0<0.49<1) / Because (0<0.49<1). If (0<a<1), then \(0<\sqrt{a}<1\); here (a=0.49). For decimal square roots, identify the bounds first.
Step 3
Exam Tip
यदि (0<a<1), तो \(0<\sqrt{a}<1\); यहाँ (a=0.49) है। दशमलव वर्गमूलों में सीमा पहले पहचानें।
The midpoint is \(\frac{-\frac{5}{2}-\frac{1}{2}}{2}=-\frac{3}{2}\). Pay attention to signs in negative fractions.
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{3}{2}\). The midpoint is \(\frac{-\frac{5}{2}-\frac{1}{2}}{2}=-\frac{3}{2}\). Pay attention to signs in negative fractions.
Step 3
Exam Tip
मध्य बिंदु \(\frac{-\frac{5}{2}-\frac{1}{2}}{2}=-\frac{3}{2}\) है। ऋणात्मक भिन्नों में संकेत पर ध्यान दें।
\(\frac{9}{7}=1+\frac{2}{7}\), so choose the second seventh part after (1). Converting an improper fraction into a mixed form is useful.
Step 2
Why this answer is correct
The correct answer is A. \(1+\frac{2}{7}\). \(\frac{9}{7}=1+\frac{2}{7}\), so choose the second seventh part after (1). Converting an improper fraction into a mixed form is useful.
Step 3
Exam Tip
\(\frac{9}{7}=1+\frac{2}{7}\), इसलिए (1) के बाद दूसरा सातवाँ भाग चुनेंगे। अपूर्णांक को मिश्र संख्या में बदलना उपयोगी है।
\(\sqrt{3}\approx1.732\), so it lies near (1.732) on the number line. Remember approximations of common square roots.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{3}\). \(\sqrt{3}\approx1.732\), so it lies near (1.732) on the number line. Remember approximations of common square roots.
Step 3
Exam Tip
\(\sqrt{3}\approx1.732\), इसलिए संख्या रेखा पर (1.732) के पास यही होगा। कुछ प्रसिद्ध वर्गमूलों के अनुमान याद रखें।
A. \(-\sqrt{13}\) और \(\sqrt{13}\)/\(-\sqrt{13}\) and \(\sqrt{13}\)
Step 1
Concept
Points at the same distance from (0) occur on both sides, so the values are \(\pm\sqrt{13}\). In distance questions, check both directions.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{13}\) और \(\sqrt{13}\) / \(-\sqrt{13}\) and \(\sqrt{13}\). Points at the same distance from (0) occur on both sides, so the values are \(\pm\sqrt{13}\). In distance questions, check both directions.
Step 3
Exam Tip
(0) से समान दूरी पर दोनों ओर बिंदु होते हैं, इसलिए मान \(\pm\sqrt{13}\) होंगे। दूरी वाले प्रश्नों में दोनों दिशाएँ जाँचें।
Since (-2.75<-2.7), (A) is to the left on the number line. In negative decimals, the more negative number is farther left.
Step 2
Why this answer is correct
The correct answer is A. (A). Since (-2.75<-2.7), (A) is to the left on the number line. In negative decimals, the more negative number is farther left.
Step 3
Exam Tip
(-2.75<-2.7), इसलिए (A) संख्या रेखा पर बाएँ है। ऋणात्मक दशमलवों की तुलना में अधिक ऋणात्मक संख्या बाएँ होती है।
Since \(-\sqrt{3}\approx-1.732\) and \(-\frac{3}{2}=-1.5\), \(-\sqrt{3}\) is farther left. For negative numbers, larger magnitude means farther left.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{3},-\frac{3}{2},0,\sqrt{2}\). Since \(-\sqrt{3}\approx-1.732\) and \(-\frac{3}{2}=-1.5\), \(-\sqrt{3}\) is farther left. For negative numbers, larger magnitude means farther left.
Step 3
Exam Tip
\(-\sqrt{3}\approx-1.732\) और \(-\frac{3}{2}=-1.5\), इसलिए \(-\sqrt{3}\) अधिक बाएँ है। ऋणात्मक संख्याओं में बड़ा परिमाण अधिक बाएँ होता है।
\(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). An irrational divided by a non-zero rational remains irrational.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{\sqrt{2}}{2}\). \(\frac{\sqrt{2}}{2}\) is irrational and its value lies between (0) and (1). An irrational divided by a non-zero rational remains irrational.
Step 3
Exam Tip
\(\frac{\sqrt{2}}{2}\) अपरिमेय है और इसका मान (0) और (1) के बीच है। अपरिमेय संख्या को परिमेय से भाग देने पर शून्येतर परिमेय के लिए अपरिमेय ही रहती है।
Because (\(\sqrt{2}\)2+12=3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{2}\) और (1) / \(\sqrt{2}\) and (1). Because (\(\sqrt{2}\)2+12=3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
Step 3
Exam Tip
क्योंकि (\(\sqrt{2}\)2+12=3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है।
Distance from (0) is (|x|), so (|x|=3.5) gives \(x=\pm3.5\). Distance is always a positive measure.
Step 2
Why this answer is correct
The correct answer is A. (-3.5) और (3.5) / (-3.5) and (3.5). Distance from (0) is (|x|), so (|x|=3.5) gives \(x=\pm3.5\). Distance is always a positive measure.
Step 3
Exam Tip
(0) से दूरी (|x|) होती है, इसलिए (|x|=3.5) के हल \(x=\pm3.5\) हैं। दूरी हमेशा धनात्मक माप होती है।
Since \(\frac{-11}{5}=-2.2\), it lies between (-3) and (-2). Be careful with position of negative decimals.
Step 2
Why this answer is correct
The correct answer is A. (-3) और (-2) / (-3) and (-2). Since \(\frac{-11}{5}=-2.2\), it lies between (-3) and (-2). Be careful with position of negative decimals.
Step 3
Exam Tip
\(\frac{-11}{5}=-2.2\), इसलिए यह (-3) और (-2) के बीच है। ऋणात्मक दशमलव में स्थान का ध्यान रखें।
Since \(2=\frac{8}{4}\), the interval from (0) to (2) has (8) fourth-parts and \(\frac{7}{4}\) is at the seventh part. Use the denominator to make equal units.
Step 2
Why this answer is correct
The correct answer is A. (8) भाग / (8) parts. Since \(2=\frac{8}{4}\), the interval from (0) to (2) has (8) fourth-parts and \(\frac{7}{4}\) is at the seventh part. Use the denominator to make equal units.
Step 3
Exam Tip
क्योंकि \(2=\frac{8}{4}\), इसलिए (0) से (2) तक (8) चौथाई भाग बनेंगे और \(\frac{7}{4}\) सातवें भाग पर होगा। हर को समान इकाई बनाने में प्रयोग करें।
The midpoint of (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). To find a midpoint on a number line, take the average.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). The midpoint of (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). To find a midpoint on a number line, take the average.
Step 3
Exam Tip
(0) और (1) का मध्य बिंदु \(\frac{0+1}{2}=\frac{1}{2}\) होता है। संख्या रेखा में मध्य निकालने के लिए औसत लें।
Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3), so \(-\sqrt{5}\) lies between (-3) and (-2). On the negative side, order reverses.
Step 2
Why this answer is correct
The correct answer is A. ((-3,-2)). Since \(2^2<5<3^2\), \(\sqrt{5}\) lies between (2) and (3), so \(-\sqrt{5}\) lies between (-3) and (-2). On the negative side, order reverses.
Step 3
Exam Tip
क्योंकि \(2^2<5<3^2\), इसलिए \(\sqrt{5}\) (2) और (3) के बीच है और ऋणात्मक मान (-3) और (-2) के बीच होगा। ऋणात्मक दिशा में क्रम उलट जाता है।
Because \(1.4^2=1.96\) and \(1.5^2=2.25\), \(\sqrt{2}\) lies between them. For hard questions, use squares of decimals.
Step 2
Why this answer is correct
The correct answer is A. \(1.4<\sqrt{2}<1.5\). Because \(1.4^2=1.96\) and \(1.5^2=2.25\), \(\sqrt{2}\) lies between them. For hard questions, use squares of decimals.
Step 3
Exam Tip
क्योंकि \(1.4^2=1.96\) और \(1.5^2=2.25\), इसलिए \(\sqrt{2}\) इनके बीच है। कठिन प्रश्नों में दशमलव के वर्ग का प्रयोग करें।
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।
