\(\sqrt{5}\approx2.236\) and \(\sqrt{2}\approx1.414\), so the difference is about (0.822). Use short approximations to locate differences of irrationals.
Step 2
Why this answer is correct
The correct answer is A. ((0,1)). \(\sqrt{5}\approx2.236\) and \(\sqrt{2}\approx1.414\), so the difference is about (0.822). Use short approximations to locate differences of irrationals.
Step 3
Exam Tip
\(\sqrt{5}\approx2.236\) और \(\sqrt{2}\approx1.414\), इसलिए अंतर लगभग (0.822) है। अपरिमेयों के अंतर का स्थान निकालने के लिए छोटे अनुमान उपयोग करें।
Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 2
Why this answer is correct
The correct answer is A. (4) और (1) / (4) and (1). Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
Step 3
Exam Tip
क्योंकि \(4^2+1^2=17\), इसलिए कर्ण \(\sqrt{17}\) होगा। संख्या रेखा पर वर्गमूल निर्माण में पाइथागोरस प्रमेय उपयोगी है।
\(\pi\approx3.14\) 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।
A. -\(\frac{5}{6}\), \(-\frac{3}{4}\), \(-\frac{2}{3}\)
Step 1
Concept
For negative numbers, the one with larger magnitude is farther left, so \(-\frac{5}{6}<-\frac{3}{4}<-\frac{2}{3}\). Compare positive values first, then reverse the order.
Step 2
Why this answer is correct
The correct answer is A. -\(\frac{5}{6}\), \(-\frac{3}{4}\), \(-\frac{2}{3}\). For negative numbers, the one with larger magnitude is farther left, so \(-\frac{5}{6}<-\frac{3}{4}<-\frac{2}{3}\). Compare positive values first, then reverse the order.
Step 3
Exam Tip
ऋणात्मक संख्याओं में अधिक परिमाण वाली संख्या अधिक बाएँ होती है, इसलिए क्रम \(-\frac{5}{6}<-\frac{3}{4}<-\frac{2}{3}\) है। पहले धनात्मक मानों की तुलना करें, फिर क्रम उलटें।
\(\sqrt{8}=2\sqrt{2}\), so the midpoint is \(\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\). Simplify first, then average.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3\sqrt{2}}{2}\). \(\sqrt{8}=2\sqrt{2}\), so the midpoint is \(\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\). Simplify first, then average.
Step 3
Exam Tip
\(\sqrt{8}=2\sqrt{2}\), इसलिए मध्य बिंदु \(\frac{\sqrt{2}+2\sqrt{2}}{2}=\frac{3\sqrt{2}}{2}\) है। सरलीकरण के बाद औसत लें।
A. बाएँ \( \frac{7}{4} \) इकाई/Left \( \frac{7}{4} \) units
Step 1
Concept
A negative number lies to the left of (0), and its distance is its absolute value \(\frac{7}{4}\). Identify direction and distance separately.
Step 2
Why this answer is correct
The correct answer is A. बाएँ \( \frac{7}{4} \) इकाई / Left \( \frac{7}{4} \) units. A negative number lies to the left of (0), and its distance is its absolute value \(\frac{7}{4}\). Identify direction and distance separately.
Step 3
Exam Tip
ऋणात्मक संख्या (0) के बाएँ होती है और दूरी उसका निरपेक्ष मान \(\frac{7}{4}\) है। दिशा और दूरी को अलग-अलग पहचानें।
A. यह (0) और (1) के बीच अपरिमेय है/It is irrational between (0) and (1)
Step 1
Concept
This decimal is non-terminating and non-repeating, so it is irrational and lies between (0) and (1). Identify rationality by the decimal pattern.
Step 2
Why this answer is correct
The correct answer is A. यह (0) और (1) के बीच अपरिमेय है / It is irrational between (0) and (1). This decimal is non-terminating and non-repeating, so it is irrational and lies between (0) and (1). Identify rationality by the decimal pattern.
Step 3
Exam Tip
यह दशमलव अनावर्ती और असांत है, इसलिए अपरिमेय है और (0) से (1) के बीच है। दशमलव पैटर्न देखकर परिमेयता पहचानें।
(|x-2|=3) means (x) is (3) units away from (2), so (x=-1,5). Place the distance on both sides of the center.
Step 2
Why this answer is correct
The correct answer is A. (-1) और (5) / (-1) and (5). (|x-2|=3) means (x) is (3) units away from (2), so (x=-1,5). Place the distance on both sides of the center.
Step 3
Exam Tip
(|x-2|=3) का अर्थ है (x), (2) से (3) इकाई दूर है, इसलिए (x=-1,5)। केंद्र से दोनों ओर दूरी लगाएँ।
Because (11<13<15), \(\sqrt{11}<\sqrt{13}<\sqrt{15}\). For positive square roots, the order of radicands is preserved.
Step 2
Why this answer is correct
The correct answer is A. \(\sqrt{13}\). Because (11<13<15), \(\sqrt{11}<\sqrt{13}<\sqrt{15}\). For positive square roots, the order of radicands is preserved.
Step 3
Exam Tip
क्योंकि (11<13<15), इसलिए \(\sqrt{11}<\sqrt{13}<\sqrt{15}\)। धनात्मक वर्गमूल में मूल संख्या का क्रम बना रहता है।
With common denominator (30), \(\frac{18}{30}<\frac{20}{30}<\frac{25}{30}\). Use a common denominator to order fractions.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3}{5},\frac{2}{3},\frac{5}{6}\). With common denominator (30), \(\frac{18}{30}<\frac{20}{30}<\frac{25}{30}\). Use a common denominator to order fractions.
Step 3
Exam Tip
समान हर (30) लेने पर \(\frac{18}{30}<\frac{20}{30}<\frac{25}{30}\)। भिन्नों का क्रम निकालने के लिए समान हर लें।
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।
