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।
C. \(\frac{1}{2},\frac{1}{3},\sqrt{\frac{1}{4}}\) जैसे अनेक बिंदु/Many points like \(\frac{1}{2},\frac{1}{3},\sqrt{\frac{1}{4}}\)
Step 1
Concept
Between (0) and (1), there are infinitely many rational and irrational numbers. Between any two real numbers, more numbers can be found.
Step 2
Why this answer is correct
The correct answer is C. \(\frac{1}{2},\frac{1}{3},\sqrt{\frac{1}{4}}\) जैसे अनेक बिंदु / Many points like \(\frac{1}{2},\frac{1}{3},\sqrt{\frac{1}{4}}\). Between (0) and (1), there are infinitely many rational and irrational numbers. Between any two real numbers, more numbers can be found.
Step 3
Exam Tip
(0) और (1) के बीच परिमेय और अपरिमेय दोनों प्रकार की अनंत संख्याएं होती हैं। किसी भी दो वास्तविक संख्याओं के बीच और संख्याएं मिलती हैं।
By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 2
Why this answer is correct
The correct answer is B. \(\sqrt{2}\). By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
Step 3
Exam Tip
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें।
(5) is positive, so it lies (5) units to the right of (0). In exams, remember right direction for positive numbers.
Step 2
Why this answer is correct
The correct answer is A. (5) इकाई दाईं ओर / (5) units to the right. (5) is positive, so it lies (5) units to the right of (0). In exams, remember right direction for positive numbers.
Step 3
Exam Tip
(5) धनात्मक है इसलिए यह (0) के दाईं ओर (5) इकाई पर होगा। परीक्षा में धनात्मक संख्या के लिए दाईं दिशा याद रखें।
\(\frac{3}{4}\) is greater than (0) and less than (1). In exams first compare the fraction value with nearby integers.
Step 2
Why this answer is correct
The correct answer is A. (0) और (1) के बीच / Between (0) and (1). \(\frac{3}{4}\) is greater than (0) and less than (1). In exams first compare the fraction value with nearby integers.
Step 3
Exam Tip
\(\frac{3}{4}\) का मान (1) से कम और (0) से अधिक होता है। परीक्षा में भिन्न की स्थिति पहले उसके मान से पहचानें।
Every point on the number line represents a real number. Between (1) and (2), both rational and irrational numbers can occur.
Step 2
Why this answer is correct
The correct answer is A. वास्तविक संख्याएँ / real numbers. Every point on the number line represents a real number. Between (1) and (2), both rational and irrational numbers can occur.
Step 3
Exam Tip
संख्या रेखा का प्रत्येक बिंदु एक वास्तविक संख्या दिखाता है। (1) और (2) के बीच परिमेय और अपरिमेय दोनों हो सकते हैं।
The middle number between (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). In exams, use the average for the midpoint.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). The middle number between (0) and (1) is \(\frac{0+1}{2}=\frac{1}{2}\). In exams, use the average for the midpoint.
Step 3
Exam Tip
(0) और (1) के बीच की मध्य संख्या \(\frac{0+1}{2}=\frac{1}{2}\) है। परीक्षा में मध्य संख्या के लिए औसत लें।
In a right triangle with legs (1) and (2), the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Such lengths come from the Pythagoras theorem.
Step 2
Why this answer is correct
The correct answer is A. (1) और (2) / (1) and (2). In a right triangle with legs (1) and (2), the hypotenuse is \(\sqrt{1^2+2^2}=\sqrt{5}\). Such lengths come from the Pythagoras theorem.
Step 3
Exam Tip
समकोण त्रिभुज में भुजाएँ (1) और (2) हों तो कर्ण \(\sqrt{1^2+2^2}=\sqrt{5}\) होता है। पाइथागोरस प्रमेय से ऐसी लंबाई मिलती है।
\(\frac{7}{4}=1.75\), so it lies between (1) and (2). Think of an improper fraction in mixed or decimal form.
Step 2
Why this answer is correct
The correct answer is A. (1) और (2) / (1) and (2). \(\frac{7}{4}=1.75\), so it lies between (1) and (2). Think of an improper fraction in mixed or decimal form.
Step 3
Exam Tip
\(\frac{7}{4}=1.75\), इसलिए यह (1) और (2) के बीच है। अशुद्ध भिन्न को मिश्रित या दशमलव रूप में सोचें।
\(\frac{1}{2}=0.5\), so it lies between (0) and (1). Thinking of a fraction as a decimal helps locate it.
Step 2
Why this answer is correct
The correct answer is A. (0) और (1) / (0) and (1). \(\frac{1}{2}=0.5\), so it lies between (0) and (1). Thinking of a fraction as a decimal helps locate it.
Step 3
Exam Tip
\(\frac{1}{2}=0.5\), इसलिए यह (0) और (1) के बीच है। भिन्न को दशमलव में सोचने से स्थान स्पष्ट होता है।
Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 2
Why this answer is correct
The correct answer is C. (15). Here \(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), and \(3\sqrt{75}=15\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value should be (12).
Step 3
Exam Tip
\(\sqrt{363}=11\sqrt{3}\), \(2\sqrt{147}=14\sqrt{3}\), और \(3\sqrt{75}=15\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) होना चाहिए।
Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4721}{1600}\). Here (\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) and (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64}). The sum is \(\frac{4096+625}{1600}=\frac{4721}{1600}\).
Step 3
Exam Tip
(\left\(\frac{5}{8}\right\)^{-2}=\frac{64}{25}) और (\left\(\frac{8}{5}\right\)^{-2}=\frac{25}{64})। योग \(\frac{4096+625}{1600}=\frac{4721}{1600}\) है।
From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 2
Why this answer is correct
The correct answer is A. \(250\sqrt{2}\). From \(\sqrt{x}=5\sqrt{2}\), (x=50), and \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\). In exams, write \(x^{\frac{3}{2}}\) as \(x\sqrt{x}\).
Step 3
Exam Tip
\(\sqrt{x}=5\sqrt{2}\) से (x=50), और \(x^{\frac{3}{2}}=x\sqrt{x}=50\cdot5\sqrt{2}=250\sqrt{2}\)। परीक्षा में \(x^{\frac{3}{2}}\) को \(x\sqrt{x}\) लिखें।
We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
Step 2
Why this answer is correct
The correct answer is C. (10). We use (x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)). Thus (60=6\left\(x+\frac{1}{x}\right\)), so the value is (10).
Step 3
Exam Tip
(x^{2}-\frac{1}{x^{2}}=\left\(x-\frac{1}{x}\right\)\left\(x+\frac{1}{x}\right\)) है। इसलिए (60=6\left\(x+\frac{1}{x}\right\)) और मान (10) है।
Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 2
Why this answer is correct
The correct answer is C. (12). Here \(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), and \(\sqrt{108}=6\sqrt{3}\). The numerator is \(12\sqrt{3}\), so the value is (12).
Step 3
Exam Tip
\(\sqrt{300}=10\sqrt{3}\), \(\sqrt{192}=8\sqrt{3}\), और \(\sqrt{108}=6\sqrt{3}\)। अंश \(12\sqrt{3}\) है, इसलिए मान (12) है।
We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
Step 2
Why this answer is correct
The correct answer is B. (47). We use (\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2). Thus \(49=x^{2}+\frac{1}{x^{2}}+2\), so the value is (47).
Step 3
Exam Tip
(\left\(x+\frac{1}{x}\right\)^{2}=x^{2}+\frac{1}{x^{2}}+2) होता है। इसलिए \(49=x^{2}+\frac{1}{x^{2}}+2\) और मान (47) है।
