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A. चिह्न और निर्देशांक क्रम की गलती/Error of sign and coordinate order
Step 1
Concept
In (\left\(7,-3\right\)), (x=7) and (y=-3). Reversing coordinates and changing sign makes the answer wrong.
Step 2
Why this answer is correct
The correct answer is A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order. In (\left\(7,-3\right\)), (x=7) and (y=-3). Reversing coordinates and changing sign makes the answer wrong.
Step 3
Exam Tip
बिंदु (\left\(7,-3\right\)) में (x=7) और (y=-3) है। निर्देशांक उलटने और चिह्न बदलने से उत्तर गलत हो जाता है।
A. चिह्न और निर्देशांक क्रम की गलती/Error of sign and coordinate order
Step 1
Concept
In (\left\(6,-2\right\)), (x=6) and (y=-2). Reversing coordinates and changing sign makes the answer wrong.
Step 2
Why this answer is correct
The correct answer is A. चिह्न और निर्देशांक क्रम की गलती / Error of sign and coordinate order. In (\left\(6,-2\right\)), (x=6) and (y=-2). Reversing coordinates and changing sign makes the answer wrong.
Step 3
Exam Tip
बिंदु (\left\(6,-2\right\)) में (x=6) और (y=-2) है। निर्देशांक उलटने से और चिह्न बदलने से उत्तर गलत हो जाता है।
B. निर्देशांक उलटे लिखना/Writing coordinates in reverse order
Step 1
Concept
A point is always written in (\left\(x,y\right\)) order. Reversing coordinates makes the solution wrong.
Step 2
Why this answer is correct
The correct answer is B. निर्देशांक उलटे लिखना / Writing coordinates in reverse order. A point is always written in (\left\(x,y\right\)) order. Reversing coordinates makes the solution wrong.
Step 3
Exam Tip
बिंदु हमेशा (\left\(x,y\right\)) क्रम में लिखा जाता है। निर्देशांक उलटे करने से हल गलत हो जाता है।
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।
Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{5}\). Here (125^{\frac{2}{3}}=(5)^{2}=25) and (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125}). The product is \(\frac{1}{5}\).
Step 3
Exam Tip
(125^{\frac{2}{3}}=(5)^{2}=25) और (25^{-\frac{3}{2}}=(5)^{-3}=\frac{1}{125})। गुणनफल \(\frac{1}{5}\) है।
Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 2
Why this answer is correct
The correct answer is B. (5). Here (25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) and (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2}). The product is (5).
Step 3
Exam Tip
(25^{\frac{3}{2}}=\(5^{2}\)^{\frac{3}{2}}=5^{3}) और (125^{-\frac{2}{3}}=\(5^{3}\)^{-\frac{2}{3}}=5^{-2})। गुणनफल (5) है।
Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2657}{784}\). Here (\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) and (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49}). The sum is \(\frac{2401+256}{784}=\frac{2657}{784}\).
Step 3
Exam Tip
(\left\(\frac{4}{7}\right\)^{-2}=\frac{49}{16}) और (\left\(\frac{7}{4}\right\)^{-2}=\frac{16}{49})। योग \(\frac{2401+256}{784}=\frac{2657}{784}\) है।
Here (49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) and (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2}). The product is \(7^{1}=7\).
Step 2
Why this answer is correct
The correct answer is A. (1). Here (49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) and (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2}). The product is \(7^{1}=7\).
Step 3
Exam Tip
(49^{\frac{3}{2}}=\(7^{2}\)^{\frac{3}{2}}=7^{3}) और (343^{-\frac{2}{3}}=\(7^{3}\)^{-\frac{2}{3}}=7^{-2})। गुणनफल \(7^{1}=7\) है।
Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{706}{225}\). Here (\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) and (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), so the sum is \(\frac{625+81}{225}=\frac{706}{225}\). In exams, invert the fraction for negative powers.
Step 3
Exam Tip
(\left\(\frac{3}{5}\right\)^{-2}=\frac{25}{9}) और (\left\(\frac{5}{3}\right\)^{-2}=\frac{9}{25}), इसलिए योग \(\frac{625+81}{225}=\frac{706}{225}\)। परीक्षा में ऋणात्मक घात पर भिन्न उलटें।
The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 2
Why this answer is correct
The correct answer is A. (0). The conjugate product is (13-3=10), and \(\sqrt{100}=10\), so the difference is (0). In exams, simplify conjugate products directly.
Step 3
Exam Tip
संयुग्म गुणनफल (13-3=10) है और \(\sqrt{100}=10\), इसलिए अंतर (0) है। परीक्षा में संयुग्म गुणनफल को तुरंत परिमेय करें।
Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{1}{2}\). Here (32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), and (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8}). The product is \(\frac{1}{2}\).
Step 3
Exam Tip
(32^{\frac{2}{5}}=\(2^{5}\)^{\frac{2}{5}}=2^{2}=4), और (4^{-\frac{3}{2}}=\(2^{2}\)^{-\frac{3}{2}}=2^{-3}=\frac{1}{8})। गुणनफल \(\frac{1}{2}\) है।
(\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) and (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), so the product is (6). In exams, invert the fraction for negative powers.
Step 2
Why this answer is correct
The correct answer is A. (6). (\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) and (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), so the product is (6). In exams, invert the fraction for negative powers.
Step 3
Exam Tip
(\left\(\frac{2}{3}\right\)^{-3}=\left\(\frac{3}{2}\right\)^{3}=\frac{27}{8}) और (\left\(\frac{9}{4}\right\)^{-1}=\frac{4}{9}), इसलिए गुणनफल (6) है। परीक्षा में ऋणात्मक घात पर भिन्न उलटें।
The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
Step 2
Why this answer is correct
The correct answer is A. \(2+2\sqrt{5}\). The first product is (7-5=2), and \(\sqrt{20}=2\sqrt{5}\), so the answer is \(2+2\sqrt{5}\). In exams, identify the conjugate product first.
Step 3
Exam Tip
पहला गुणनफल (7-5=2) है और \(\sqrt{20}=2\sqrt{5}\), इसलिए उत्तर \(2+2\sqrt{5}\) है। परीक्षा में पहले संयुग्म गुणनफल पहचानें।
Here (\left\(\frac{2}{3}\right\)0=1) and (\left\(\frac{1}{2}\right\)2=\frac{1}{4}). Therefore the sum is \(\frac{5}{4}\).
Step 2
Why this answer is correct
The correct answer is C. \(\frac{5}{4}\). Here (\left\(\frac{2}{3}\right\)0=1) and (\left\(\frac{1}{2}\right\)2=\frac{1}{4}). Therefore the sum is \(\frac{5}{4}\).
Step 3
Exam Tip
(\left\(\frac{2}{3}\right\)0=1) और (\left\(\frac{1}{2}\right\)2=\frac{1}{4}) है। इसलिए योग \(\frac{5}{4}\) है।
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2-b-2) का रूप है। चरण 2: (\(\sqrt{13}\)2-\(\sqrt{5}\)2=13-5=8)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a+b)(a-b)=a-2-b-2) का रूप है। चरण 2: (42-\(\sqrt{7}\)2=16-7=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2-b-2) का रूप है। चरण 2: (\(\sqrt{11}\)2-\(\sqrt{2}\)2=11-2=9)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
In conjugate multiplication, directly use difference of squares. चरण 1: यह ((a+b)(a-b)=a-2-b-2) का रूप है। चरण 2: (32-\(\sqrt{5}\)2=9-5=4)। चरण 3: संयुग्म गुणन में वर्गों का अंतर सीधे लगाएं।
In conjugate multiplication, directly use the difference of squares. चरण 1: यह ((a-b)(a+b)=a-2-b-2) का रूप है। चरण 2: (\(\sqrt{7}\)2-\(\sqrt{3}\)2=7-3=4)। चरण 3: संयुग्म गुणन में सीधे वर्गों का अंतर लगाएं।
For conjugate products, difference of squares gives the answer quickly. चरण 1: यह ((a+b)(a-b)=a-2-b-2) का रूप है। चरण 2: (22-\(\sqrt{3}\)2=4-3=1)। चरण 3: संयुग्म रूप वाले गुणन में वर्गों का अंतर जल्दी उत्तर देता है।
A. सत्यजीत राय फिर ए पी जे अब्दुल कलाम फिर एम एस सुब्बुलक्ष्मी फिर अमर्त्य सेन/Satyajit Ray then A P J Abdul Kalam then M S Subbulakshmi then Amartya Sen
Step 1
Concept
Satyajit Ray was honoured in 1992 Kalam in 1997 Subbulakshmi in 1998 and Amartya Sen in 1999. Remember the sequence of years.
Step 2
Why this answer is correct
The correct answer is A. सत्यजीत राय फिर ए पी जे अब्दुल कलाम फिर एम एस सुब्बुलक्ष्मी फिर अमर्त्य सेन / Satyajit Ray then A P J Abdul Kalam then M S Subbulakshmi then Amartya Sen. Satyajit Ray was honoured in 1992 Kalam in 1997 Subbulakshmi in 1998 and Amartya Sen in 1999. Remember the sequence of years.
Step 3
Exam Tip
सत्यजीत राय 1992 कलाम 1997 सुब्बुलक्ष्मी 1998 और अमर्त्य सेन 1999 में सम्मानित हुए। परीक्षा में लगातार वर्षों का क्रम याद रखें।
Satyajit Ray Maulana Azad and J R D Tata were all honoured in 1992. Remember the 1992 group together.
Step 2
Why this answer is correct
The correct answer is A. जे आर डी टाटा / J R D Tata. Satyajit Ray Maulana Azad and J R D Tata were all honoured in 1992. Remember the 1992 group together.
Step 3
Exam Tip
सत्यजीत राय मौलाना आजाद और जे आर डी टाटा तीनों 1992 में सम्मानित हुए। परीक्षा में 1992 का समूह साथ याद रखें।
D. प्रकाश तर्क असंगत है/Light logic is inconsistent
Step 1
Concept
Highlight should face light and shadow should fall opposite. Exam tip: check light consistency.
Step 2
Why this answer is correct
The correct answer is D. प्रकाश तर्क असंगत है / Light logic is inconsistent. Highlight should face light and shadow should fall opposite. Exam tip: check light consistency.
Step 3
Exam Tip
हाइलाइट प्रकाश की ओर और छाया विपरीत दिशा में होनी चाहिए। परीक्षा में light consistency जांचें।
Ghaghara Gandak and Kosi are left bank Himalayan tributaries of Ganga. For exams remember Son as a right bank tributary.
Step 2
Why this answer is correct
The correct answer is C. घाघरा गंडक और कोसी / Ghaghara Gandak and Kosi. Ghaghara Gandak and Kosi are left bank Himalayan tributaries of Ganga. For exams remember Son as a right bank tributary.
Step 3
Exam Tip
घाघरा गंडक और कोसी गंगा की बाएं तट की हिमालयी सहायक नदियां हैं। परीक्षा में सोन को दाहिने तट की नदी याद रखें।
Substituting (\left\(\frac{5}{2},-\frac{3}{2}\right\)) makes both \(2x+y=\frac{7}{2}\) and \(x-2y=\frac{11}{2}\) true. Check the intersection point in both equations.
Step 2
Why this answer is correct
The correct answer is A. \(2x+y=\frac{7}{2}\), \(x-2y=\frac{11}{2}\). Substituting (\left\(\frac{5}{2},-\frac{3}{2}\right\)) makes both \(2x+y=\frac{7}{2}\) and \(x-2y=\frac{11}{2}\) true. Check the intersection point in both equations.
Step 3
Exam Tip
(\left\(\frac{5}{2},-\frac{3}{2}\right\)) रखने पर \(2x+y=\frac{7}{2}\) और \(x-2y=\frac{11}{2}\) दोनों सत्य हैं। प्रतिच्छेद बिंदु को दोनों समीकरणों में जांचें।
Substituting (\left\(-\frac{3}{2},4\right\)) makes both (2x+y=1) and \(x+2y=\frac{13}{2}\) true. The intersection point should be checked in both equations.
Step 2
Why this answer is correct
The correct answer is A. (2x+y=1), \(x+2y=\frac{13}{2}\). Substituting (\left\(-\frac{3}{2},4\right\)) makes both (2x+y=1) and \(x+2y=\frac{13}{2}\) true. The intersection point should be checked in both equations.
Step 3
Exam Tip
(\left\(-\frac{3}{2},4\right\)) रखने पर (2x+y=1) और \(x+2y=\frac{13}{2}\) दोनों सत्य हैं। प्रतिच्छेद बिंदु को दोनों समीकरणों में जांचना चाहिए।
Substituting (\left\(\frac{7}{2},-\frac{1}{2}\right\)) makes (x-y=4) and \(2x+y=\frac{13}{2}\) true. Check the point in both equations.
Step 2
Why this answer is correct
The correct answer is A. (x-y=4), \(2x+y=\frac{13}{2}\). Substituting (\left\(\frac{7}{2},-\frac{1}{2}\right\)) makes (x-y=4) and \(2x+y=\frac{13}{2}\) true. Check the point in both equations.
Step 3
Exam Tip
(\left\(\frac{7}{2},-\frac{1}{2}\right\)) रखने पर (x-y=4) और \(2x+y=\frac{13}{2}\) सत्य हैं। विकल्पों में बिंदु को दोनों समीकरणों में जांचें।
Here \(x+y=\frac{7}{2}+\frac{9}{2}=\frac{16}{2}=8\). Values of (x) and (y) are read directly from the intersection point.
Step 2
Why this answer is correct
The correct answer is A. (8). Here \(x+y=\frac{7}{2}+\frac{9}{2}=\frac{16}{2}=8\). Values of (x) and (y) are read directly from the intersection point.
Step 3
Exam Tip
यहाँ \(x+y=\frac{7}{2}+\frac{9}{2}=\frac{16}{2}=8\)। प्रतिच्छेद बिंदु से (x) और (y) के मान सीधे पढ़े जाते हैं।
\(3.75=\frac{15}{4}\) and \(-2.5=-\frac{5}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{15}{4},-\frac{5}{2}\right\)). \(3.75=\frac{15}{4}\) and \(-2.5=-\frac{5}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 3
Exam Tip
\(3.75=\frac{15}{4}\) और \(-2.5=-\frac{5}{2}\)। दशमलव निर्देशांक को सरल भिन्न में बदलना बेहतर रहता है।
The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.
Step 2
Why this answer is correct
The correct answer is B. \(x=-\frac{5}{2},\ y=3\). The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.
Step 3
Exam Tip
बिंदु में पहला निर्देशांक (x) और दूसरा (y) होता है। ऋण भिन्न निर्देशांक पढ़ते समय क्रम न बदलें।
Here \(x+y=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\). Values of (x) and (y) are read directly from the intersection point.
Step 2
Why this answer is correct
The correct answer is A. (6). Here \(x+y=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\). Values of (x) and (y) are read directly from the intersection point.
Step 3
Exam Tip
यहाँ \(x+y=\frac{5}{2}+\frac{7}{2}=\frac{12}{2}=6\)। प्रतिच्छेद बिंदु से (x) और (y) के मान सीधे पढ़े जाते हैं।
\(2.25=\frac{9}{4}\) and \(-1.5=-\frac{3}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{9}{4},-\frac{3}{2}\right\)). \(2.25=\frac{9}{4}\) and \(-1.5=-\frac{3}{2}\). It is better to convert decimal coordinates into simplified fractions.
Step 3
Exam Tip
\(2.25=\frac{9}{4}\) और \(-1.5=-\frac{3}{2}\)। दशमलव निर्देशांक को सरल भिन्न में बदलना बेहतर रहता है।
The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.
Step 2
Why this answer is correct
The correct answer is B. \(x=-\frac{3}{2},\ y=4\). The first coordinate of a point is (x) and the second is (y). Do not change order while reading negative fraction coordinates.
Step 3
Exam Tip
बिंदु में पहला निर्देशांक (x) और दूसरा (y) होता है। ऋण भिन्न निर्देशांक पढ़ते समय क्रम न बदलें।
A. बिंदु (\left\(3.5,2.5\right\))/Point (\left\(3.5,2.5\right\))
Step 1
Concept
\(\frac{7}{2}=3.5\) and \(\frac{5}{2}=2.5\). While reading a graph, understand the relation between fraction and decimal forms.
Step 2
Why this answer is correct
The correct answer is A. बिंदु (\left\(3.5,2.5\right\)) / Point (\left\(3.5,2.5\right\)). \(\frac{7}{2}=3.5\) and \(\frac{5}{2}=2.5\). While reading a graph, understand the relation between fraction and decimal forms.
Step 3
Exam Tip
\(\frac{7}{2}=3.5\) और \(\frac{5}{2}=2.5\)। ग्राफ पढ़ते समय भिन्न और दशमलव रूप का संबंध समझें।
(\left\(0,0\right\)) satisfies both (2x-y=0) and (x+3y=0). To check origin, put (x=0,\ y=0).
Step 2
Why this answer is correct
The correct answer is B. (2x-y=0) और (x+3y=0) / (2x-y=0) and (x+3y=0). (\left\(0,0\right\)) satisfies both (2x-y=0) and (x+3y=0). To check origin, put (x=0,\ y=0).
Step 3
Exam Tip
(\left\(0,0\right\)) दोनों समीकरणों (2x-y=0) और (x+3y=0) को संतुष्ट करता है। मूलबिंदु की जाँच में (x=0,\ y=0) रखें।
\(4.5=\frac{9}{2}\) and \(1.5=\frac{3}{2}\). Write decimal coordinates read from a graph as simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. (\left\(\frac{9}{2},\frac{3}{2}\right\)). \(4.5=\frac{9}{2}\) and \(1.5=\frac{3}{2}\). Write decimal coordinates read from a graph as simplified fractions.
Step 3
Exam Tip
\(4.5=\frac{9}{2}\) और \(1.5=\frac{3}{2}\)। ग्राफ से मिले दशमलव निर्देशांक को सरल भिन्न में लिखें।
In the point (\left\(-4,3\right\)), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.
Step 2
Why this answer is correct
The correct answer is A. (x=-4,\ y=3). In the point (\left\(-4,3\right\)), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.
Step 3
Exam Tip
बिंदु (\left\(-4,3\right\)) में पहला निर्देशांक (x) और दूसरा (y) होता है। ऋण निर्देशांक में क्रम न बदलें।
In ( \left\(-3,2\right\) ), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.
Step 2
Why this answer is correct
The correct answer is B. (x=-3,\ y=2). In ( \left\(-3,2\right\) ), the first coordinate is (x) and the second is (y). Do not change order with negative coordinates.
Step 3
Exam Tip
( \left\(-3,2\right\) ) में पहला निर्देशांक (x) और दूसरा (y) होता है। ऋण निर्देशांक में क्रम नहीं बदलना चाहिए।
\(3.5=\frac{7}{2}\) and \(2.5=\frac{5}{2}\). Write decimal coordinates read from a graph as simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. ( \left\(\frac{7}{2},\frac{5}{2}\right\) ). \(3.5=\frac{7}{2}\) and \(2.5=\frac{5}{2}\). Write decimal coordinates read from a graph as simplified fractions.
Step 3
Exam Tip
\(3.5=\frac{7}{2}\) और \(2.5=\frac{5}{2}\)। ग्राफ से मिले दशमलव निर्देशांक को सरल भिन्न में लिखें।
In the point ( \left\(-2,5\right\) ), the first coordinate is (x) and the second is (y). Do not change the order while reading negative coordinates.
Step 2
Why this answer is correct
The correct answer is B. (x=-2,\ y=5). In the point ( \left\(-2,5\right\) ), the first coordinate is (x) and the second is (y). Do not change the order while reading negative coordinates.
Step 3
Exam Tip
बिंदु ( \left\(-2,5\right\) ) में पहला निर्देशांक (x) और दूसरा (y) है। ऋण निर्देशांक पढ़ते समय क्रम न बदलें।
\(2.5=\frac{5}{2}\) and \(1.5=\frac{3}{2}\). When reading decimals from a graph, write them as simplified fractions.
Step 2
Why this answer is correct
The correct answer is A. ( \left\(\frac{5}{2},\frac{3}{2}\right\) ). \(2.5=\frac{5}{2}\) and \(1.5=\frac{3}{2}\). When reading decimals from a graph, write them as simplified fractions.
Step 3
Exam Tip
\(2.5=\frac{5}{2}\) और \(1.5=\frac{3}{2}\)। ग्राफ से दशमलव बिंदु पढ़ने पर सरल भिन्न में लिखें।
\(\sqrt{6}\approx2.45\), so \(-\sqrt{6}\approx-2.45\), which is left of (-2). The smaller number lies to the left.
Step 2
Why this answer is correct
The correct answer is A. \(-\sqrt{6}\). \(\sqrt{6}\approx2.45\), so \(-\sqrt{6}\approx-2.45\), which is left of (-2). The smaller number lies to the left.
Step 3
Exam Tip
\(\sqrt{6}\approx2.45\), इसलिए \(-\sqrt{6}\approx-2.45\) है जो (-2) से बाईं ओर है। छोटी संख्या बाईं ओर होती है।
Moving left decreases the number, so the point is \(2-\frac{3}{4}\). In exams, choose addition or subtraction according to direction.
Step 2
Why this answer is correct
The correct answer is B. \(2-\frac{3}{4}\). Moving left decreases the number, so the point is \(2-\frac{3}{4}\). In exams, choose addition or subtraction according to direction.
Step 3
Exam Tip
बाईं ओर जाने पर संख्या घटती है, इसलिए बिंदु \(2-\frac{3}{4}\) होगा। परीक्षा में दिशा के अनुसार जोड़ या घटाव चुनें।
(p\left\(\frac{1}{3}\right\)=3\left\(\frac{1}{3}\right\)2+2\left\(\frac{1}{3}\right\)-1=0). Use brackets while substituting fractions.
Step 2
Why this answer is correct
The correct answer is A. (0). (p\left\(\frac{1}{3}\right\)=3\left\(\frac{1}{3}\right\)2+2\left\(\frac{1}{3}\right\)-1=0). Use brackets while substituting fractions.
Step 3
Exam Tip
(p\left\(\frac{1}{3}\right\)=3\left\(\frac{1}{3}\right\)2+2\left\(\frac{1}{3}\right\)-1=0) है। भिन्न रखते समय कोष्ठक लगाएँ।
We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{4x^{3}y^{4}}{5}\). We get (\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{4x^{3}y^{4}}{5}\).
Step 3
Exam Tip
(\left\(\frac{125x^{-9}}{64y^{12}}\right\)^{\frac{1}{3}}=\frac{5x^{-3}}{4y^{4}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{4x^{3}y^{4}}{5}\) मिलता है।
Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
Step 2
Why this answer is correct
The correct answer is A. \(9r^{6}s^{-8}\). Inside, \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\). Raising to (-1) gives \(9r^{6}s^{-8}\).
Step 3
Exam Tip
अंदर \(\frac{9r^{-4}s^{3}}{81r^{2}s^{-5}}=\frac{1}{9}r^{-6}s^{8}\) है। (-1) घात लेने पर \(9r^{6}s^{-8}\) मिलता है।
Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{2}}{y^{2}z^{2}}\). Inside, \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), so its reciprocal is \(x^{4}y^{-5}z^{-2}\). Multiplying by \(\frac{y^{3}}{x^{2}z^{4}}\) gives \(\frac{x^{2}}{y^{2}z^{6}}\).
Step 3
Exam Tip
अंदर \(\frac{x^{-4}y^{5}}{z^{-2}}=x^{-4}y^{5}z^{2}\), इसलिए उल्टा \(x^{4}y^{-5}z^{-2}\) है। \(\frac{y^{3}}{x^{2}z^{4}}\) से गुणा करने पर \(\frac{x^{2}}{y^{2}z^{6}}\) मिलता है।
Since (\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125}). In exams, take the square root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{343}{125}\). Since (\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125}). In exams, take the square root first.
Step 3
Exam Tip
(\left\(\frac{25}{49}\right\)^{\frac{1}{2}}=\frac{5}{7}), इसलिए (\left\(\frac{25}{49}\right\)^{-\frac{3}{2}}=\left\(\frac{5}{7}\right\)^{-3}=\frac{343}{125})। परीक्षा में पहले वर्गमूल निकालें।
Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). Inside, \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), and its square is \(16x^{-16}y^{12}\). Multiplying by \(\frac{x^{16}}{16y^{12}}\) gives (1).
Step 3
Exam Tip
अंदर \(\frac{8x^{-3}y^{2}}{2x^{5}y^{-4}}=4x^{-8}y^{6}\), इसका वर्ग \(16x^{-16}y^{12}\) है। फिर \(\frac{x^{16}}{16y^{12}}\) से गुणा करने पर (1) मिलता है।
Since (\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the fourth root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{64}{27}\). Since (\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27}). In exams, take the fourth root first.
Step 3
Exam Tip
(\left\(\frac{81}{256}\right\)^{\frac{1}{4}}=\frac{3}{4}), इसलिए (\left\(\frac{81}{256}\right\)^{-\frac{3}{4}}=\left\(\frac{3}{4}\right\)^{-3}=\frac{64}{27})। परीक्षा में पहले चौथा मूल निकालें।
Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x}{4}\). Here \(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), so its reciprocal is \(\frac{x^{5}}{4}\), and multiplying by \(x^{-4}\) gives \(\frac{x}{4}\). In exams, simplify the bracket first.
Step 3
Exam Tip
\(\frac{4x^{-2}}{x^{3}}=4x^{-5}\), इसलिए व्युत्क्रम \(\frac{x^{5}}{4}\) है और \(x^{-4}\) से गुणा करने पर \(\frac{x}{4}\) मिलता है। परीक्षा में पहले कोष्ठक को सरल करें।
We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{3x^{2}y^{3}}{4}\). We get (\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}}). The power \(-\frac{1}{3}\) gives the reciprocal \(\frac{3x^{2}y^{3}}{4}\).
Step 3
Exam Tip
(\left\(\frac{64x^{-6}}{27y^{9}}\right\)^{\frac{1}{3}}=\frac{4x^{-2}}{3y^{3}})। \(-\frac{1}{3}\) घात लेने पर व्युत्क्रम \(\frac{3x^{2}y^{3}}{4}\) मिलता है।
Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 2
Why this answer is correct
The correct answer is A. \(7r^{5}s^{-6}\). Inside, \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\). Raising to (-1) gives \(7r^{5}s^{-6}\).
Step 3
Exam Tip
अंदर \(\frac{7r^{-3}s^{2}}{49r^{2}s^{-4}}=\frac{1}{7}r^{-5}s^{6}\) है। (-1) घात लेने पर \(7r^{5}s^{-6}\) मिलता है।
Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).
Step 2
Why this answer is correct
The correct answer is A. \(\frac{z}{xy^{2}}\). Inside, \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), so its reciprocal is \(x^{2}y^{-4}z^{-3}\). Multiplying by \(\frac{y^{2}}{x^{3}z^{2}}\) gives \(\frac{1}{xy^{2}z^{5}}\).
Step 3
Exam Tip
अंदर \(\frac{x^{-2}y^{4}}{z^{-3}}=x^{-2}y^{4}z^{3}\), इसलिए उल्टा \(x^{2}y^{-4}z^{-3}\) है। \(\frac{y^{2}}{x^{3}z^{2}}\) से गुणा करने पर \(\frac{1}{xy^{2}z^{5}}\) मिलता है।
Since (\left\(\frac{16}{81}\right\)^{\frac{1}{4}}=\frac{2}{3}), (\left\(\frac{16}{81}\right\)^{-\frac{3}{4}}=\left\(\frac{2}{3}\right\)^{-3}=\frac{27}{8}). In exams, take the fourth root first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{27}{8}\). Since (\left\(\frac{16}{81}\right\)^{\frac{1}{4}}=\frac{2}{3}), (\left\(\frac{16}{81}\right\)^{-\frac{3}{4}}=\left\(\frac{2}{3}\right\)^{-3}=\frac{27}{8}). In exams, take the fourth root first.
Step 3
Exam Tip
(\left\(\frac{16}{81}\right\)^{\frac{1}{4}}=\frac{2}{3}), इसलिए (\left\(\frac{16}{81}\right\)^{-\frac{3}{4}}=\left\(\frac{2}{3}\right\)^{-3}=\frac{27}{8})। परीक्षा में चौथा मूल पहले निकालें।
Inside, \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), and its square is \(4x^{-12}y^{8}\). Multiplying by \(\frac{x^{12}}{4y^{8}}\) gives (1).
Step 2
Why this answer is correct
The correct answer is A. (1). Inside, \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), and its square is \(4x^{-12}y^{8}\). Multiplying by \(\frac{x^{12}}{4y^{8}}\) gives (1).
Step 3
Exam Tip
अंदर \(\frac{6x^{-2}y^{3}}{3x^{4}y^{-1}}=2x^{-6}y^{4}\), इसका वर्ग \(4x^{-12}y^{8}\) है। फिर \(\frac{x^{12}}{4y^{8}}\) से गुणा करने पर (1) मिलता है।
Since (\left\(\frac{64}{125}\right\)^{\frac{1}{3}}=\frac{4}{5}), (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}=\left\(\frac{4}{5}\right\)^{-2}=\frac{25}{16}). In exams, take the cube root first.
Step 2
Why this answer is correct
The correct answer is B. \(\frac{25}{16}\). Since (\left\(\frac{64}{125}\right\)^{\frac{1}{3}}=\frac{4}{5}), (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}=\left\(\frac{4}{5}\right\)^{-2}=\frac{25}{16}). In exams, take the cube root first.
Step 3
Exam Tip
(\left\(\frac{64}{125}\right\)^{\frac{1}{3}}=\frac{4}{5}), इसलिए (\left\(\frac{64}{125}\right\)^{-\frac{2}{3}}=\left\(\frac{4}{5}\right\)^{-2}=\frac{25}{16})। परीक्षा में पहले घनमूल निकालें।
Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}). In exams, subtract the inner exponents first.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{x^{6}}{4}\). Inside, \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\), so (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4}). In exams, subtract the inner exponents first.
Step 3
Exam Tip
अंदर \(\frac{2x^{-3}}{x^{2}}=2x^{-5}\) है, इसलिए (\left\(2x^{-5}\right\)^{-2}x^{-4}=\frac{x^{10}}{4}x^{-4}=\frac{x^{6}}{4})। परीक्षा में पहले अंदर की घातें घटाएं।
We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.
Step 2
Why this answer is correct
The correct answer is A. \(\frac{2xy^{2}}{3}\). We get (\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), so the power \(-\frac{1}{3}\) gives its reciprocal \(\frac{2xy^{2}}{3}\). In exams, treat the negative fractional power as a reciprocal after rooting.
Step 3
Exam Tip
(\left\(\frac{27x^{-3}}{8y^{6}}\right\)^{\frac{1}{3}}=\frac{3x^{-1}}{2y^{2}}), इसलिए \(-\frac{1}{3}\) घात देने पर उसका व्युत्क्रम \(\frac{2xy^{2}}{3}\) है। परीक्षा में भिन्न घात के बाद ऋणात्मक संकेत को व्युत्क्रम मानें।
