30 results found for "construct-ap" in Class 10.
C. \( \sqrt{5} \)
पायथागोरस से कर्ण \( \sqrt{2^2+1^2}=\sqrt{5} \) होगा। वर्गमूल निर्माण में समकोण त्रिभुज उपयोगी है। / By Pythagoras, the hypotenuse is \( \sqrt{2^2+1^2}=\sqrt{5} \). Right triangles help in square-root construction.
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Which legs of a right triangle are most suitable to construct \(\sqrt{17}\) on the number line?
A. (4) और (1)(4) and (1)
क्योंकि \(4^2+1^2=17\), इसलिए कर्ण \(\sqrt{17}\) होगा। संख्या रेखा पर वर्गमूल निर्माण में पाइथागोरस प्रमेय उपयोगी है। / Because \(4^2+1^2=17\), the hypotenuse will be \(\sqrt{17}\). Pythagoras theorem is useful for square-root construction on the number line.
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Which legs of a right triangle can be useful to construct \(\sqrt{3}\) on the number line?
A. \(\sqrt{2}\) और (1)\(\sqrt{2}\) and (1)
क्योंकि (\(\sqrt{2}\)2+12=3), अतः कर्ण \(\sqrt{3}\) होगा। क्रमिक वर्गमूल बनाने में पाइथागोरस का प्रयोग होता है। / Because (\(\sqrt{2}\)2+12=3), the hypotenuse is \(\sqrt{3}\). Successive square roots are constructed using Pythagoras.
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A. (5) और (2)(5) and (2)
\(5^2+2^2=29\), इसलिए कर्ण \(\sqrt{29}\) होगा। भुजाओं के वर्गों का योग जांचें। / \(5^2+2^2=29\), so the hypotenuse will be \(\sqrt{29}\). Check the sum of squares of the sides.
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B. \(\sqrt{17}\)
कर्ण \(=\sqrt{4^2+1^2}=\sqrt{17}\) होगा। ऐसी रचना में पाइथागोरस प्रमेय लगाएं। / The hypotenuse is \(\sqrt{4^2+1^2}=\sqrt{17}\). Use Pythagoras in this type of construction.
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C. \(\sqrt{13}\)
कर्ण \(=\sqrt{3^2+2^2}=\sqrt{13}\) होता है। समकोण त्रिभुज में पाइथागोरस प्रमेय लागू करें। / The hypotenuse is \(\sqrt{3^2+2^2}=\sqrt{13}\). Apply Pythagoras in a right triangle.
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B. \(\sqrt{2}\)
पाइथागोरस से कर्ण \(=\sqrt{1^2+1^2}=\sqrt{2}\) होगा। ऐसी रचनाओं में वर्गों का योग जरूर देखें। / By Pythagoras the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). In such constructions always add the squares of perpendicular sides.
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Which legs are useful to construct \(\sqrt{5}\) on the number line?
A. (1) और (2)(1) and (2)
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। / 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.
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Which right triangle hypotenuse is useful to construct \(\sqrt{2}\) on the number line?
A. भुजाएँ (1) और (1)Legs (1) and (1)
समकोण त्रिभुज की भुजाएँ (1) और (1) होने पर कर्ण \(\sqrt{1^2+1^2}=\sqrt{2}\) होता है। इसी लंबाई को संख्या रेखा पर चाप से रखते हैं। / If the legs are (1) and (1), the hypotenuse is \(\sqrt{1^2+1^2}=\sqrt{2}\). This length is transferred to the number line with an arc.
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If (a=15) and (d=-4), what are the first four terms?
B. (15,11,7,3)
पहला पद (15) है और हर बार (4) घटता है। इसलिए पद (15,11,7,3) होंगे। / The first term is (15), and (4) is subtracted each time. Hence the terms are (15,11,7,3).
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Which sequence is formed with (a=9) and (d=-2)?
B. \(9,7,5,3,\ldots\)
पहला पद (9) है और हर बार (2) घटता है। इसलिए अनुक्रम \(9,7,5,3,\ldots\) बनेगा। / The first term is (9) and (2) is subtracted each time. Hence the sequence is \(9,7,5,3,\ldots\).
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A. (12,17,22,27)
पहला पद (12) है और हर बार (5) जोड़ना है। इसलिए पद (12,17,22,27) होंगे। / The first term is (12) and (5) is added each time. Hence the terms are (12,17,22,27).
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If (a=11) and (d=6), what are the first four terms of the arithmetic progression?
C. (11,17,23,29)
पहला पद (11) है और हर बार (6) जोड़ना है। इसलिए पहले चार पद (11,17,23,29) हैं। / The first term is (11) and (6) is added each time. Therefore, the first four terms are (11,17,23,29).
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Which sequence starts with (a=4) and (d=6)?
A. \(4,10,16,22,\ldots\)
पहला पद (4) है और हर बार (6) जोड़ना है। इसलिए सही अनुक्रम \(4,10,16,22,\ldots\) है। / The first term is (4), and (6) is added each time. Hence the correct sequence is \(4,10,16,22,\ldots\).
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If (a=3) and (d=8), what are the first three terms?
B. (3,11,19)
पहला पद (3) है और हर बार (8) जोड़ना है। इसलिए पहले तीन पद (3,11,19) होंगे। / The first term is (3), and (8) is added each time. Therefore, the first three terms are (3,11,19).
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Which sequence starts with (a=9) and (d=2)?
A. \(9,11,13,15,\ldots\)
पहला पद (9) है और हर बार (2) जोड़ना है। इसलिए सही अनुक्रम \(9,11,13,15,\ldots\) है। / The first term is (9), and (2) is added each time. Hence the correct sequence is \(9,11,13,15,\ldots\).
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If (a=2) and (d=5), what are the first three terms?
B. (2,7,12)
पहला पद (2) है और हर बार (5) जोड़ना है। इसलिए पहले तीन पद (2,7,12) हैं। / The first term is (2), and (5) is added each time. Therefore, the first three terms are (2,7,12).
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A. \(x^2-4x+1\)
योग (4) और गुणनफल (1) है। अतः मोनिक बहुपद \(x^2-4x+1\) होगा। / The sum is (4) and the product is (1). Therefore the monic polynomial is \(x^2-4x+1\).
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Which polynomial has zeroes (-1) and (4)?
A. \(x^2-3x-4\)
शून्यकों से बहुपद ((x+1)(x-4)=x-2-3x-4) बनता है। चिन्ह बदलते समय सावधानी रखें। / The polynomial is ((x+1)(x-4)=x-2-3x-4). Be careful with signs while forming factors.
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A. \(x^2-8x+15\)
नए शून्यकों का योग \(\alpha+\beta+2=8\) और गुणनफल \(\alpha\beta+\alpha+\beta+1=15\) है। अतः बहुपद \(x^2-8x+15\) है। / The new sum is \(\alpha+\beta+2=8\) and product is \(\alpha\beta+\alpha+\beta+1=15\). Thus the polynomial is \(x^2-8x+15\).
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If the sum of zeroes is (-4) and product is (7), which is the monic quadratic polynomial?
A. \(x^2+4x+7\)
\(मोनिक द्विघात (x^2-(\)योग)x+गुणनफल) होता है। इसलिए (x-2-(-4)x+7=x-2+4x+7) है। \(/ A monic quadratic is (x^2-(\)sum)x+product\(). Hence (x^2-(-4)x+7=x^2+4x+7).\)
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If the zeroes of a quadratic polynomial are (2) and (5), what is one monic polynomial?
A. \(x^2-7x+10\)
शून्यक (2) और (5) होने पर बहुपद ((x-2)(x-5)=x-2-7x+10) है। मोनिक बहुपद में \(x^2\) का गुणांक (1) होता है। / With zeroes (2) and (5), the polynomial is ((x-2)(x-5)=x-2-7x+10). A monic polynomial has leading coefficient (1).
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If (4) and (9) are roots of a quadratic equation, which equation can it be?
A. \(x^2-13x+36=0\)
मूल (4) और (9) हों तो ((x-4)(x-9)=0), यानी \(x^2-13x+36=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं। / If roots are (4) and (9), then ((x-4)(x-9)=0), that is \(x^2-13x+36=0\). In exams, form factors with opposite signs of roots.
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If (3) and (7) are roots of a quadratic equation, which equation can it be?
A. \(x^2-10x+21=0\)
मूल (3) और (7) हों तो ((x-3)(x-7)=0), यानी \(x^2-10x+21=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं। / If roots are (3) and (7), then ((x-3)(x-7)=0), that is \(x^2-10x+21=0\). In exams, form factors with opposite signs of roots.
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If (2) and (5) are roots of a quadratic equation, which equation can it be?
A. \(x^2-7x+10=0\)
मूल (2) और (5) हों तो समीकरण ((x-2)(x-5)=0) यानी \(x^2-7x+10=0\) है। परीक्षा में मूलों के विपरीत चिन्ह से गुणनखंड बनाएं। / If the roots are (2) and (5), the equation is ((x-2)(x-5)=0), that is \(x^2-7x+10=0\). In exams, form factors with opposite signs of roots.
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If a number gives quotient 27 and remainder 111 when divided by 112, what is the number?
A. 3135
संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। / Number (=) divisor \(\times\) quotient (+) remainder.
\(112\times27+111=3024+111=3135\)। / \(112\times27+111=3024+111=3135\).
शेषफल 111, भाजक 112 से छोटा है, इसलिए रूप वैध है। / The remainder 111 is less than divisor 112, so the form is valid.
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If a number gives quotient 31 and remainder 95 when divided by 96, what is the number?
A. 3071
संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। / Number (=) divisor \(\times\) quotient (+) remainder.
\(96\times31+95=2976+95=3071\)। / \(96\times31+95=2976+95=3071\).
शेषफल 95, भाजक 96 से छोटा है, इसलिए रूप वैध है। / The remainder 95 is less than divisor 96, so the form is valid.
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If a number gives quotient 24 and remainder 75 when divided by 76, what is the number?
A. 1899
संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। / Number (=) divisor \(\times\) quotient (+) remainder.
\(76\times24+75=1824+75=1899\)। / \(76\times24+75=1824+75=1899\).
शेषफल 75, भाजक 76 से छोटा है, इसलिए यह रूप वैध है। / The remainder 75 is less than divisor 76, so this form is valid.
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If a number gives quotient 16 and remainder 44 when divided by 45, what is the number?
A. 764
संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। / Number (=) divisor \(\times\) quotient (+) remainder.
\(45\times16+44=720+44=764\)। / \(45\times16+44=720+44=764\).
शेषफल 44, भाजक 45 से छोटा है, इसलिए रूप वैध है। / The remainder 44 is less than divisor 45, so the form is valid.
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If a number gives quotient 15 and remainder 31 when divided by 32, what is the number?
A. 511
संख्या \(=भाजक\timesभागफल+शेषफल\) होती है। / Number (=) divisor \(\times\) quotient (+) remainder.
\(32\times15+31=480+31=511\)। / \(32\times15+31=480+31=511\).
शेषफल 31, भाजक 32 से छोटा है, इसलिए रूप वैध है। / Remainder 31 is less than divisor 32, so the form is valid.
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