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This section includes InterviewSolutions, each offering curated multiple-choice questions to sharpen your knowledge and support exam preparation. Choose a topic below to get started.
| 7601. |
2-√3/2+√3 |
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Answer» 7-4√3 2.8660254037844 |
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| 7602. |
Tell me the easy way of step writing in construction |
| Answer» Nothing is easy in maths | |
| 7603. |
Name two metal which melt in palm |
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Answer» Gallium and cesium.They have low melting point. Sodium and potassium |
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| 7604. |
(a-b)2 |
| Answer» 2a-2b | |
| 7605. |
a² + b²=??? |
| Answer» (a+b)2\xa0-2ab | |
| 7606. |
Iif sin A3÷4find cos and tana |
| Answer» We have,{tex} \\sin A = \\frac { \\text { Perpendicular } } { \\text { Hypotenuse } } = \\frac { 3 } { 4 }{/tex}So, we draw a triangle ABC, right angled at B such that,BC = 3 units ,\xa0AC = 4\xa0units.By Pythagoras theorem, we have{tex} A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }{/tex}{tex} \\Rightarrow \\quad 4 ^ { 2 } = A B ^ { 2 } + 3 ^ { 2 }{/tex}{tex} \\Rightarrow \\quad A B ^ { 2 } = 4 ^ { 2 } - 3 ^ { 2 }{/tex}{tex} \\Rightarrow \\quad A B ^ { 2 } = 7{/tex}{tex} \\Rightarrow \\quad A B = \\sqrt 7 {/tex}When we consider the trigonometrical-ratios of\u200b{tex} \\angle A{/tex}\u200b\u200b\u200b\u200b\u200b\u200b , we have :-Base = AB = 4 units, Perpendicular = BC = 3 units, Hypotenuse = AC = 5 units{tex} \\therefore \\quad \\cos A = \\frac { \\text { Base } } { \\text { Hypotenuse } } = \\frac { 4 } { \\sqrt 7}{/tex}\xa0and,\xa0{tex}\\quad \\tan A = \\frac { \\text { Perpendicular } } { \\text { Base } } = \\frac { 3 } { 4 }{/tex} | |
| 7607. |
Two numbers differ by 3 and their product is 504.find the number |
| Answer» \xa0Let the first number be xAnd second number be x-3x(x-3)=504x2-3x-504=0x2-24x+21x-504=0x(x-24)+21(x-24)=0(x+21) (x-24)=0x=-21andx=24 | |
| 7608. |
Factorise 2x square minus 89x plus 210 |
| Answer» | |
| 7609. |
From which book the question come in mathematics paper in board |
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Answer» Exam will come from maths book not read vut understand. D section me 3 question hataker baki ncert se NCERT |
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| 7610. |
Find the sum of all two digit natura number whish are divisible bye 3 and 6 ? |
| Answer» Ap=102,108,114,------------996an=a+(n-1)d996=102+6n-66n=900n=150Sn=n÷2(2a+(n-1)d | |
| 7611. |
Kane gives 3 chockh slam to demon king finn balor |
| Answer» | |
| 7612. |
Bad news hhh is returning |
| Answer» | |
| 7613. |
New sheild forms dean amberose seth rollins and replacer of roman reigns kurt angle |
| Answer» | |
| 7614. |
Maximum value of sin0 |
| Answer» 0 | |
| 7615. |
Sin^2A+3CosA-2=0 then find Cos^3+Sin^3=???????? |
| Answer» | |
| 7616. |
math kyo bani |
| Answer» Math is for for genius | |
| 7617. |
How I recognize the difference between linear equation sum and quadratic equation sum |
| Answer» Linear equations have heighest degree 1.on the other hand quadratic equations have heighest degree 2. | |
| 7618. |
prove that area of two similar triangle is equal to the ratio oftheir corresponding side |
| Answer» Areas of Similar Triangles NCERT Solutions Grade 10 Given: △ABC ~ △DEF. AP is the median to side BC of △ABC and DQ is the median to side EF of △DEF.ACDF=BCEF {Corresponding sides of similar triangles are proportional}⇒ACDF=2PC2QF=PCQF (1){P is the mid-point of BC and Q is the mid-point of EF} To Prove: ar(△ABC)ar(△DEF)=AP2DQ2 Proof: ar(△ABC)ar(△DEF)=BC2EF2{The ratio of the areas of two similar triangles is equal to the square of the ratio of their corresponding sides}⇒ar(△ABC)ar(△DEF)=(2PC)2(2QF)2=PC2QF2 (2) In △APC and △DQFACDF=PCQF from (1)And, ∠C=∠F {Corresponding angles of similar triangles are equal} Therefore, by SAS similarity criterion, △APC ~ △DQFTherefore, APDQ=PCQF (3) Putting (3) in (2), we getar(△ABC)ar(△DEF)=AP2DQ2 Hence Proved | |
| 7619. |
What is set definition of linear equation |
| Answer» | |
| 7620. |
Mco |
| Answer» | |
| 7621. |
Find lcm and hcf of 100 &190 |
| Answer» HCF\xa0{tex}\\times{/tex}LCM = one number {tex}\\times{/tex} another number=\xa0100 {tex}\\times{/tex}190 = 19000 | |
| 7622. |
If x²- 3kx + 5 -9 is completely divisible by x-3 then find the value of k |
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Answer» K = 5/9 No |
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| 7623. |
Prove that root 5 is a irrational numbet |
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Answer» No 1. This irrationality proof for the square root of 5 uses Fermat\'s method of infinite descent: Suppose that √5 is rational, and express it in lowest possible terms (i.e., as a fully reduced fraction) as mn for natural numbers m and n. Then √5 can be expressed in lower terms as 5n − 2mm − 2n, which is a contradiction. |
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| 7624. |
2+4=--- |
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Answer» I will not send because I am reading in class 10 not in U. K. G. 6 6 6 |
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| 7625. |
a/x+b/y=1/2a/√x+b/√y=1/√2Find x and y |
| Answer» | |
| 7626. |
If ax2+bx+c=a(x-p)² then find the relation between a ,b,c |
| Answer» | |
| 7627. |
if alpha and beta are zeroes |
| Answer» Then take alpha as a 1 zero and beta as a one zero put it in ratio and then you got answee | |
| 7628. |
Solve x/2x+1 + 2x+1/x = 29 |
| Answer» | |
| 7629. |
When was board paper start |
| Answer» Central Board of Secondary Education (CBSE) will conduct the class 10 board examination in the month of March, contrary to the ongoing rumours about exam being held in February. Students, who will be appearing for the examination, next year must start their preparation, now onwards. The Board has released sample question papers and marking scheme for all the main subjects for which the board exam will be held. CBSE class 10 sample question paper and marking scheme for exam 2017-2018 will be available at the official website of the Board at cbse.nic.in or cbseacademic.in. | |
| 7630. |
Find your answer in AP whose |
| Answer» What we have to findMention it here | |
| 7631. |
About the sign convention in briefly |
| Answer» | |
| 7632. |
Find middle term of the AP 10,7,4,......(_62) |
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Answer» -26 There is 25 terms in this series. Therefore 13th term will be the middle term. -23 |
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| 7633. |
Who wrote the ncert book of mathematics |
| Answer» | |
| 7634. |
23+34 |
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Answer» Wrong answer right answer is 57 67 |
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| 7635. |
Euclid lemma division |
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Answer» let a and b be any positive integer. then exist other integer q and r,such that a = bq +r where 0_ |
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| 7636. |
Hjk |
| Answer» | |
| 7637. |
2divide4 |
| Answer» 0.5 | |
| 7638. |
Who has taken sanskrit as main subject |
| Answer» | |
| 7639. |
Sir new patterns ka paper solve paper nhi hai kea |
| Answer» Mr sir hu ate you | |
| 7640. |
Find the value of 60 degree geometrically ? |
| Answer» | |
| 7641. |
Solve for X ,Y give X |
| Answer» | |
| 7642. |
Ra |
| Answer» | |
| 7643. |
In a leap year 53 Mondays come. What is the probability of it ? |
| Answer» 0 | |
| 7644. |
Find the sum of first 32th terms of list of number whose nth term is given by an=3+2/3n |
| Answer» | |
| 7645. |
The dimension of an cone is 950 cm3 and it\'s area of curve is 50 cm2. What is the height of cone ? |
| Answer» 19 | |
| 7646. |
What is tanA and cosA what is the answer if they are divided |
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Answer» Sin/cos×cos These are trigonometric ratios and after division we will get sinA/cos2A |
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| 7647. |
Please solve ex.8.4 in easy method |
| Answer» | |
| 7648. |
Explain therom 10.2 |
| Answer» The lengths of tangents drawn from an external point to a circle are equal. | |
| 7649. |
If x=a sin A +b cos A and y =a cos A+b sin A. Prove thatx^2+y^2=a^2+b^2 |
| Answer» We have, {tex}a\\cos \\theta - b\\sin \\theta = x{/tex}....(i)and {tex}a\\sin \\theta + b\\cos \\theta = y{/tex}...(ii)Squaring Eq. (i) and (ii) and then adding, we get{tex}{x^2} + {y^2} = {(a\\cos \\theta - b\\sin \\theta )^2} + {(a\\sin \\theta + b\\cos \\theta )^2}{/tex}{tex} \\Rightarrow {x^2} + {y^2} = {a^2}{\\cos ^2}\\theta + {b^2}{\\sin ^2}\\theta{/tex}{tex} - 2ab\\cos \\theta \\sin \\theta + {a^2}{\\sin ^2}\\theta + {b^2}{\\cos ^2}\\theta + 2ab\\sin \\theta co\\operatorname{s} \\theta {/tex} [{tex}\\because {/tex} (a\xa0+ b)2 = a2 + b2 + 2ab\xa0and (a\xa0- b)2 = a2 + b2 - 2ab]{tex} \\Rightarrow {x^2} + {y^2} = {a^2}{\\cos ^2}\\theta + {b^2}{\\sin ^2}\\theta + {a^2}{\\sin ^2}\\theta + {b^2}{\\cos ^2}\\theta {/tex}{tex} \\Rightarrow {x^2} + {y^2} = {a^2}({\\cos ^2}\\theta + {\\sin ^2}\\theta ) + {b^2}({\\sin ^2}\\theta + {\\cos ^2}\\theta ){/tex}{tex}[\\because \\sin^2\\theta+\\cos^2\\theta=1]{/tex}{tex}\\Rightarrow x^2+y^2=a^2+b^2{/tex}Hence proved, LHS = RHS | |
| 7650. |
Prove euclid division leema |
| Answer» | |