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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.
| 32701. |
Are any two Isosceles triangles similar? |
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Answer» if their sides $angles are similar then they also..but if they not then might be no how can you say it so ? please give reason for your answer No Yes they can be |
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| 32702. |
the product of ria\'s age 3 years ago with her age 5 years later is 240 find the current age |
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Answer» 15 Her current age is 15yrs |
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| 32703. |
Prove pgt tgoerem |
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| 32704. |
Show that n3-n is divisible by 6 where n is a positive integer. |
| Answer» put n=6q and n=6q+1 | |
| 32705. |
In angel OP |
| Answer» What is trigonometry ratio? | |
| 32706. |
Abour rectangle |
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Answer» it is 2D figure it has length & breadthit\'s area is lb&perimeter is 2(l sum b )here:l is length& b is breadth..it\'s all about rectangle Answer |
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| 32707. |
If n is an odd Integer then prove that (n square - 1) is divisible by 8. |
| Answer» Let n = 4q + 1 (an odd integer){tex}\\therefore \\quad n ^ { 2 } - 1 = ( 4 q + 1 ) ^ { 2 } - 1{/tex}{tex}= 16 q ^ { 2 } + 1 + 8 q - 1 \\quad \\text { Using Identity } ( a + b ) ^ { 2 } = a ^ { 2 } + 2 a b + b ^ { 2 }{/tex}{tex}= 16{q^2} + 8q{/tex}{tex}= 8 \\left( 2 q ^ { 2 } + q \\right){/tex}= 8m, which is divisible by 8. | |
| 32708. |
What is odd numbers |
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Answer» The numbers which are divisible by one or itself are called odd numbers. Like 2,3,5,7,11,etc 2and3 The no. which is divisisible by one and itself. |
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| 32709. |
Solve for x X plus 1 /X minus 1 +X minus 2 /X plus 2 =4 minus 2x plus 3 /X minus 2 |
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| 32710. |
What is AP ? Explain with examples |
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Answer» The series of such a number in which the common difference of all the consecutive term are same. Like a, a+d,a+2d --------------------a+nd-d where a is any real number and d is the common difference. A series of a number is called an AP 1,3,5,7,9 It is an AP because it has diffrence of 2 3-1=2 a2 -a 1=d |
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| 32711. |
What is Brock lesnar? |
| Answer» Work | |
| 32712. |
Why we make a equation of given sentence |
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| 32713. |
what is olfactory indicators |
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Answer» Those substances which indicates the presence of acid and base by the smell called as olfactory indicator Those substances whose smell(or odour) changes in acidic or basic solutions are called olfactory indicators. |
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| 32714. |
Definition of series , sequence and progression |
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| 32715. |
Sin^2+cos^2 |
| Answer» 1 | |
| 32716. |
2237+777788 |
| Answer» Co² +H²o | |
| 32717. |
Chapter 9 ki exercise ka 15 question |
| Answer» Check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 32718. |
2x2-7x+3=0 find the roots ,by the method of completing the square |
| Answer» We have 2x2 - 7x + 3 = 0{tex}\\implies2( x^2 - {7 \\over 2}x + {3\\over 2}) = 0{/tex}{tex}\\implies\u200b\u200b x^2 - {7 \\over 2}x + {49 \\over 16} = {-3 \\over 2} +{ 49 \\over 16}{/tex} (Adding 49/16 to both sides){tex}\\implies x^2 -2 \\times x \\times {7 \\over 4} + ({7 \\over 4})^2 = {-24 +49 \\over 16}{/tex}{tex}\\implies (x-{7\\over4})^2 = {25 \\over 16}{/tex}{tex}\\implies x-{7\\over 4}= \\pm \\sqrt({25 \\over 16}){/tex}{tex}\\implies x={7\\over 4} \\pm {5 \\over 4}{/tex}{tex}\\implies x={7\\over 4} + {5 \\over 4}\\, and \\,x={7\\over 4} - {5 \\over 4}{/tex}{tex}\\implies x=3\\, and \\,{1\\over 2}{/tex}{tex}\\therefore{/tex}the roots of the given equation are {tex}3{/tex} and {tex}1\\over 2{/tex}. | |
| 32719. |
Will the periodic test marks included in board percentage? |
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Answer» The marks of best 2 periodic tests will be converted in to % out of 20%. .. This will be internal marks which will be sent to the board. And board will add ip those percentage ti your percentage of final exam.That\'s the new format!???????? Yes?✒ |
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| 32720. |
what is important in 2 chapter |
| Answer» Division algorithms Quadratic polynomial formulaCubic polynomial formula | |
| 32721. |
Show that any positive odd integers is of the form 6q+1,or 6q+3,or6q+5where q is some integers |
| Answer» Let a be any positive integer and b = 6∴ by Euclid’s division lemmaa = bq + r, 0≤ r and q be any integer q ≥ 0∴ a = 6q + r,where, r = 0, 1, 2, 3, 4, 5If a is even then then remainder by division of 6 is 0,2 or 4Hence r=0,2,or 4or A is of form 6q,6q+2,6q+4As, a = 6q = 2(3q), ora = 6q + 2 = 2(3q + 1), ora = 6q + 4 = 2(3q + 2).If these 3 cases a is an even integer.but if the remainder is 1,3 or 5 then r=1,3 or 5or A is of form 6q+1,,6q+3 or,6q+5Case 1:a = 6q + 1 = 2(3q) + 1 = 2n + 1,\xa0Case 2: a = 6q + 3 = 6q + 2 + 1,= 2(3q + 1) + 1 = 2n + 1,\xa0Case 3: a = 6q + 5 = 6q + 4 + 1= 2(3q + 2) + 1 = 2n + 1This shows that odd integer is of the form 6q + 1, or 6q + 3, or 6q + 5, where q is some integer. | |
| 32722. |
Ex-6.5 Q..1 |
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| 32723. |
Exercise6.5 ka ques.3 samajh nii aa raha kya kare |
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Answer» 3 part me ? ACD and BAD 2 part me ? ACB and ACD I part me ? ABC and ? ABD le kar dono ko similar dikhayenge |
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| 32724. |
Class 6 chapter 4 |
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| 32725. |
Theorem 6.8 |
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| 32726. |
How should I present my spinning wheel for my exibition ? Pls help me . |
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| 32727. |
....,8,12,16......,160a=?,n=? |
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Answer» a=4,n=40 So an = a+(n-1) d D = 8 - 12 = 4 |
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| 32728. |
(2,3),(4,1) |
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| 32729. |
3x+10x+8=0 |
| Answer» 13x+8=013x=-8X=-8/13 | |
| 32730. |
For some integer m, every odd integer is of the form (a)m. (b)m+1. (c)2m. (d)2m+1 |
| Answer» 2m as when m=1 then ans wil 2 which is even similarly when m=2 then ans will 4 which is even and so on. | |
| 32731. |
You ask |
| Answer» No thanks....?? | |
| 32732. |
1+cosΦ+sinΦ|1+cosΦ+sinΦ=1+sinΦ|cosΦ |
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| 32733. |
Factorise x^2+x+30 |
| Answer» x2 + x - 30 = x2 + 6x - 5x + 30= (x +6) (x -5)x = -6 , x = 5 | |
| 32734. |
If one of the zero of the polynomial ax^2+bx+c is double the other then prove that 2b^2=9ac |
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| 32735. |
How can we use step deviation method if class interval is not equal |
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| 32736. |
How can we use step deviation method of class interval are the equal |
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| 32737. |
Express each number as a product of its prime factors (1) 140 |
| Answer» So, the prime factors of 140 = 2 {tex}\\times{/tex}\xa02 {tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07 = 22\xa0{tex}\\times{/tex}\xa05 {tex}\\times{/tex}\xa07. | |
| 32738. |
2x+7y=0 |
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| 32739. |
SinQ÷1-cosQ= |
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| 32740. |
What is convex quadrilateral? |
| Answer» A convex quadrilateral is a four sided polygon that has interior angles that measure less than 180 degrees each. | |
| 32741. |
3.5x-9=2.4x+3 |
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| 32742. |
Io |
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| 32743. |
A v |
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| 32744. |
How do we find root 2 |
| Answer» 1.1441 | |
| 32745. |
Given that root 3 is an irrational number prove that 2+3 root 3 is an irrational number. |
| Answer» To Prove: 2+{tex}\\sqrt3{/tex}\xa0is an irratinal number.Given:\xa0{tex}\\sqrt3{/tex}\xa0is irrational number.Proof: Let 2 +\xa0{tex}\\sqrt{3}{/tex}\xa0be a rational number.{tex}\\Rightarrow{/tex}\xa02 +\xa0{tex}\\sqrt{3}{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}, p, q\xa0{tex}\\in{/tex}\xa0I, q\xa0{tex}\\ne{/tex}\xa00\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\sqrt{3}{/tex}\xa0=\xa0{tex}\\frac{p}{q}{/tex}\xa0- 2 =\xa0{tex}\\frac{p - 2q}{q}{/tex} =\xa0{tex}\\frac{integer}{integer}{/tex}\xa0{tex}\\implies{/tex}{tex}\\sqrt{3}{/tex}\xa0is rational number\xa0{tex}\\Rightarrow{/tex} which is a contradiction to the fact that\xa0{tex}\\sqrt{3}{/tex}\xa0is a rational\xa0hence 2 +\xa0{tex}\\sqrt{3}{/tex} is irrational number. | |
| 32746. |
Solve for x 1÷a+b+x=1÷a+1÷b+1÷x |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + x ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { x }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + x ) } - \\frac { 1 } { x } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { x - ( a + b + x ) } { x ( a + b + x ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { x ( a + b + x ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { x ( a + b + x ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0x(a + b + x) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0x2 + ax + bx + ab = 0{tex}\\Rightarrow{/tex}\xa0x(x +a) + b(x +a) = 0{tex}\\Rightarrow{/tex}\xa0(x\xa0+ a) (x + b) = 0{tex}\\Rightarrow{/tex}\xa0x + a = 0 or x + b = 0{tex}\\Rightarrow{/tex}\xa0x = -a or x = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 32747. |
2{4×4} |
| Answer» 64 | |
| 32748. |
Prove tan 2A = cot (A-18) |
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Answer» A= 36 tan 2A =cot(A-18)tan(90-2A) =cot(A-18)Cot(90-2A)=cot(A-18)90-2A=A-1890+18 =A+2A108= 3AA=108\\3A=36 |
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| 32749. |
0/0=2 |
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| 32750. |
Find the value of (-1) + (-1)^2n + (-1)^2n+1 + (-1)^4n+1Where n is any positive integer |
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