Saved Bookmarks
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.
| 10951. |
Boy theorem proof |
| Answer» | |
| 10952. |
356678+56779 |
| Answer» | |
| 10953. |
What is probabilities |
| Answer» Sambhavna | |
| 10954. |
What is C.P.C.T ?? Pls explain briefly |
|
Answer» Corresponding Parts of Congruent Triangles\xa0let ABC and DEF are two congruent triangles such that\xa0their corresponding sides are equal i.e AB = DE ,\xa0BC=EF , CA=FD or their corresponding angles are equal i.e LA = LD , LB = LE , LF = LC No bro it is CORRESPONDING PARTS OF CONGRUENT TRIANGLES Congruent parts of Congruent triangles |
|
| 10955. |
Where odd integer is form 2q+1,2q,2q+3 where q is some integers prove by Euclid theorm |
| Answer» | |
| 10956. |
Theorem 5 on triangles |
| Answer» | |
| 10957. |
The sum the first n terms of an AP is (3n²+6n). Find the n th terms and the 15 th terms of this AP. |
| Answer» Here it is given that sum of first n terms\xa0is equal to Sn\xa0= 3n2\xa0+ 6n\xa0Now,\xa0Sn-1\xa0= 3(n - 1)2\xa0+ 6(n - 1)= 3(n2\xa0- 2n + 1) + 6n - 6= 3n2\xa0- 6n + 3 + 6n - 6= 3n2\xa0- 3Therefore nth term, Tn\xa0= Sn\xa0- Sn-1\xa0= 3n2\xa0+ 6n - 3n2\xa0+ 3= 6n + 3And, 15th term T15\xa0=\xa06(15) + 3= 90 + 3 = 93Therefore nth\xa0term is 6n+3 and 15th\xa0term is 93 | |
| 10958. |
Evaluate the following Sin60 cos30+ sin30 cos60 |
| Answer» | |
| 10959. |
Find the distance between the following pair of points.(2,3)(4,1) |
| Answer» w | |
| 10960. |
How can we prove Pythagoras theorem |
| Answer» | |
| 10961. |
By graphical method. 2x+y-5=0X+y-3=0 |
| Answer» | |
| 10962. |
If two vertices of parallelogram are given then how we found other two |
| Answer» | |
| 10963. |
Pls tell the course structure of class10 of PA 2 |
| Answer» | |
| 10964. |
Exercise 13 ,1 |
| Answer» | |
| 10965. |
Now I am solving rd sharam is it good? |
| Answer» I too solvind rd sharam it is too good | |
| 10966. |
Using Euclid division algorithm, find the HCF of 156and504 |
| Answer» | |
| 10967. |
Find a point on the y-axis which is equidistant from the points A(6,5) and B(-4,3). |
| Answer» We have to find a point\xa0on the y-axis which is equidistant from the points A(6,\xa05) and B (- 4, 3).We know that a point on y-axis is of the form (0, y). So, let the required point be P (0, y).Then,PA = PB{tex}\\Rightarrow \\sqrt { ( 0 - 6 ) ^ { 2 } + ( y - 5 ) ^ { 2 } } = \\sqrt { ( 0 + 4 ) ^ { 2 } + ( y - 3 ) ^ { 2 } }{/tex}{tex}\\Rightarrow{/tex}\xa036 + (y - 5)2 = 16 + (y - 3)2{tex}\\Rightarrow{/tex}\xa036 + y2\xa0- 10y + 25 = 16 + y2 - 6y + 9{tex}\\Rightarrow{/tex}\xa04y = 36{tex}\\Rightarrow{/tex}\xa0y = 9So, the required point is (0, 9). | |
| 10968. |
2x+3y=34xy |
| Answer» Divide throughout by xy. | |
| 10969. |
What is the geometrical meaning of zeroes of a polynomial? |
| Answer» | |
| 10970. |
Which term of the AP 3,8,13,18.......is 88 |
| Answer» 17th term of ap is 88 | |
| 10971. |
If the sum of first n,2n and 3n terms of Ap be s1,s2&s3 respectively than prove that s3=3 (s2-s1) |
| Answer» Let a be the first term and d be the common difference of the given AP. Then,S1 = sum of first n terms of the given AP,S2 = sum of first 2n terms of the given AP,S3 = sum of first 3n terms of the given AP.S1 ={tex}\\frac{n}{2}{/tex}{tex}\\cdot{/tex}{2a+(n-1)d}, S2={tex}\\frac{{2n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(2n-1)d}, and S3= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}{2a+(3n-1)d}{tex}\\Rightarrow{/tex}3(S2-S1) = 3{tex}\\cdot{/tex}[{2na+n(2n - 1)d}\xa0- {na+{tex}\\frac{1}{2}{/tex}n(n-1)d}]= 3{tex}\\cdot{/tex}[na + {tex}\\frac{3}{2}{/tex}n2d-{tex}\\frac{1}{2}{/tex}nd] = {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+3nd - d]= {tex}\\frac{{3n}}{2}{/tex}{tex}\\cdot{/tex}[2a+(3n-1)d}=S3.Hence, S3=3(S2-S1). | |
| 10972. |
Real numbera |
| Answer» | |
| 10973. |
If angle a to=40 find angle aob |
| Answer» | |
| 10974. |
1 divide by 17 complete ans |
| Answer» | |
| 10975. |
How to find a value of k in polynomial |
| Answer» | |
| 10976. |
Solve for x and y :2(ax + by) + (a + 4b) = 0;2(bx + ay) + (b-4a) = 0 |
| Answer» ?? | |
| 10977. |
Is there any natural number which cannot be divided by 1 |
| Answer» No | |
| 10978. |
Find two number whose sum is 27 and product is 182 |
| Answer» 13 and 14 | |
| 10979. |
If sec A = x+1/x, prove that sec A *tan A= 2x or 1/2x |
| Answer» | |
| 10980. |
What is linear combination and how to express a term in linear combination? |
| Answer» | |
| 10981. |
what is the value of any number with power -1 |
|
Answer» That numbers itself in subtraction 1 |
|
| 10982. |
In the figure, angle B=90° AD=CD=26m, BD=10m & AE=CD |
| Answer» | |
| 10983. |
\'5\'6\'6\'5\'=3\'[email\xa0protected]!5!2!5 |
| Answer» | |
| 10984. |
what do aaa criteria say |
|
Answer» According to similarties of triangle if the corresponding angle of two triangle are equal then we can say the triangle must be similiar Aaa criteria says all three angles are equal it is is use to prove similarity of triangles When three angles of two triangle are equal then we say it is aaa criteria |
|
| 10985. |
The sum s of first five even natural number is given s=n(n+1) . find n if sum is 420 |
| Answer» s=n(n+1420 = n(n+1)n2+n- 420=0n2+21n - 20n - 420 =0n(n+21) - 20(n+21)=0n-20 = 0 or n+21=0 but n cannot be -vethus n=20check your answer s = 20(20+1)=21x20=420 | |
| 10986. |
If the roots of the equation (b-c)x^2 + (c-a)x +(a-b) =0 are equal then prove that 2b = a+c |
| Answer» {tex} (b-c)x^2 + (c-a)x +(a-b) =0{/tex}roots of te given equation are equal so\xa0{tex}{b^2-4ac}=0{/tex}{tex}\\implies (c-a)^2-4 (b-c)(a-b)=0{/tex}{tex}\\implies c^2-2ac+a^2-4(ab-b^2-ac+bc)=0{/tex}{tex}\\implies c^2-2ac+a^2-4ab+4b^2+4ac-4bc=0{/tex}{tex}\\implies a^2+4b^2+c^2-4ab-4bc+2ac=0{/tex}{tex}\\implies (a-2b+c)^2=0{/tex}{tex}\\implies (a-2b+c)=0{/tex}{tex}\\implies a+c=2b{/tex}{tex}2b= a+c\\; \\; Hence\\; proved{/tex}\xa0 | |
| 10987. |
If A,B,C are the angles of a ∆ABC ,show that sin( B+C/2= cosA/2 |
| Answer» Using the angle sum property of a triangle, we get:{tex}sin(\\frac{B+C}{2}){/tex}\xa0=\xa0{tex}sin(\\frac{180^\\circ - A}{2} ){/tex}\xa0[{tex}\\,\\because A + B + C = 180^\\circ{/tex}]or {tex}sin(\\frac{B+C}{2}){/tex}\xa0=\xa0{tex}sin(90^\\circ - \\frac{A}{2}){/tex}But sin and cos are complementary ratios; {tex}sin(90^\\circ-\xa0\\omega) = cos\\,\\omega{/tex}\xa0{tex}\\therefore sin(\\frac{B+C}{2}){/tex}\xa0= {tex}cos(\\frac{A}{2}){/tex}\xa0 | |
| 10988. |
If tan thita + 1/tan thita= 2, find the value of tan square thita + 1/tan square thita |
| Answer» {tex}given \\; \\; \\tan \\theta + {1\\over \\tan \\theta}= 2{/tex}to find{tex}\\tan^2 \\theta + {1\\over \\tan^2 \\theta}{/tex}Solution: {tex} \\tan \\theta + {1\\over \\tan \\theta}= 2{/tex}Squaring both side{tex}\\implies( \\tan \\theta+ {1\\over \\tan \\theta})^2= (2)^2{/tex}{tex}\\implies \\tan^2 \\theta + {1\\over \\tan^2 \\theta}+2\\tan \\theta\\times {1\\over \\tan \\theta}=4{/tex}{tex}\\implies \\tan^2 \\theta + {1\\over \\tan^2 \\theta}+2=4{/tex}{tex}\\implies \\tan^2 \\theta + {1\\over \\tan^2 \\theta}=4-2{/tex}{tex}\\implies \\tan^2 \\theta + {1\\over \\tan^2 \\theta}=2{/tex}\xa0\xa0 | |
| 10989. |
1+cosA + sinA / cosA + sinA= secA + TanA |
| Answer» | |
| 10990. |
If a+b=(a/b)+(b/a). Then find a^2+b^2 in terms of digits |
| Answer» | |
| 10991. |
Fghhxx |
|
Answer» it states that the line integral of magmetic field over the closed path is directly proportional to the u0\xa0times threading the closed path\xa0 it states that the line integral of magmetic field over the closed path is directly proportional to the u0\xa0times threading the closed path |
|
| 10992. |
What is the diagnal? |
| Answer» a straight line joining two opposite corners of a square, rectangle etc\xa0 | |
| 10993. |
Arithmetic progression |
| Answer» Arithmetic Progression =A\xa0sequence of numbers in\xa0which each differs from the preceding one\xa0by a constant quantity e.g 1, 2, 3, 4, ....... etc , 9, 7, 5, 3, ....... etc | |
| 10994. |
How many multiple of 4 flying between 10 and 250 |
| Answer» The multiples of 4 that lie between 10 and 250 are:12, 16, 20, 24, ...., 248a2 - a1 = 16 - 12 = 4a3 - a2 = 20 - 16 = 4a4 - a3 = 24 - 20 = 4As ak+1 - ak is the same for k = 1, 2, 3, etc.The above list of numbers forms an\xa0AP with the first term a\xa0= 12and the common difference d = 4Last term (l) = 248Let there be n term s in this AP. Then, nth term = l{tex} \\Rightarrow {/tex}\xa0a + (n - 1)d = 248{tex} \\Rightarrow {/tex}\xa012 + (n - 1)4 = 248{tex} \\Rightarrow {/tex}\xa0(n - 1)d = 248 - 12{tex} \\Rightarrow {/tex}\xa0(n - 1) = 236{tex} \\Rightarrow n - 1 = \\frac{{236}}{4}{/tex}{tex} \\Rightarrow {/tex}\xa0n - 1 = 59{tex} \\Rightarrow {/tex}\xa0n = 59 + 1{tex} \\Rightarrow {/tex}\xa0n = 60Hence, 60 multiples of 4 lie between 10 and 250. | |
| 10995. |
What is the formula to know the highest out comes when more than 2 coins are tosed. |
|
Answer» 1 when (1) one coin is tossed we get 2 outcomes either head (H) or\xa0tail (T) i.e. 21\xa0= 22 when (2) two coins are tossed we get 4 outcomes HH, TT, HT, or TH i.e 22\xa0= 43 when (3),three coins ,,,,,,,,,,,,,,,,,,,,,, 8 ,,,,,,,,,, HHH, HHT, HTH, HTT, THH, THT, TTH, TTT i.e 23\xa0= 8 and so on\xa0\xa0 2 raised to power \'the no. of coins tossed\' |
|
| 10996. |
Prove that √7 is an irratonal number |
| Answer» Suppose {tex}\\sqrt7{/tex} is a rational number. That is , {tex}\\sqrt7{/tex} = {tex}\\frac{p}{q}{/tex} for some p{tex}\\in{/tex}Z and q {tex}\\in{/tex}Z. We can assume the fraction is in lowest fraction, That is p and q shares no common factors.Then {tex}\\sqrt7q=p{/tex}Squaring both side we get,{tex}7q^2=p^2{/tex}So {tex}p^2{/tex} is a multiple of 7,let\'s assume {tex}p=7m{/tex}Then, {tex}7q^2=\\left(7m\\right)^2{/tex}{tex}7q^2=49m^2{/tex}Or {tex}q^2=7m^2{/tex}So {tex}q^2{/tex} is a multiple of 7,{tex}\\therefore{/tex} q is multiple of 7Thus p and q shares a common factor.This is contradiction.{tex}\\Rightarrow {/tex}{tex}\\sqrt { 7}{/tex} is an irrational number. | |
| 10997. |
2π |
| Answer» | |
| 10998. |
10 board exams will start from |
| Answer» | |
| 10999. |
To get goods marks tell science and math books Please |
|
Answer» Read the chapter from s.chand And for science learn NÇERT solution from s.chand In order to get good marks in maths you must read any of the follwing booksR.D. Sharma Books, Reliable Books, And Together with mathematics by Rachna Sagar\xa0 |
|
| 11000. |
Find the H.C.F and L.C.M of 18 and 24 by prime factorisation method |
|
Answer» factors of 18 = 2x3x3 24 = 2x2x2x3common factors in both are 2 and 3 onlythus HCF = 2x3 = 6please check your answer divide 18 by 6 the remainder is zero 24 by 6 the remainder is zero if we have to do the LCM and HCF of this by prime factorization we have to divide these two numbers by a number which they are both diesel divisible if we divide this number by 2 we get 9 and 12 after this wedding this by 3 week at 3 and 4 again 3 and we get 1 and 4 and the four finally we get the answer one that the LCM of these two numbers is 2 x 3 multiply 3 and again by 4 18=2×3×324=2×2×2×3Common factor=2×3Hcf=6 |
|