InterviewSolution
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InterviewSolution
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.
| 38151. |
(a-b-c)(a-b-c) |
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| 38152. |
How to prove that root 6 is irrational |
| Answer» First let root 6 is a rational no.Now √ 6= a\\b√6b=a Here a is a rational no and a= √6b Here √6b is a rational no. So our contradiction is wrong so it is proved that √6is irrational no | |
| 38153. |
prove mid-point theorem by using Basic Proportionality theorem(BPT) |
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| 38154. |
Types of subsets |
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| 38155. |
Gjio |
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| 38156. |
6 ohm restance wire is doble up by folding calculate the new restance |
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| 38157. |
0.5 ÷0.25 |
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Answer» 2 2 Its answer is 2 2 |
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| 38158. |
2+2-9-9×2 |
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Answer» -23 By BODMAS 23 Know .. |
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| 38159. |
2^x-2^x^-^1=4 |
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| 38160. |
How can we make equation of class 10 |
| Answer» By formula | |
| 38161. |
Exersise 3.5 |
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| 38162. |
Find root by square method |
| Answer» Give full question please | |
| 38163. |
Find roots by square method |
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| 38164. |
Why p is not equal to 0 |
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Answer» In form of p/q Where |
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| 38165. |
Okk |
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| 38166. |
How to solve the quadratic equation by the method of completing square ? |
| Answer» Say we have a simple expression like x2 + bx. Having x twice in the same expression can make life hard. What can we do?Well, with a little inspiration from Geometry we can convert it, like this:Completing the Square GeometryAs you can see x2 + bx can be rearranged nearly into a square ...... and we can complete the square with (b/2)2In Algebra it looks like this:x2 + bx\t+ (b/2)2\t=\t(x+b/2)2 "Complete the Square" So, by adding (b/2)2 we can complete the square.And (x+b/2)2 has x only once, which is easier to use. | |
| 38167. |
Root 5 is irrational |
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Answer» Yes ,it is It is proof by contradiction method |
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| 38168. |
146 prime factors |
| Answer» 1,2,73 | |
| 38169. |
Prove that:sin square A + cos square A=1 |
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| 38170. |
Mathematics activity book name devjyoti |
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| 38171. |
Where to buy CBSE previous years papers of class 10 all subjects |
| Answer» In penda | |
| 38172. |
Route to |
| Answer» What root to | |
| 38173. |
In triangle abc ab = 6 and de parallel to bc such that ae =1/4ac then the length of ad |
| Answer» Hii | |
| 38174. |
Prove that the 2 + 1 /root2 is irrational no |
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Answer» Root 2 is irrational And when rational is divided by irrational it will alwz give irrational ☺ I mean Contradiction method Solve this question using the assumption and conyradiction method as solved in NCERT. |
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| 38175. |
3ab (3ab) |
| Answer» 9a^2b^2 | |
| 38176. |
Chapter8 how to learn all formulas |
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Answer» practise it... Write in the notebook Ratlo ji |
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| 38177. |
Which subject are u choosing |
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Answer» Mathematics Maths |
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| 38178. |
Solve for x and y by elimination:- 2x+y=5 and 10x-3y=27 |
| Answer» x=21/8 and y=-1/4 | |
| 38179. |
If the hcf(210,55)is expressible in the form 210×5-55y, find y |
| Answer» Find HCf of 210 and 55. .i.e 5 and then equate in the equation 210×5 - 55y=5 which is equal to 19 | |
| 38180. |
Determine the ratio in which 2x+3y-30=0 divides the join of A(3,4) and B(7,8) and at what point |
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| 38181. |
2÷24 |
| Answer» 0.0833333333333 | |
| 38182. |
If the HCF of 65 and 117 is expressible in the form 65 m dash 117 then the value of M is |
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Answer» O M=1.8 |
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| 38183. |
write the largest number. it comes after the other while counting.3 5 9 8 ____ |
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| 38184. |
✓2 |
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| 38185. |
Using prime factorisation find the hcf and lcm of 36,84 |
| Answer» HCF is 12 and LCM 252 | |
| 38186. |
Different between length and breadth is 23m and it\'s perimeter is 206m than what is its area |
| Answer» Let lenght be = xLet breadth be = y =( x-23)Perimeter = 2(l+b) 206 =2(x+y) 206 = 2[x+(x-23)] 206 =2(2x-23) 206 =4x-46 206-46= 4x 160=4x 160/4 = x 40= xTherefore lenght = 40mBreadth =40-23=17m | |
| 38187. |
What is no solution |
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Answer» When a1is not equal to b1. similar a2 and b2. When lines on graph are parallel it is said to be no solution. |
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| 38188. |
What is the difference between a and b? |
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| 38189. |
The sum of two numbers is1000 and the difference between their square is256000 . Find the number |
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Answer» a+b=1000a2-b2=256000(a+b)(a-b)=256000(a-b)=256000÷1000=256a+b=10002a=1000+256=1256a=1256÷2=628 (i).... a+b=1000 and (ii)... (a+b) (a-b) =256000 and then put the value of a+b from (i) in (ii) and you will get a-b=256 it will be eq (iii) again from (iii) a=256+b put this value in (i) and you will get b=372 now again put this value in (i) and you will get a= 628. Okay, it is the solution along with answer |
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| 38190. |
How to factories 6x²-3-7x |
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Answer» (3x+1) (2x-3) Arrange it in descending order Beacuse by middle term splitting the exact answer is not coming 6x^2 - 3 - 7xI thik your question is wrong Factorise using factorisation method (3x+1) (2x-3) |
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| 38191. |
If a+b+c=5 |
| Answer» As we know,\xa0{tex}a ^ { 3 } + b ^ { 3 } + c ^ { 3 } - 3 a b c ={/tex}{tex}( a + b + c ) \\left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - a b - b c - c a \\right){/tex}{tex}= ( a + b + c ) \\left[ a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ( a b + b c + c a ) \\right]{/tex}{tex}= 5 \\left\\{ a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - ( a b + b c + c a ) \\right\\}{/tex}{tex}= 5 \\left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - 10 \\right){/tex}Now,\xa0{tex} a + b + c = 5{/tex}Squaring both sides, we get{tex}( a + b + c ) ^ { 2 } = 5 ^ { 2 }{/tex}{tex}\\Rightarrow a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 ( a b + b c + c a ) = 25{/tex}{tex}\\therefore a ^ { 2 } + b ^ { 2 } + c ^ { 2 } + 2 ( 10 ) = 25{/tex}{tex}\\Rightarrow a ^ { 2 } + b ^ { 2 } + c ^ { 2 } = 25 - 20 = 5{/tex}Now,\xa0{tex}a ^ {3} + b ^ {3} + c ^ { 3 } - 3 a b c = 5 \\left( a ^ { 2 } + b ^ { 2 } + c ^ { 2 } - 10 \\right){/tex}{tex}= 5 ( 5 - 10 ) = 5 ( - 5 ) = - 25{/tex}Hence, proved. | |
| 38192. |
t-2t3-15t |
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| 38193. |
if cos 9a =sina and 9a |
| Answer» Cos 9a =sin a,sin (90-9a)=sina,90-9a=a,90=10a,a=9 so,tan 5×9=tan 45=1 | |
| 38194. |
Explain theorem 1.3 |
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| 38195. |
2x+y=o |
| Answer» 2x+y=02(1)+y=02+y=0Y=-2 | |
| 38196. |
Find zeros of the quadratic polynomial x²+88x+125 |
| Answer» F | |
| 38197. |
Question 5 ex_2.4 |
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| 38198. |
Marking system of 10th cbse 2019 |
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| 38199. |
If 2p,p+10,3p+2 are in A.P then find the value of p |
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Answer» a2+a2=a3+a12a2=a3+a1putting the values2(p+10)=3p+2+2p2p+20=5p+22p-5p=2-20-3p= -18p=-18/ -3here minus cuts minus so we are left with 18/3 =6 therefore p=6 ?? jvjjv 2p+p+10=3p+10=P+10+3p+2=4p+12=3p-4p=12-10=-p=2=p=-2 |
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| 38200. |
All circles are .( Congruent, similar) |
| Answer» Similar | |