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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.
| 39401. |
(1÷alfa)×(1÷alfa)+(1÷bita)(1÷bita)= |
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| 39402. |
Tringal |
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Answer» ??? What. to do??? |
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| 39403. |
How to do any digit number square in few seconds |
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Answer» But mam/ sir you didn\'t give me My answer please i need this answer Okkk.. mre log gd night Koi jinda bhi h ya nhi By the help of trick |
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| 39404. |
If 1+1=2 then √1+√1=? |
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Answer» 2 2 2 |
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| 39405. |
A triangle ABC is given such that AB=15 ,BC=27 |
| Answer» What is to be find out?? | |
| 39406. |
If x=2/3 and x=-3 are the roots of the equation ax2+7x+b=0, find the values of a and b |
| Answer» X=2/3. X=-33x-2=0. X+3=0(3x-2)(x+3)3x2+9x-2x-63x2+7x-6So as we can see that on comparing both equation we get that A=3B=-6 | |
| 39407. |
Show that only one of the numbers n, n+2 and n+4 is divisible by 3 |
| Answer» Let the number be (3q + r){tex}n = 3 q + r \\quad 0 \\leq r < 3{/tex}{tex}\\text { or } 3 q , 3 q + 1,3 q + 2{/tex}{tex}\\text { If } n = 3 q \\text { then, numbers are } 3 q , ( 3 q + 1 ) , ( 3 q + 2 ){/tex}{tex}3 q \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 1 \\text { then, numbers are } ( 3 q + 1 ) , ( 3 q + 3 ) , ( 3 q + 4 ){/tex}{tex}( 3 q + 3 ) \\text { is divisible by } 3{/tex}.{tex}\\text { If } n = 3 q + 2 \\text { then, numbers are } ( 3 q + 2 ) , ( 3 q + 4 ) , ( 3 q + 6 ){/tex}{tex}( 3 q + 6 ) \\text { is divisible by } 3{/tex}.{tex}\\therefore \\text { out of } n , ( n + 2 ) \\text { and } ( n + 4 ) \\text { only one is divisible by } 3{/tex}. | |
| 39408. |
The 4th term of an AP is zero.prove that its 25th term is triple its 11th term. |
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Answer» 4th term is 0. i.e., a+3d=0therefore a=-3d25th term is a+24dputting the value of a in 25th term= -3d+24d= 21dand, 11th term is a+10d putting the value of a in 11th term= -3d+10d= 7d21d=3(7d)25th term = 3 of 11th termhence proved..... a+3d=0 a= -3d a25=a+24d=3(a11=a+10d)Put a=-3d-3d+24d=3(-3d+10d)21d=3(7d)21d=21d |
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| 39409. |
How to easily know to values of trigonometry |
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Answer» Ya it\'s one of the tricks. Take 0 1 2 3 4 divide them by 4 and then square root all of them. The values of sin0,30,45,60,90 then cos will be reverse of this and rest of the values can be calculated by using identities There is a easy trick for it but sorry it can\'t be explained here. |
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| 39410. |
What do you mean by priam |
| Answer» Prism is\xa0a solid geometric figure whose two ends are similar, equal, and parallel rectilinear figures, and whose sides are parallelograms. | |
| 39411. |
A circle can have _____parallel tangents at the most |
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Answer» 2 2 |
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| 39412. |
15cot A ,=5/18 find cosA |
| Answer» Cos A = 1/√2917 | |
| 39413. |
Value of k for which the eqution x |
| Answer» Complete the question | |
| 39414. |
Prove that √sec²A+cos²a=tanA+cotA |
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Answer» Ru mad Ranjiv Then do solution Firstly rationlise the lhs side |
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| 39415. |
The sum of first n terror an A. P is 6n+4n.find the 25th term of this AP. |
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Answer» Diksha Singh if you don\'t know then don\'t comment dumb here. 25th term is 10 Google naam ki bhi koi cheez hoti hai Please give me answer |
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| 39416. |
Prove that sinA-cosA+1/sins+cosA-1=1/secA-tanA |
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| 39417. |
SinA-2sin³÷2cos³-cosA=tanA prvove |
| Answer» Take sin common in the. Numerator and cos common in the denominator.use the identity sin^2A + cos^2 A | |
| 39418. |
You have ACP MATHS |
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| 39419. |
An equilateral triangle has two vertices at the point 3 |
| Answer» Let ABC be the equilateral triangle such that,A = (3,0), B=(6,0) and C=(x,y)Distance between:{tex}\\sqrt {( x_{2}-x_{1})^2+(y_{2} -y_{1})^{2} }{/tex}we know that,AB=BC=ACBy distance formula we get,AB=BC=AC=3unitsAC=BC{tex}\\sqrt{(3-x)^2+y^2}=\\sqrt{(6-x)^2+y^2}{/tex}{tex}9+x^2-6 x+y^2=36+x^2-12 x+y^2{/tex}{tex}6 x=27{/tex}{tex}x=27 / 6=9 / 2{/tex}BC = 3 units{tex}\\sqrt{(6-\\frac{27}{6})^2+y^2}=3{/tex}{tex}(\\frac{(36-27)}{6})^2+y^2=9{/tex}{tex}(\\frac{9}{6})^2+y^2=9{/tex}{tex}(\\frac{3}{2})^2+y^2=9{/tex}{tex}\\frac{9}{4}+y^2=9{/tex}{tex}9+4 y^2=36{/tex}{tex}4 y^2=27{/tex}{tex}y^2=\\frac{27}{4}{/tex}{tex}y=\\sqrt{(\\frac{27}{4})}{/tex}{tex}y=3 \\sqrt{\\frac{3}{2}}{/tex}{tex}(x, y)=(9 / 2,3 \\sqrt{\\frac{3}{2}}){/tex}Hence third vertex of equilateral triangle = C =\xa0{tex}(9 / 2,3 \\sqrt{\\frac{3}{2}}){/tex} | |
| 39420. |
Why m1:m2= k:1 |
| Answer» Correct | |
| 39421. |
2x-5x+3=0 |
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Answer» 1 2x - 5x + 3 = 0-3x = -3x = 3/3x = 1 |
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| 39422. |
3.12/8 |
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Answer» 0.39 0.39 0.39 |
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| 39423. |
125372828.7272 divided by 124616 |
| Answer» 1006. 0732 | |
| 39424. |
Under root 2 + under root 3 is irrational prove |
| Answer» Let root 2 + root3 be rational no . So it will be equal to a upon b. Make an equation and square both sides after that you will have your ans | |
| 39425. |
Show that any positive odd integer is of the from6q+1 or 6q+3or 6q+5 where q is some integer |
| Answer» Some positive numbers between 1-6= 1q+2q+3q+4q+5q+6q 2, 4, 6 are even number they divisible by 2 so they cannot be odd positive integer therefpre 6q+1, 6q+3, 6q+5 are odd positive integers | |
| 39426. |
abx2 -(a+b)2 (x-1)=0 solve by quadretic formula |
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| 39427. |
3000+30p divided by 39+p |
| Answer» P = -393000+ 30p30(100+1p)30(100-39)30*611830 | |
| 39428. |
The 17th term of an ap exceeds its 10th term by .7 find the common difference |
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Answer» 7 a +(17-1)d=a+(10-1)d+7a+16d=a+9d+7a-a+16d-9d=77d=7d=7/7d=1 d =1 7d= 7 a+16d = a+9d + 7 |
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| 39429. |
if 5sina + 3cosa = 4,find the value of 3sina-5cosa |
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Answer» +3√2 snd -3√2 -21/5 is the correct answer i think |
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| 39430. |
Chapter 4 exercise 5;2 |
| Answer» You can check NCERT Solutions here :\xa0https://mycbseguide.com/ncert-solutions.html | |
| 39431. |
When we get rd sharma full solution |
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| 39432. |
Cbse class 10th paper 2016 outside Delhi. |
| Answer» You can check last year papers here :\xa0https://mycbseguide.com/cbse-question-papers.html | |
| 39433. |
3+2 under root 5 is irrational prove that |
| Answer» We will prove this by contradiction.Let us suppose that (3+2 {tex}\\sqrt { 5 }{/tex}) is rational.It means that we have co-prime integers a and b such that{tex}\\frac { a } { b } = 3 + 2 \\sqrt { 5 } \\quad \\frac { a } { b } - 3 = 2 \\sqrt { 5 }{/tex}{tex}\\Rightarrow \\frac{{a - 3b}}{b} = 2{\\sqrt 5 \\,{ \\Rightarrow }}\\frac{{a - 3b}}{{2b}} = \\sqrt 5 {/tex} ....(1)a and b are integers.It means L.H.S of (1) is rational but we know that {tex}\\sqrt { 5 }{/tex} is irrational. It is not possible. Therefore, our supposition is wrong. (3+2 {tex}\\sqrt { 5 }{/tex}) cannot be rational.Hence, (3+2 {tex}\\sqrt { 5 }{/tex}) is irrational. | |
| 39434. |
(-6,7),(-1,-1-) |
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| 39435. |
For what value of k _2 a zero of a polynomial 3x2+4x+2k |
| Answer» X=-5/2 | |
| 39436. |
4/x-3=5/2x+3 |
| Answer» By cross multiplaction we get,5x-15=8x-125x-8x=-12+15-3x=27x=-9 | |
| 39437. |
Ex - 5.2 que no. 1 |
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Answer» Aage put kar a~n=a+(n-1)d |
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| 39438. |
Define the theory of \' half chord\' |
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| 39439. |
Find the sum off first five multiples of five |
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Answer» 5,10,15,20,25Sn=n/2(2a(n-1)d)=5/2(2×5(5-1)×5)5/2(10+20)5/2×205×1575 75 |
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| 39440. |
Prove in triangle ABC AD=BE, angle B=angleC and DE parallel to AB |
| Answer» Bro write your qus. Properly. I think its incomplete | |
| 39441. |
Solved theorem of chapter 1 of class 10th |
| Answer» See example no. 3 on page no.6 From which school you ?? | |
| 39442. |
If the second term of ap is 8 and 5th term is 17. find its 19th term. |
| Answer» Case 1,A+(2-1)d=8A+d=8........(1)Case 2,A+(5-1)d=17A+4d=17........ (2)By subtracting eq(1)from eq (2) we get,D=3Putting value in eq1or2we get,A=5n19th term =a+(19-1)d =5+(18×3) =5+54 =59 àńś'} | |
| 39443. |
What is an ap series? |
| Answer» An ap is the series of numbers in which the common difference of between two numbers is same | |
| 39444. |
in the given figure ^a=90°, ^b=90° ob=4.5cm oa=6cm & ad=4cm find qb |
| Answer» In {tex}\\triangle{/tex}PAO and {tex}\\triangle{/tex}QBO{tex}\\angle{/tex}A = {tex}\\angle{/tex}B = 90o{tex}\\angle{/tex}POA = {tex}\\angle{/tex}QOB (Vertically Opposite Angle){tex}\\triangle{/tex}PAO ~ {tex}\\triangle{/tex}QBO,(by AA criteria){tex}\\frac { O A } { O B } = \\frac { P A } { Q B }{/tex}or,\xa0{tex}\\frac { 6 } { 4.5 } = \\frac { 4 } { Q B }{/tex}or,\xa0{tex}Q B = \\frac { 4 \\times 4.5 } { 6 }{/tex}Therefore, QB = 3 cm | |
| 39445. |
How to make a mix surd to pure surd |
| Answer» Mixed surds can be expressed in the form of pure surds. Because if we make rational co-efficient under radical sign, it will become a pure surd.2 √7 = √22 × 7 = √4 × 7 = √28 | |
| 39446. |
8/(2x+3y) +21/(2x-3y)=11; 5/(2x+3y) +7/(2x-3y)=-6; |
| Answer» Given equations are{tex}\\frac{8}{2x - 3y}{/tex}\xa0+\xa0{tex}\\frac{21}{2x + 3y}{/tex}\xa0= 11...(i){tex}\\frac{5}{2x - 3y}{/tex}\xa0+\xa0{tex}\\frac{7}{2x + 3y}{/tex}\xa0= 6 ....(ii)Putting\xa0{tex}\\frac{1}{2x - 3y}{/tex}\xa0= A and\xa0{tex}\\frac{1}{2x + 3y}{/tex}\xa0= B in equation (i) & (ii) so that we may get the pair of linear equations in variables A & B as following :-8A + 21B = 11 ...(iii). and 5A + 7B = 6...(iv)Multiplying eq. (iv) by 3 & then subtracting eq. (iii) from it , we get ;{tex}\\Rightarrow{/tex}\xa0A = 1\xa0Substituting A = 1 in eq. (iii) ,8\xa0{tex}{/tex}× 1 + 21B = 11{tex}\\Rightarrow{/tex}\xa021B = 3{tex}\\Rightarrow{/tex}\xa0B =\xa0{tex}\\frac{1}{7}{/tex}Since, A = 1{tex}\\Rightarrow{/tex}\xa0{tex}\\frac{1}{2x - 3y}{/tex}\xa0= 1{tex}\\Rightarrow{/tex}\xa02x - 3y = 1...(vi)Where B =\xa0{tex}\\frac{1}{7}{/tex}{tex}\\Rightarrow{/tex}{tex}\\frac{1}{2x + 3y}{/tex}\xa0=\xa0{tex}\\frac{1}{7}{/tex}{tex}\\Rightarrow{/tex}2x + 3y = 7...(vii)Adding (vi) and (vii), we get{tex}\\Rightarrow{/tex}x = 2Substituting x = 2 in eq.(vi),\xa02{tex}{/tex}× 2 - 3y = 1{tex}\\Rightarrow{/tex}\xa0-3y = -3{tex}\\Rightarrow{/tex}\xa0y = 1{tex}\\therefore{/tex}\xa0x = 2, y = 1. | |
| 39447. |
(a-b)^2 |
| Answer» a^2+b^2-2ab | |
| 39448. |
All in one is not good book |
| Answer» Ya all in one is one of the best reference book ...... | |
| 39449. |
Bhhdo |
| Answer» | |
| 39450. |
Exercise 14.2 full parts answer |
| Answer» Get it through this app. | |