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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.
| 22051. |
Find value of which one root of quadratic equations kxsquare -14x+18=0 is 2 |
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Answer» K=5 Value of k |
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| 22052. |
SinA-cosA+1/sinA+cosA-1=tanA-1+secA/tanA+1-secA |
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Answer» Q.SinA-cosA+1/sinA+cosA-1=tanA-1+secA/tanA+1-secA Answer: |
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| 22053. |
Empheric relationship between to find central tendency |
| Answer» Mode=3median-2mean | |
| 22054. |
Formulas of surface area volume |
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| 22055. |
Whats the formula of vol. Of frustum??? |
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Answer» 1/3 PIE ×HEIGHT (R ^2+ r^2+Rr) 1/3pie h ( r2 + R 2 + r.R ) Ťhanks? 1/3πh(R^2 + r^2 +Rr) |
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| 22056. |
If the quadratic equation of px square -2root 5 px +15 =0 has two equal roots then find p |
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Answer» Now solve your self For equal roots , b^2 -4ac =0 |
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| 22057. |
a3 +b3 |
| Answer» a+b^3-3ab(a+b) | |
| 22058. |
Surface area and volume rd sharma |
| Answer» Surface area and volume rd sharma mein kya nahi aara hai??? | |
| 22059. |
Show that 5-2√3irrantion number |
| Answer» assume that 5-2 root 3 is rational5-2 root 3 = a/b , where a and b are integers .\xa0-2 root 3 = a/b - 52 root 3 = 5 - ab\xa02 root 3 = 5b/b - a/broot 3 = 5b - a / 2bwe know that a, b , 2 and 5 are integers and they are also rational\xa0therefore root 3 will be rational\xa0but we know that root 3 is irrationalthere is a contradictionso, 5 - 2 root 3 is an irrational number. Thank you?? | |
| 22060. |
Find the 20th term from the last term of ap 3,8,13........253 |
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Answer» 158 Ap 253,248,243A is equal to 253 d is equal to 5 158...... |
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| 22061. |
Find the sum of all two digits odd positive numbers |
| Answer» 2475..... | |
| 22062. |
(a+b)square |
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Answer» a square + b square + 2 ab a square +b square +ab square ab2 |
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| 22063. |
If the sum of the first 14 term of an ap is 1050 and its first term is 10 find the 20th term |
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Answer» It should be 100...bcuz d=10...so a+9d=10+90=100...hope this is ryt 200 |
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| 22064. |
The 17th term of an AP exceeds it\'s 10th term by 7 find the common difference |
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Answer» How Common difference = 1 |
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| 22065. |
If a×a,b×b,c×c are in A.P. ,then prove that-a÷(b+c),b÷(a+c),c÷(a+b) are in A.P. |
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| 22066. |
what is symmetrical frequency distribution? |
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| 22067. |
Find the value of k for which 9x square +9kx +25=0 has no real root |
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Answer» K < 10/3 If D= b^2-4ac is negative then it have no real rootD<0b^2-4ac<0(9k)^2-4(9)(25)<0 sovle this ilinear equation. ilinear equation is not in our syllabus 10/3 |
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| 22068. |
TanA/1-cotA+cotA/1-tanA=1+tanA+cotA |
| Answer» We have,{tex}\\mathrm { LHS } = \\frac { \\tan A } { 1 - \\cot A } + \\frac { \\cot A } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan A } { 1 - \\frac { 1 } { \\tan A } } + \\frac { \\frac { 1 } { \\tan A } } { 1 - \\tan A }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan A } { \\frac { \\tan A -1 } { \\tan A } } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } + \\frac { 1 } { \\tan A ( 1 - \\tan A ) }{/tex}{tex}\\Rightarrow \\quad \\text { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A - 1 } - \\frac { 1 } { \\tan A ( \\tan A - 1 ) }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 3 } A - 1 } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[Taking LCM]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { ( \\tan A - 1 ) \\left( \\tan ^ { 2 } A + \\tan A + 1 \\right) } { \\tan A ( \\tan A - 1 ) }{/tex}\xa0[{tex}\\because{/tex}\xa0a3\xa0- b3\xa0= ( a - b )(a2\xa0+ ab + b2)]{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A + \\tan A + 1 } { \\tan A }{/tex}{tex}\\Rightarrow \\quad \\mathrm { LHS } = \\frac { \\tan ^ { 2 } A } { \\tan A } + \\frac { \\tan A } { \\tan A } + \\frac { 1 } { \\tan A }{/tex}{tex}\\Rightarrow{/tex}\xa0LHS = tanA + 1 + cotA = RHS\xa0[ since (1/tanA) =cotA ] | |
| 22069. |
If a square is inscribed in a circle . What is the ratio of the area of circle and square |
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Answer» If a square is inscribed in a circle, then the diagonals of the square are diameters of the circle.Let the diagonal of the square be d cm.Thus, we have:Radius, r = |
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| 22070. |
Evaluate 3sin 3A +2cos (5+10){the whole divided by} root 3 tan 3A -cosec(5A-20), when A=10. |
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| 22071. |
Is AAS is the ceiteria for similarity |
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Answer» No i m taking ur test if it is about AAS criteria it is wrong becaise AAS is not any criteria of similarity But u r right AA is sufficirnt but u dont hive answer according to my quesyion try again True else only AA is sufficient |
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| 22072. |
How to derive all the formula of frustum |
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| 22073. |
In given figure |
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Answer» Figure?? Ans |
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| 22074. |
Solve (-5-x)^2 |
| Answer» (-5-x)2=(-5)2+(-x)2+2(-5)(-x)=25+x\u200b\u200b\u200b\u200b\u200b\u200b2\u200b\u200b\u200b\u200b\u200b+10x | |
| 22075. |
Guys today is my maths exam, l need ur wishes Plz koi all the best boldo ? |
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| 22076. |
In which case mean =median=mode? |
| Answer» 3 MEDIAN = MODE + 2 MEAN | |
| 22077. |
Find the area of the largest triangle that can be inscribed in a semi-circle of radius r unit. |
| Answer» As base of triangle inscribed in a semi circle is 2rand altitude is r then area of two right angled triangle with base and height with r is 2r.... Thank you??? | |
| 22078. |
Or and and concept from probability |
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| 22079. |
Prove that :[ {cosA - sinA + 1} ÷ {cosA + sinA - 1} ] = cosecA + cotA...??? |
| Answer» It\'s ncert example and already solved | |
| 22080. |
If tan if tan theta |
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Answer» What????? Please complete your question first... |
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| 22081. |
How to find n term on ap in question how to recognize from the question |
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| 22082. |
Explain why (7×6×5×4×3×2×1×5) is a composite number |
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Answer» Let u take common then it will be 5 not 7 but again no need to take common Hello kiran no need to take common here clearly (7×2×3×5×2×2×3×2×1×5) has factors of primes and this factorisation is unique To take 7as common from this number And the no is 25200and if u divide its 7 u gets 3600 and its compelete by 7 So this no is composite number. |
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| 22083. |
Practical work |
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| 22084. |
basic formula for trigonometry |
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| 22085. |
Alpha and beta are zeroes of polynomial x²-6x+ a. Find the value of a, if 3alpha +2beta = 20 |
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Answer» p(x)=x2-6x+aa=3alpha+2beta=20.....(1)from the quadratic expressionalpha+beta=6.....(2)and alpha×beta=a.....(3)multiply eq.. (2)by 2and subtract from(1)a=20-6×2a=20-12=8alphasubstitute this value in (2)beta=6-8=-2then( -2)is beta and( 8)is alpha then=alpha×beta =8×-2=-16 answer Find sum of zeroes alpha+beta=-b/a u will get a value then find the value of alp.and beta with the help of given information and then find product of zroz =c/a u will get the value of a How u solve this explain -16 |
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| 22086. |
what is the basic formula for understand trignometry |
| Answer» Just learn the identities And change question in sin ans cos | |
| 22087. |
(1/x+4)-(1/x-11)=11/30Send by steps if someone is giving answer for this question |
| Answer» we have to Solve the equation,\xa0{tex}\\frac { 1 } { x + 4 } - \\frac { 1 } { x - 7 } = \\frac { 11 } { 30 } , x \\neq - 4{/tex}\xa0,{tex}7{/tex}\xa0for x.Given equation,\xa0{tex}\\frac { 1 } { x + 4 } - \\frac { 1 } { x - 7 } = \\frac { 11 } { 30 }{/tex}or,\xa0{tex}\\frac { x - 7 - x - 4 } { ( x + 4 ) ( x - 7 ) } = \\frac { 11 } { 30 }{/tex}or,\xa0{tex}( - 11 ) \\times 30 = 11 ( x + 4 ) ( x - 7 ){/tex}or,\xa0{tex}x ^ { 2 } - 3 x + 2 = 0{/tex}or,\xa0{tex}x ^ { 2 } - 2 x - x + 2 = 0{/tex}or,\xa0(x - 1)(x - 2 ) = 0{tex}\\therefore {/tex}\xa0x = 1, 2 | |
| 22088. |
please.,can you send me guess paper of maths 2019 for revision of exams |
| Answer» You should try support material provided in govt. School. | |
| 22089. |
Full solution |
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| 22090. |
Can anyone give me blueprint of mathematics for 2k19 boards class 10 |
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| 22091. |
Model test paper of mathematics |
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| 22092. |
Is there will be choice in maths paper for the students of 2018-2019? |
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Answer» No..... No but an internal choice is provided pf questions from same lessons No, from the batch of 2019-20 this scheme works.... No |
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| 22093. |
The root of the equation |
| Answer» Which equation?? | |
| 22094. |
Given= BD=1/3BCTO PROVE= 9(AD²)=7(AB²) |
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Answer» Ncert book It\'s question no 15 of exercise 6.5 Abc ia an equilateral triangle |
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| 22095. |
0÷0=2 proof |
| Answer» 0/0=2 100-100/100-100 10²-10²/10(10-10) (10-10) (10+10)/(10) (10-10) (10+10)/10 20/10 2 | |
| 22096. |
Is 7/75 repeating or non repeating decimal expression |
| Answer» 7/75 is a non terminating, and repeating decimal expression. | |
| 22097. |
Can 4n square+8 be the nth term of an AP ?Give reasons. |
| Answer» No, because nth term of an a.p can only be an linear relation, i.e an=a+ (n-1)d. | |
| 22098. |
Determine the A.p whose third term is 5 and the 7th term is 9 |
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Answer» a3 = a + (3 – 1)d = a + 2d = 5 ... (1)a7 = a + (7 – 1)d = a + 6d = 9 ... (2)Solving the pair of linear equations (1) and (2),a = 3, d = 1Hence, the required AP is 3, 4, 5, 6, 7,\xa0... Ap is 3,4,5,6,7........ 2,3,4,5,6,7..... . |
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| 22099. |
Cos a+Sec a=1 |
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| 22100. |
Solve for x : 1/(a+b+x)=1/a+1/b+1/x |
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Answer» It is very easy to solve . Let it take the 1/x in the L.H.S and the solve it by taking LCM . Then, we get the value of x=. -a and -b 8210649693 my Watts app number U can seek help whenever I Will ready to help Answer will be x=-a ,-b Bhai x ka square nhi ho raha hai yaha (X-a-b-x)/ax+bx +x square = (a+b)/ab =1/(a+b+x)- 1/x = ( b+a)/ab |
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