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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.
| 11351. |
Drive the formula for CSA of frustum of a cone. |
| Answer» Read RS aggarwal for its derivation\xa0 | |
| 11352. |
A sector of a circle of radius 10.5cm then find the perimeter of sector where angle is 60 |
| Answer» r = 10.5 cm , θ = 60oPerimeter of the sector = 2r +\xa0{tex}\\frac { 2 \\pi r \\theta } { 360 ^ { \\circ } }{/tex}{tex}= 10.5 \\times 2 + 2 \\times \\frac { 22 } { 7 } \\times \\frac { 10.5 \\times 60 } { 360 }{/tex}{tex}= 21 + 2 \\times22 \\times1.5 \\times \\frac { 1 } { 6 }{/tex}{tex}= 21 + 22 \\times1.5 \\times \\frac { 1 } { 3}{/tex}{tex}= 21 + 11 = 32{/tex} cm | |
| 11353. |
If semiperimeter od triangle ABC=23cm then find AF+BD+CE. Where f lies on ab d on bc and e on ac.. |
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| 11354. |
The ratio of first n terms of two ap is 7n+1:4n+27 find the rario of their mth terms\xa0 |
| Answer» Let a, and A be the first terms and d and D be the common difference of two A.PsThen, according to the question,{tex}\\frac { S _ { n } } { S _ { n } ^ { \\prime } } = \\frac { \\frac { n } { 2 } [ 2 a + ( n - 1 ) d ] } { \\frac { n } { 2 } [ 2 A + ( n - 1 ) D ] } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,\xa0{tex}\\frac { 2 a + ( n - 1 ) d } { 2 A + ( n - 1 ) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}or,{tex}\\frac { a + \\left( \\frac { n - 1 } { 2 } \\right) d } { A + \\left( \\frac { n - 1 } { 2 } \\right) D } = \\frac { 7 n + 1 } { 4 n + 27 }{/tex}Putting,\xa0{tex}\\frac { n - 1 } { 2 } = m - 1{/tex}{tex}n-1 = 2m - 2{/tex}{tex}n= 2m - 2 + 1{/tex}or, {tex}n = 2m - 1{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 7 ( 2 m - 1 ) + 1 } { 4 ( 2 m - 1 ) + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 7 + 1 } { 8 m - 4 + 27 }{/tex}{tex}\\frac { a + ( m - 1 ) d } { A + ( m - 1 ) D } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex}Hence,\xa0{tex}\\frac { a _ { m } } { A _ { m } } = \\frac { 14 m - 6 } { 8 m + 23 }{/tex} | |
| 11355. |
a/ax-1+b/bx-1=a+b |
| Answer» | |
| 11356. |
PQ is a chord of a circle with centre O and PT is a tangent. If Angle QPT = 60, find Angle PRQ |
| Answer» Ans.\\(\\angle OPT = 90\\)\\(\\angle QPT = 60 \\space \\space \\space \\space(Given)\\)\\(\\angle OPQ = \\angle OPT - \\angle APT\\)\\(\\angle OPQ = 90-60 = 30 \\)\\(in \\triangle OPQ, OP = OQ\\)So,\\(\\angle OPQ = \\angle OQP = 30 \\)\\(\\angle POQ = 180-30-30 =120\\)\\(\\angle PLQ = 60 \\)PLQR is cyclic Quadrilateral, so\xa0\\(\\angle PRQ = 120 \\)\xa0 | |
| 11357. |
Perimeter of sector is 31cm and radius is 6.5 .Find ar.of sector.\xa0 |
| Answer» Radius = 6.5 cmPerimeter of circle = 31 cm2r + l = 312 × 6.5 + l = 3113 + l = 31l = 31 - 13l = 18 cmAreas of sector =\xa0\\(\\)\\(1/2\\) × l × r = 1/2 × 18 × 6.5 = 9 × 6.5 = 58.5 sq. cm | |
| 11358. |
If p term of ap is 1/q and qterm is1/p then prove pq term is 1 |
| Answer» ap\xa0= 1/qa+ (p-1) d = 1/q -(i)aq\xa0= 1/pa+ (q-1)d = 1/p -(ii)Subtracting (i) & (ii)d (p-1-q+1) = 1/q- 1/pd(p-q) = (p-q)/ pqd = 1/pq - (iii)Putting it in eqn (i)a = 1/pq -(iv)apq\xa0= a + (pq-1) dPutting (iii) & (iv)apq\xa0= 1/pq + (pq-1) / pqapq\xa0= 1\xa0 | |
| 11359. |
3x+2y=12 if y=3 |
| Answer» Ans. 3x+2y =12\xa0put value of y =33x + 2×3 =12=> 3x+6=12=>3x= 12-6=> 3x=6=> x =2 | |
| 11360. |
If\xa0PQ=PR prove that PL=LQ\xa0\xa0 |
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| 11361. |
if roots of equation (a-b)x2+(b-c)x+(c-a)=0,prove that b+c=2a\xa0 |
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Answer» ttttttttjj from The question must be, if roots of equation (a-b)x2+(b-c)x+(c-a)=0 are equal , prove that b+c = 2a.Using Discriminant,D = B2-4AC as compared with the general quadratic equation Ax2+Bx+C=0so, A = a-b\xa0B = b-cC = c-aFor roots to be equal, D=0(b-c)2\xa0- 4(a-b)(c-a) =0b2+c2-2bc -4(ac-a2-bc+ab) =0b2+c2-2bc -4ac+4a2+4bc-4ab=04a2+b2+c2+2bc-4ab-4ac=0(2a-b-c)2=0i.e. 2a-b-c =02a= b+c |
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| 11362. |
Find the sum of series 3+5+7+6+9312+9+13+17+....to 3n terms.\xa0\xa0 |
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| 11363. |
Find the sum of 24 terms of the series 12- 22\xa0+ 32-42+52-62.......\xa0 |
| Answer» Series : (12-22) + (32-42) + (52-62) +....= (1-2) (1+2) + (3-4)(3+4) + (5-6)(5+6)+...= -3, -7, -11, .... forms an APa = -3 ; d = -7+3 = 4 ; n = 24Sn\xa0=\xa0\\({\\ n\\over 2 }\\)\\({ [2a+(n-1)d]}\\)= 12[ -6+23 X 4]12[ -6+ 92]= 12 X 86 = 1032 | |
| 11364. |
If I got a cgpa of 8.6 in the first term can I get a 10cgpa in the final result? |
| Answer» You will must get, but only then, when you will double your study time:-):-):-)That means increase in study & decrease in passing times..... | |
| 11365. |
what is 8/15 of an hour ( in minutes )\xa0 |
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Answer» 8/15 of an hour= 8/15 of 60 min= 8/15 × 16 min= 8 × 4 min= 32 minutes 32minuts |
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| 11366. |
If 3+5+7+........ to n terms/ 5+8+11+.........to 10 terms = 7. Find \'n\' ? |
| Answer» Both the numerator and denominator are the sums of two series up to \'n\' terms:-Since the sum = (n/2)(2a+(n-1)d)For series in the numerator :-a=3, d=2, n=nTherefore numerator = (n/2)*(6+(n-1)2) = (n/2)*(2n+4)And for the denominator series:-a=5, d=3, n=10Therefore denominator = (10/2)*(10+(10-1)3) = (n/2)*(3n+7) = 185Therefore Acc. to question:-(n/2)*(2n+4) / (185) = 7\xa0= n2\xa0+2n- 1295 = 0\xa0= n2\xa0+ 37n - 35n - 1295 = 0=n=35 ( Since ncan not be equal to 37) | |
| 11367. |
(x÷x+1)2. -. 5(x÷x+1). +. 6 = 0 |
| Answer» Ans.\xa0\\(({x \\over (x+1)})^2 - 5 ({x \\over (x+1)}) + 6 = 0\\)\\(Put \\space ({x \\over (x+1)}) = y, We \\space get \\)=>\xa0\\(y^2 -5y +6 =0\\)=>\xa0\\(y^2 - 3y -2y +6 = 0\\)=>\xa0\\(y(y-3) -2(y-3) = 0\\)=>\xa0\\((y-3)(y-2) = 0\\)=> y = 3 , 2Now,\xa0\\({x\\over (x+1) } = 3 \\space \\space or \\space \\space {x\\over (x+1) } = 2\\)=> x = 3x +3 or x = 2x + 2=> x = -3/2 or x = -2\xa0Values of x = -3/2, -2\xa0 | |
| 11368. |
9x2 -. 9(a+b)x. +. (2a2 . +.\xa05ab. +. 2b2) |
| Answer» Ans.\xa0\\(9x^2 - 9(a+b)x + (2a^2 + 5ab+2b^2) = 0\\)=>\xa0\\(9x^2 - 3(3a+3b)x + (2a^2 + 4ab+ab +2b^2 ) = 0\\)=>\xa0\\(9x^2 - 3[(2a+b)+(a+2b)]x + (2a^2 + 4ab+ab +2b^2 ) = 0\\)=>\xa0\\(9x^2 - 3[(2a+b)+(a+2b)]x + [2a(a + 2b)+b(a +2b)] = 0\\)=>\xa0\\(9x^2 - 3[(2a+b)+(a+2b)]x + [(2a+b)(a +2b)] = 0\\)=>\xa0\\(9x^2 - 3(2a+b)x-3(a+2b)x + (2a+b)(a +2b) = 0\\)=>\xa0\\([3x-(a+2b)][3x - (2a+b)] = 0\\)=>\xa0\\([3x-(a+2b)] = 0 \\space \\space or \\space \\space [3x - (2a+b)] = 0\\)=>\xa0\\(3x=(a+2b) \\space \\space or \\space \\space 3x = (2a+b) \\)=>\xa0\\(x= {(a+2b) \\over 3} \\space \\space or \\space \\space x = {(2a+b) \\over 3}\\) | |
| 11369. |
The product of two successive integral multiples of 5 is 300. Determine the multiples.\xa0 |
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Answer» 15 and 20 Ans. Let First Number = 5x [because it is multiple of 5.]then Next Number = 5x + 5\xa0According to ques.5x ( 5x + 5 ) = 300=> 25x2\xa0+ 25x = 300=> 25x2 + 25x - 300 = 0divide equation by 25, we get\xa0=> x2 + x - 12 = 0=> x2\xa0+ 4x - 3x - 12 = 0=> x( x+4) - 3(x+4) = 0=> (x+4) (x-3) = 0Either (x + 4 ) = 0 or (x-3) = 0=> x = -4 or x = 3\xa0\xa0So, Multiple of 5 are : 15, 20 |
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| 11370. |
Solve the equation:22x+3=65(2x-2)+122. |
| Answer» Ans. Equation is in the Form = 22x+3\xa0\xa0= 65 (2x\xa0- 2\xa0) + 122=> 22x\xa0. 23\xa0= 65 ( 2x\xa0- 2\xa0) + 122=> 8(2x)2\xa0= 65 ( 2x\xa0- 2) + 122Put 2x\xa0= y , We get\xa0=> 8y2\xa0= 65(y-2) + 122=> 8y2 = 65y - 130 + 122=> 8y2\xa0- 65y + 8 = 0=> 8y2 - 64y - y + 8 = 0=> 8y ( y - 8) - 1(y-8) = 0=> (8y-1) ( y - 8) = 0Either 8y -1 = 0 or y - 8 = 0So, y = 1/8 or 8=> 2x\xa0= 1/8 or 2x\xa0= 8=> 2x\xa0= (2)-3 or 2x\xa0= (2)3On comparing, We get\xa0x = -3, 3 | |
| 11371. |
If Sn\xa0denotes the sum of first n terms of A.P. Prove that S30\xa0= 3(S20- S10). |
| Answer» R.H.S.3[20/2{2a + (20 - 1)d} - 10/2{2a + (10 - 1)d]= 3[10{2a + 19d - 5{2a + 9d}]= 15[4a + 38d - 2a - 9d]= 30/2[2a + 29d]= 30/2[2a + (30 - 1)d]= S30 | |
| 11372. |
(a2+b2)x2-2b(a+c)x+(b2+c2)=0For equal roots ...To find b2=???? |
| Answer» Ans. On comparing with standard form.of quadratic equation\xa0Ax2\xa0+ Bx + C = 0, We getA = (a2\xa0+ b2)B = -2b(a+c)C = (b2\xa0+ c2)as equation has equal roots,SoD = 0=> B2\xa0- 4AC = 0=> [-2b(a+c)]2\xa0- 4(a2\xa0+ b2)(b2\xa0+c2) =0=> 4b2(a2\xa0+ c2+2ac) = 4(a2b2\xa0+ a2c2\xa0+ b4\xa0+ b2c2)divide by 4 both sides,=> a2b2\xa0+ b2c2\xa0- 2acb2\xa0- a2b2\xa0+ a2c2\xa0+ b4\xa0- b2c2\xa0= 0=> (ac)2\xa0+ (b2)2\xa0- 2(ac)(b2) =0=> (ac - b2)2\xa0=0squareroot both side=> ac - b2\xa0= 0=> b2\xa0= acHence proved | |
| 11373. |
I got B1 in maths in SA-1, so what I have to do to get 10 gp in final result |
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Answer» Yes, do as much practice you can and try to score A1 in SA-2. All the best! Score complete marks in SA-2 Maths and you will get A1 as the final result.\xa0Also FA-3 and FA-4 should also have A1 grade.\xa0 |
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| 11374. |
a÷x-a. +. b÷x-b. =. 2c÷x-c |
| Answer» Ans. a/ x-a + b/x-b = 2c/x-ctaking (x-a) (x-b) LCM=> (ax-ab+bx-ab)/(x-a)(x-b) = 2c/x-cCross multiply, we get=> ax2-2abx+ bx2\xa0-acx +2abc-bcx = 2cx2-2acx -2bcx + 2abc=> ax2\xa0+ bx2\xa0-2cx2\xa0= 2abx -acx -bcx=> (a+b-2c)x2\xa0- (2ab -ac -bc)x = 0=> x [(a+b-2c)x - (2ab-ac-bc)] = 0Either x = 0\xa0Or (a+b-2c)x - (2ab-ac-bc) =0=> x =(2ab-ac-bc)/(a+b-2c)\xa0 | |
| 11375. |
for what value of k will be k+9 2k-1 amd 2k+7 are the consecutive terms of an AP? |
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Answer» If these are consecutive terms then c.d should be same...2k-1-(k+9)=2k+7-(2k-1)2k-1-k-9=2k+7-2k+1k-10=8k=18 If given terms are in AP, then2k - 1 = (k + 9 + 2k + 7)/24k - 2 = 3k + 16k = 18 |
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| 11376. |
Why volume of cone is 1/3 of cylinder volume |
| Answer» Because cone is 1/3 part of a cylinder | |
| 11377. |
3x2-23x-110=0 by using quadratic equation\xa0 |
| Answer» 3x2-23x-110=03x2-33x+10x-110=03x(x-11)+10(x-11)=0(x-11)(3x+10)=0x-11=0, 3x+10=0x = 11, -10/3\xa0 | |
| 11378. |
Find the diffrence between the points (a,b) (-a,-b) |
| Answer» Difference\xa0between\xa0(a,b)\xa0and\xa0(-a,-b)=-a-a2+-b-b2=-2a2+-2b2=4a2+4b2=4(a2+b2)=2a2+b2 | |
| 11379. |
The distance between the points (2,-2)&(-1,x) is 5 then one of the value of x is\xa0 |
| Answer» Given : A (2,-2) and B(-1,x) and Distance b/w these two is 5according to distance formula(x2-x1)2 + (y2-y1)2 = d2=> (-1-2)2 + (x+2)2 = 25=> 9 + x2 + 4 +4x = 25=> x2 + 4x - 12 = 0=> x2 + 6x -2x - 12 = 0=> x (x+6) -2(x+6) = 0\xa0=> (x+6) (x-2) = 0either x + 6 = 0 or x-2=0so x = -6, 2\xa0 | |
| 11380. |
Show that the points (7,3) (3,0) (0,-4) (4,-1)are vertices of rhombus\xa0\xa0 |
| Answer» Let A(7,3) , B(3,0) , C(0,-4) and D(4,-1) represent the points.To prove that these points are vertices of a rhombus we have to show that all sides are equal. We may calculate it using distance formula.AB=\xa0(3-7)2+(0-3)2=\xa016+9=25=\xa05\xa0unitsBC=(0-3)2+(-4-0)2=9+16\xa0=25=5\xa0unitsCD=(4-0)2+(-1+4)2=16+9\xa0=25\xa0=\xa05unitsDA=(7-4)2+(3+1)2=9+16=25=5\xa0unitsThe diagonals are not equal.\xa0AC=\xa0(7-0)2+(3+4)2=49+49=98=72unitsBD=(4-3)2+(-1-0)2\xa0=\xa01+1\xa0=2unitsThis shows that these points are vertices of rhombus. | |
| 11381. |
Find the sum of all 3 digit no. Which leave a remainder 3 when divided by 5 . |
| Answer» First 3 Digit Number that leaves remainder 3 when divided by 5 is 103.and last will be 998A.P. = 103, 108, 113, ..................., 998first term a = 103Common Difference d = 5nth term = 998n = ?as we knownth term = a + ( n-1)d998 = 103 + ( n-1)5895/ 5 = n -1179 = n -1n = 180\xa0now find sum of 180 terms of APS180 =\xa0180/ 2 ( 103 + 998 )=> S180 = 90\xa0* 1001=> S180 = 900090 | |
| 11382. |
If mth\xa0term of an A.P. is 1/n and nth\xa0term is 1/m .show that its (mn)th term is 1 |
| Answer» Let a and d are first term and common difference of AP | |
| 11383. |
Find k so that 15,k,-1are in A.P. series\xa0 |
| Answer» As they are in APso 2b = a + c2k = 15 - 1\xa02k = 14k = 7\xa0\xa0 | |
| 11384. |
Solve 4x2-2(a2+b2)x+a2b2=0 find the value of x\xa0 |
| Answer» a= 4b= -2(a2+b2)c= a2b2Using Discriminant formula,D= b2-4ac[-2(a2+b2)]2-4 x4xa2b24(a4+b4+2a2b2)-16a2b2(2a2-2b2)2\xa0x=\xa0(-b±D)/\xa02ax= [2(a2+b2)+2(a2-b2)]/8 = 4a2/8 = a2/2x= [2(a2+b2)-2(a2-b2)]/8 = 4b2/8 = b2/2 | |
| 11385. |
If ax2+bx+c=0has equal roots then proove that c=b2/4a |
| Answer» For equal roots, discriminant D=0i.e. b2-4ac=0b2=4acSo, c= b2/4a | |
| 11386. |
The difference between the circumference and radius of acircle is 37cm. Find the area of the circle. |
| Answer» According to question,2 x\xa022/7 x r - r = 37r(44/7 - 1) = 37r x 37/7 = 37r = 37 x 7 / 37r = 7 cmArea of circle = 22/7 x (7)2 = 22/7 x 7 x 7 = 154 cm2 | |
| 11387. |
Find the sum of first odd 10 natural numbers?\xa0 |
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Answer» The 1st 10 odd natural numbers are 1,3,5,7.....Here, a = 1, d = 2We know, Sn\xa0= n/2 {2a + (n-1)d}S10\xa0= 10/2 {2×1 + (10-1)×2} = 5 (2 + 9×2) = 5 (2 + 18) = 5\xa0×\xa020 = 100 (Answer) Sum of First N Natural Odd Numbers = N2Sum of First N Natural Even Numbers = N2\xa0+ Nwhere n is number of terms.\xa0for example : sum of first 5 even numbers = (5)2\xa0+ 5 = 30\xa0sum of first 5 ODD\xa0numbers = ( 5)2 = 25\xa0 Here the AP is 1, 3, 5, ......, 19a = 1, d = 1, n = 10, l = 19Sn = (n/2)(a + l)S10\xa0= (10/2)(1 + 19) = 5 × 20 = 100 You can get your doubt cleared with teachers from IIT and AIIMS on HashLearn. Download the HashLearn app here http://bit.ly/2fDYZtb |
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| 11388. |
Can you please help me solve the first question? |
| Answer» Please write the question for which you are seeking an answer. We would be happy to help you. | |
| 11389. |
important tips to solve word problems in quadratic equations. |
| Answer» \tRead the statements carefully\ttry to understand the meaning\xa0\tthen convert the statements in mathematical equation\tas soon as you are able to understand meaning of statements, no one ll stop you solving sums | |
| 11390. |
how to find angle of elevation when two sides are given |
| Answer» If given two sides are Perpendicular and Base, then Angle of elevation can be find by tan x = Perpendicular / Base OR cot x = Base / PerpnedicularIf given two sides are Perpendicular and Hypotenuse, then Angle of elevation can be find by sin x = Perpendicular /\xa0Hypotenuse OR cosec x =\xa0Hypotenuse / PerpnedicularIf given two sides are Base and Hypotenuse, then Angle of elevation can be find by cos x = Base /\xa0Hypotenuse OR sec x = Hypotenuse / Base | |
| 11391. |
Tanθ + tan(90-θ)=secθ + sec(90-θ) |
| Answer» We have to prove that: tan θ + tan (90° – θ) = sec θ sec (90° – θ)Here, LHS = tan θ + tan (90° – θ)= tan θ + cot θ{tex}= \\frac { \\sin \\theta } { \\cos \\theta } + \\frac { \\cos \\theta } { \\sin \\theta }{/tex}{tex}= \\frac { \\sin ^ { 2 } \\theta + \\cos ^ { 2 } \\theta } { \\sin \\theta \\cos \\theta }{/tex}{tex}= \\frac { 1 } { \\sin \\theta \\cos \\theta }{/tex}\xa0[∵ sin2\xa0θ + cos2\xa0θ = 1]= cosec θ·sec θ⇒ LHS = cosec θ·secθRHS = sec θ sec θ (90° - θ)= sec θ cosec θ⇒ LHS = RHSHence, verified. | |
| 11392. |
obtain all the zeroes of the polynomial x4+6x3+x2-24x-20, if two zeroes are 2 and -5\xa0 |
| Answer» ACQ, Factors are (x - 2) and (x + 5) => (x - 2)(x + 5) => x2+3x-10on dividing given polynomial by x2+3x-10, we get the quotient x2+3x+2Therefore, x4+6x3+x2-24x-20 = (x2+3x-10)(x2+3x+2)=> x4+6x3+x2-24x-20 = (x2+3x-10)(x2+2x+x+2)=> x4+6x3+x2-24x-20 = (x2+3x-10)[x(x+2)+1(x+2)]=> x4+6x3+x2-24x-20 = (x2+3x-10)(x+2)(x+1)Therefore, all other zeroes are -2 and -1. | |
| 11393. |
sin 60 cos30 +sin30 cos60 |
| Answer» sin60cosc30+sin30cos60 = (3^2/2 * 3^2/2) + (1/2 * 1/2) =3/4\xa0+ 1/4\xa0= 1 | |
| 11394. |
a+b=4 find sin theta + cos theta |
| Answer» | |
| 11395. |
1/SECX-TANX=SECX+TANX |
| Answer» (1/SECX-TANX)(SECX+TANX) = SECX + TANX=> (SECX + TANX)/(SEC2X - TAN2X) = SECX + TANX=> SECX + TAN X = SECX + TAN X [SINCE SEC2X - TAN2X = 1] | |
| 11396. |
how to convert more than type frequency into original frequency\xa0 |
| Answer» To convert more than type frequency into original frequency, subtract frequency from its previous frequncy. Watch the following table:\tClass MarksMore than frequencyOriginal Frequency45 - 5050 - 5555 - 6060 - 6565 - 7070 - 7510098907854162812243816\t\xa0 | |
| 11397. |
In triangle ABC, DEIIBC and CDIIEF. Prove that AD2 = AF * AB |
| Answer» In triangle ABC, using BPT,AD/AB = AE/AC ...........eq.(i)In triangle ADC, using BPT,AF/AD = AE/AC. ...........eq.(ii)From eq.(i) and (ii), we getAD/AB = AF/AD=>. AD2 = AF x AB proved | |
| 11398. |
Find the roots of below quadratic equation:x2-6x+9=0 |
| Answer» x2\xa0- 3x - 3x + 9 = 0x(x - 3) - 3(x - 3) = 0(x - 3)(x - 3) = 0x - 3 = 0 and x - 3 = 0x = 3 and x = 3Therefore two roots are 3 and 3. | |
| 11399. |
if the sum of first m tern of an A.P is am^2+bm, find its common difference |
| Answer» Taking m = 1, we have S1 = T1 =\xa0a + bTaking m = 2, we have S2 = T1 +\xa0T2 =\xa04a + 2bTaking m = 3, we have S3 = T1 +\xa0T2\xa0+ T3\xa0=\xa09a + 3bTherefore, T2 = S2 - S1 = 4a + 2b - a - b = 3a + bAlso, T3 = S3 - S2 = 9a + 3b - 4a - 2b = 6a + bTherefore, Common difference = 6a + b - 3a - b = 3a\xa0 | |
| 11400. |
probability\xa0 |
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Answer» The extent to which an event is likely to occur, measured by the ratio of the favourable cases to the whole number of cases possible. You can get your doubts cleared instantly, 24/7 with IIT/AIIMS tutors on HashLearn. Download here: http://bit.ly/2fDYZtb |
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