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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.
| 11951. |
Find LHS |
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| 11952. |
Cos 3 |
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| 11953. |
When we have 6 box which sum is equel 20 one letter no repte |
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| 11954. |
Simplify(1-tanA)^2/1-cotA)^2 = tan^2A |
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| 11955. |
Class6. Ex 7.6 q no 1. =4/9+2/7 |
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| 11956. |
Triangles |
| Answer» Three sided figure | |
| 11957. |
I want last years question with solution then what is the procedure to dwnld that paper |
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| 11958. |
2*5=? |
| Answer» 10 | |
| 11959. |
Represent the equation x-1=-1;3x+2y=12 on gragh |
| Answer» On graph | |
| 11960. |
Quadric equation |
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| 11961. |
Prove that square of any two side of triangle is greater than third side |
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| 11962. |
Sin 30 value |
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Answer» ½ 1/2 1/2 1/2 |
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| 11963. |
a10-a16 ,a=3,d=90 |
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Answer» a10 = a + (10-1)da10=3+9x90a10=813 ................ ia16=3+15x90a16=1353 .................... iia10 - a16 = 813 - 1353 = - 540 a10 = a + (10-1)da10=3+9x90a10=813 ................ ia16=3+15x90a16=1353 .................... iia10 - a16 = 813 - 1353\xa0 = - 540 |
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| 11964. |
Proof of basic proportionality theorem |
| Answer» basic proportionality theoremin a triangle ABC if line DE II BC\xa0such that it intersects AB in D and AC in E thenAD/DB = AE/ECif a line is drawn parallel to one side intersecting the other two sides , then it divides the two sides in the same ratio\xa0\xa0 | |
| 11965. |
prove that 5 is irrational |
| Answer» 5 is not an irratonal. it rational number because we can write 5 in the form of\xa0{tex}p\\over q{/tex} | |
| 11966. |
√3xsquare-8x+4√3 by factorisation method |
| Answer» √3xsquare-6x-2x+4√3=√3(x-2√3)-2(x-2√3)=(x-2√3)(√3x-2) | |
| 11967. |
If the common difference is 3 then find a20- a15 |
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| 11968. |
Why sin30=1/2 |
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Answer» It\'s reason is given in the text book please read it carefully. Construct right angle triangle ABC1. Draw an arbitrarily long 2. Mark any point namely A on it and use protractor to mark 30° from A and draw another long line 3. Mark point C on the initial line that you drew and again, use the protractor to mark 90° to AC. 4. Extend the 30° from line A and the 90° from line C until they meet 5. Let the point of intersection B and measure the ratio of BC/AB and will be 0.5 or 1/2 |
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| 11969. |
Prove that root over 5 is irrational |
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Answer» Use contradict method Don\'t be lazy to ask these simple question |
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| 11970. |
How to solve the crossfication |
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| 11971. |
2x-3y=5 |
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| 11972. |
Prove bpt by paper cutting method |
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Answer» Take a coloured paperDraw a triangle on the coloured paper (of reasonable size)Name it as ABCCut\xa0the triangle ABCNow draw a line parallel to BC such that it intersects other two sides AB in D\xa0and AC in E i.e DE is parallel to BCMeasure the lengths of AD , DB, AE, EC with the help of a scale (ruler)You will find that AD/DB = AE/EC\xa0 take a coloured paperdraw a triangle on the coloured paper (of reasonable size)name it as ABCCut\xa0the triangle ABCnow draw a line parallel to BC such that it intersects other two sides AB in D\xa0and AC in E i.e DE is parallel to BCmeasure the lengths of AD , DB, AE, ECYou will find that AD/DB = AE/EC\xa0 |
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| 11973. |
math chart |
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| 11974. |
Method of how to multiple in decimals |
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| 11975. |
2+8+6 |
| Answer» 16 | |
| 11976. |
State Fundamental theorem of algorithm. |
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Answer» a=bq+r For any positive integer a we have integer b which uniquely divides a to give quotient q and remainder |
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| 11977. |
Which term of the AP:121,117,113,_ _ _is its first negative terms |
| Answer» Here the given AP is :\xa0121,117,113........a=121 and d=117-121 = -4The general term of an AP is given byan\xa0= a+(n-1)dLet nth term is negative{tex}\\Rightarrow{/tex}\xa0a+(n-1)d<0{tex}\\Rightarrow{/tex}\xa0121+(n-1)(-4)<0{tex} \\Rightarrow {/tex}\xa0121-4n+4<0{tex} \\Rightarrow {/tex}\xa0-4n+125 < 0{tex} \\Rightarrow {/tex}\xa0-4n< -125Hence, 4n>125{tex} \\Rightarrow {/tex}\xa0n >{tex}\\frac{{125}}{4}{/tex}So, n > 31.25Hence, the 32nd term is first negative term. | |
| 11978. |
Which term of the AP :121,117,113_ _ _ is the first negative term |
| Answer» Here the given AP is :\xa0121,117,113........a=121 and d=117-121 = -4The general term of an AP is given byan\xa0= a+(n-1)dLet nth term is negative{tex}\\Rightarrow{/tex}\xa0a+(n-1)d<0{tex}\\Rightarrow{/tex}\xa0121+(n-1)(-4)<0{tex} \\Rightarrow {/tex}\xa0121-4n+4<0{tex} \\Rightarrow {/tex}\xa0-4n+125 < 0{tex} \\Rightarrow {/tex}\xa0-4n< -125Hence, 4n>125{tex} \\Rightarrow {/tex}\xa0n >{tex}\\frac{{125}}{4}{/tex}So, n > 31.25Hence, the 32nd term is first negative term. | |
| 11979. |
Datesheet of 10th ka URL do koi! |
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| 11980. |
Impotant question gor quadratic equation |
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| 11981. |
RIMSHA????? |
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| 11982. |
If 5th term of an ap is 0 show that 33rd term is two times its 19th term |
| Answer» So easy ?? | |
| 11983. |
RIMSHA plzzz come |
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| 11984. |
How do we find root |
| Answer» By long division method | |
| 11985. |
Find the area of the circle that can be inscribed in a square of side 8cm |
| Answer» side of square=8diameter=8radius=4area=pi x r2 = 22/7 x 16 = 3.14x16=50.24 cm2 | |
| 11986. |
(1/cosecA-cotA)-(1/sinA)=(1/sinA)-(1/cosecA+cotA) |
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| 11987. |
Find the roots of the equation 5x ×x+2x+5 |
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Answer» It has no real roots. -5/3 vvjjhdhxhz |
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| 11988. |
Rd sharma chapter 8 examples no. 8 |
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| 11989. |
2x _5+8=0 |
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Answer» 2x+3=0X=-3/2 2x _5+ 8 |
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| 11990. |
Alfa and bita are zeros of x2-5x+k such that Alfa -bita =1 |
| Answer» (Alfa+bita)=5 and alfa ×bita=K. Also alfa - bita =1(given) (alfa +bita)2-(alfa - bita)2 =4alfa×bita 5ka square - 1ka square =4alfa×bita 4alfa ×bita =(25-1)=24Alfa ×bita =6Hence, k=6. | |
| 11991. |
How can i read ch triangle i dont understand this |
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| 11992. |
The n term of an ap is 7-4n find common difrence |
| Answer» an = 7 - 4n {tex}\\Rightarrow{/tex}\xa0a1\xa0= 7 - 4 {tex}\\times{/tex}\xa01\xa0= 3a2 = 7 - 4 {tex}\\times{/tex} 2 = -1 and a3 = 7 - 4 {tex}\\times{/tex}\xa03 = -5{tex}\\therefore{/tex}\xa0Common difference = a2\xa0- a1 = -1 - 3 = -4. | |
| 11993. |
Who to solve trigonometry questions |
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| 11994. |
54+6 |
| Answer» 60 | |
| 11995. |
Prove that sinA-cosA+1/ sinA+cosA-1=1/secA-tanA using this identity secA-tanA |
| Answer» 1)divide numerator and denominator by cosA in left hand side and try to solve it | |
| 11996. |
Sin u +cos u= m and sec u + cosec u=n then prove that n(m²-1)=2m |
| Answer» Given,\xa0sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex}\xa0= m and sec{tex}\\theta{/tex}\xa0+ cosec{tex}\\theta{/tex}\xa0= nConsider,\xa0n(m2 - 1) = (sec{tex}\\theta{/tex}\xa0+ cosec{tex}\\theta{/tex})[(sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex})2\xa0-\xa01]{tex}= \\left( \\frac { 1 } { \\cos \\theta } + \\frac { 1 } { \\sin \\theta } \\right){/tex}[(sin2{tex}\\theta{/tex}\xa0+ cos2{tex}\\theta{/tex}) + 2sin{tex}\\theta{/tex}cos{tex}\\theta{/tex}\xa0- 1]{tex}= \\frac { ( \\sin \\theta + \\cos \\theta ) } { \\sin \\theta \\cos \\theta } \\cdot [2 \\sin \\theta \\cos \\theta]{/tex}= 2(sin{tex}\\theta{/tex}\xa0+ cos{tex}\\theta{/tex})= 2m | |
| 11997. |
If check whether 1-tan2A/1+tan2A=cos2A-sin2A or not |
| Answer» 1-tan2A/sec2A1-sin2A/cos2A/sec2A=cos2A-sin2A/cos2A/1/cos2A=cos2A-sin2S | |
| 11998. |
Prove that :1/tanA +cotA 1/1+sinA +1/1-sinA=2sec2A |
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| 11999. |
X+=14X-y=4Solve in substitution method |
| Answer» x+y=14x=14-y(1)put it in equation 2x-y=414-y-y=4 (from1)14-2y=4-2y=-10y=5x=14-yx=9 | |
| 12000. |
What is the solution of the pair of equations y=0 and y=-3 |
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