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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.
| 7251. |
Find the a p of the term |
| Answer» | |
| 7252. |
2-1 |
| Answer» 1 | |
| 7253. |
Check whether 301 is a term of the lists of no. 5,11,17,32 |
| Answer» a=5d=6let us assume that it is the nth term of APan=a+(n-1)d = 3015+(n-1)6=3015+6n-6=301n=302/6=50.3 but it is coming in decimaltherefore it is not a term of AP | |
| 7254. |
x÷x+1 +x+1÷x =34÷15 |
| Answer» x=-5\\2 or x=3\\2 | |
| 7255. |
SirTan;÷????? |
| Answer» | |
| 7256. |
(1+ Ten²A) (1-SinA) (1+CosA) |
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| 7257. |
If cot A=12_5 verify that tan²A -Sin²A= Sin¾Sec²A |
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| 7258. |
√2-√8 |
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| 7259. |
Find SinA+cosecA |
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Answer» 0 0 |
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| 7260. |
34+32+30+...+10 |
| Answer» 20 | |
| 7261. |
Sec39/cosec51 -2/√(3)[tan17 tan38 tan60 tan52 tan73 - 3(sin^2 31 + sin^2 59)] |
| Answer» Given,\xa0{tex}\\frac { \\sec 39 ^ { \\circ } } { \\operatorname { cosec } 51 ^ { \\circ } } + \\frac { 2 } { \\sqrt { 3 } }{/tex}{tex}tan17° tan38° tan60°tan52°tan73° -3(sin^231° + sin^259°){/tex}=\xa0{tex}\\frac { \\sec 39 ^ { \\circ } } { \\operatorname { cosec } \\left( 90 ^ { \\circ } - 39 ^ { \\circ } \\right) } + \\frac { 2 } { \\sqrt { 3 } }{/tex}tan17°tan 38° tan 60°tan (90°-38°) tan (90°-17°) -3(sin231° + sin2(90° - 31°)){tex}= \\frac { \\sec 39 ^ { \\circ } } { \\sec 39 ^ { \\circ } } + \\frac { 2 } { \\sqrt { 3 } } {/tex}tan 17° tan 38°\xa0{tex}\\times{/tex}\xa0{tex}\\sqrt{3}{/tex}\xa0{tex}\\times{/tex}\xa0cot 38° {tex}\\times{/tex}\xa0cot 17° -3(sin231° + cos231°)= 1 +\xa0{tex}\\frac { 2 } { \\sqrt { 3 } } \\times \\sqrt { 3 }\\times1\\times1{/tex}\xa0- 3 [{tex}\\because tan\\theta . cot \\theta = 1{/tex}]{tex}= 1 + 2 - 3\\\\ = 0{/tex} | |
| 7262. |
ax2+bx+c=0 find the value of C |
| Answer» C=1 | |
| 7263. |
Sir what the formula of ap |
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| 7264. |
Proove that relation between the zeroes and coefficient of a quadratic polynomial |
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| 7265. |
Pythagoras thermos |
| Answer» H^2=P^2+B^2. Here H is hypotenuse, P is perpendicular and B is base of triangle | |
| 7266. |
How can I delete a comment??? |
| Answer» | |
| 7267. |
find the roots of the equation by completing square method ....5x^2-6x-2=0 |
| Answer» Devide this eqation by 5 and then the midlle term of eqation multiply by half and square of .then the no. Coming add the midlle term and substract the last term and the first three terms make (a^2-b^2) and the last two terms take LCM then the no. Has given by last two no.come on the R.H.S.the rhs is take to unroot in rhs then the roots have find | |
| 7268. |
Theorem 6.2 |
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| 7269. |
How to find factors in factorization method |
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| 7270. |
Aa |
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| 7271. |
Theoram 6.6 |
| Answer» Answer | |
| 7272. |
Simplify:2root45+3root20/2root3 and say whether rational or irrational |
| Answer» I think irrational | |
| 7273. |
Which term of the ap 24 ,21,18,15, is first negative term |
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Answer» a=24d= - 3an=<0an=a+(n-1)d24 -3n+3<027 - 3n <027<3n9 |
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| 7274. |
Gdgssvsvsvzvdb |
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| 7275. |
Find the value of x,y if the distance of the point x,y from -3,0 as well as from 3,0 are 4 |
| Answer» let point are A , B , CA (-3,0), B(3,0) , C (X,Y)AC=BC=4 given(x+3)2 + (y-0)2 = (x-3)2 + (y)2x2+9+6x +y2 = x2+9- 6x+y2x2+6x+9 + y2=4 .... ix2-6x+9 + y2=4............ iifrom i and ii12x=0x=0, y= root -5\xa0 | |
| 7276. |
finđ distance between points (cos,sin) (sin,cos) |
| Answer» | |
| 7277. |
The AM of a and b is {a^(n+1)+b^(n+1)}÷{(a^n)+(b^n)} |
| Answer» Not understannd | |
| 7278. |
New sample question paper of all subject of 2017 -2018 of CBSE which help in final board |
| Answer» Yes,sample papers gives all information about new board papers. | |
| 7279. |
If 1 is a zero of polynomial p(x) = ax²-3(a-1)x-1 then find the value of a. |
| Answer» A=1 | |
| 7280. |
The difference between the circumference and the diameter is 30cm . Find the radius of the circle |
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Answer» circumference=2pirdiameter=2r2 pi r - 2r = 302r (pi - 1) = 30r = 15 / 22/7 -\xa01 = 15 x 7 / 15 = 7 cm 30 cm If u take radius as x then diameter will be 2x.So 2x-x=30. So x is 30Radius is 30 cm |
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| 7281. |
What is formula of area at cube |
| Answer» Area of cube = 6a2 | |
| 7282. |
Which term of AP 92,88,84,80 ........................ Is 0 |
| Answer» a=92d= - 4let nth term = 0an=a+(n-1)d0=92+4-4nn=96/4=2424th term = 0check ur answer a24=92+23x-4 = 92-92=0 ok | |
| 7283. |
Arithmetic progression In this AP find the missing term5,_,_,9 1/2 |
| Answer» a=5 and a+3d=9 so put the value of a in 2eqn | |
| 7284. |
(a-b)(x+y)=a)2+b)2 |
| Answer» | |
| 7285. |
The roots of a quadratic equation are 5and -2 then the equation is? |
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Answer» 5x^2-2^x+1=0 X^2-3x-10 |
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| 7286. |
What is the maximum and minimum value of sinA |
| Answer» Maximum is sin 90 that is 1 n minimum is sin 0 that is 0 | |
| 7287. |
Two x square +ax-a square |
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| 7288. |
√25 |
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Answer» 5 5 |
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| 7289. |
Post science samples paper of 2018 |
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| 7290. |
Write a rational number between root 2 and root 5 |
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| 7291. |
Find the sum of the first 25terms of an ap whose nth term is given by tn =7-3n |
| Answer» Given, {tex}n^{th}{/tex}\xa0term of an AP,\xa0{tex}{a_n} = 7-3n{/tex}{tex}\\therefore a= 7-3(1)=4{/tex}{tex}a_{25}=7-3(25)=-68{/tex}Sum of\xa0{tex}n{/tex}\xa0terms of an AP,{tex}S_n=\\frac{n}{2}(a+l){/tex}, where\xa0{tex}a {/tex}\xa0is the first term and\xa0{tex}l{/tex}\xa0is the last term ,here\xa0{tex}l=a_{25}{/tex}Clearly, sum of the first 25 terms, (S25){tex} = \\frac{{25}}{2}(a + {a_{25}}) {/tex}{tex}= \\frac{{25}}{2}[4 + (-68)]{/tex}{tex}= \\frac{25}{2}(-64){/tex}{tex}=25\\times(-32){/tex}{tex} = -800{/tex} | |
| 7292. |
Find the value of k, for which the point are collinear (8,1),(k, -4),(2,-5) |
| Answer» (8, 1), (k, -4), (2, -5)Since, the given points are collinear, it means the area of triangle formed by them is equal to zero.Area of Triangle = {tex}\\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right] = 0{/tex}⇒ ½ [8 {−4−(−5)} +k(−5−1)+2{1−(-4)}]⇒ ½ (8 − 6k + 10 )= 0⇒ ½ (18 − 6k) = 0⇒ 18 − 6k = 0⇒ 18 = 6k⇒ k = 3 | |
| 7293. |
1+2 |
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Answer» 3 3 3 |
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| 7294. |
Using Euclid’s division algorithm, find the HCF of 2160 and 3520. |
| Answer» HCF is 3 | |
| 7295. |
What rational |
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Answer» A number which can be written in the form of p/q where p and q are integers. P/q That can be written in the form of p/q |
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| 7296. |
In construction chapter is it important to write justification of each question |
| Answer» | |
| 7297. |
2018 board question paper is tuff or not |
| Answer» No it will be normal | |
| 7298. |
sin 25\xa0sin 65 + cos 25 sin 65 |
| Answer» 1 | |
| 7299. |
What is the standard form |
| Answer» | |
| 7300. |
What is the value of pie |
| Answer» 22/7 or3.14 | |