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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.
| 4251. |
Prove that root 2 is irrational |
| Answer» See rs aggrawal book chapter no. 1 | |
| 4252. |
About question trignometry |
| Answer» Which question? | |
| 4253. |
Area of isoselis triangle |
| Answer» {tex}A = \\frac { b h _ { b } } { 2 }{/tex} | |
| 4254. |
Trianglr |
| Answer» ..?? | |
| 4255. |
If sinA=cosA find the value of A |
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Answer» Hi r u in brainy SinA=sin(90-A)A=90-A2A=90A=45 |
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| 4256. |
How many of have taken holiday frm school?? |
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Answer» Randi sab kya kaha rahi h ham online bol kiya bolna h adahik bolagi to randi to samaj lana kiya karanga tumahara sath Hey do anybody of u know about khusbu khatun Ok i may talk to him when he will be online No Anjali dont u have her contact Yes annanya u r right His Acha us raunak ne tumhe tang kiya isliye u dont want to listrn hos name U knw what anaanya i m missing my friend anjali Mana aj K ha Me too Me annanya |
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| 4257. |
Circumference is 282 find volumesh of spehre |
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Answer» no ...it will be 538.85 approx 22,892,971.5 is the volume of sphere use formula v =c^3\\pie ^2 |
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| 4258. |
100*100 |
| Answer» 10000 | |
| 4259. |
Practicals of class 10 maths |
| Answer» | |
| 4260. |
T ratios of particular angles |
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Answer» U have to learn to make chart of sin(-)For first root 0/4. (0degree) 2nd root 1/4. (30\'d) 3rd. Root 2/4. 45\' 4th. Root. 3/4. 60\' 5th. Root 4/4. 90\'Do calculation if you really want to learn. Cos (-) reverse of sinTan0, 1/√3, 1,√3,~ Cot is reverse of tan.Cosec opposite to every degree of sinLike first 0 converted to 1/0 means infiniteSame in sec opposite to every degree of cos. see textbook |
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| 4261. |
Find the area of rhombus each side of which measures 40cm and one of whose diagonal is 48cm. |
| Answer» Diagonals of Rhombus bisect each other *48×1/2=24cmDiagonals of Rhombus bisect each other at right angle By Pythagoras theorem = 24^2 + AO^2 = 40^2 =576+ AO^2 =1600 =AO^2=. 1600-576= 1024=√1024=32cmD1=48cm (given)D2=32x2=64cm* Area of Rhombus=1/2 x D1 x D2=1/2x48x64= 1536 | |
| 4262. |
Be//ab ad//2x dc=x+3 be=2x-1 ce=x |
| Answer» | |
| 4263. |
Find the coordinates of the point on y-axis which is nearest to the point (–2, 5). |
| Answer» You should take points be(0,y) And solve the problem | |
| 4264. |
The sum of product of zero P(X)=63x-7x-9 |
| Answer» | |
| 4265. |
(5+6)⅔ |
| Answer» | |
| 4266. |
Find the perpendicular distance of A (5,12) from y-axis |
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Answer» Its 5 5cm from perpendicular from y-axis 12 |
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| 4267. |
What is the pH value of Acid rain. |
| Answer» less than 5.6... | |
| 4268. |
Define empirical relation |
| Answer» mode = 3median - 2mean.. | |
| 4269. |
How many polynomials will have their zeroes as -2&5 |
| Answer» more than 3 | |
| 4270. |
Find area of quadrilateral whose vertices are _4;2 _3;_5 3;_2 2;3 |
| Answer» Let A(-4, -2), B(-3, -5), C(3, -2) and D(2, 3)Area of Quadrilateral ABCD = Area of Triangle ABD + Area of Triangle BCD .................... (1)Using formula to find area of triangle:\xa0Area of {tex}\\triangle{/tex}ABD{tex} = \\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right]{/tex}{tex} = \\frac{1}{2}\\left[ { - 4( - 5 - 3) - 3\\left\\{ {3 - ( - 2)} \\right\\} + 2\\left\\{ { - 2 - ( - 5)} \\right\\}} \\right]{/tex}{tex} = \\frac{1}{2}(32 - 15 + 6) = \\frac{1}{2}(23) = 11.5sq\\;units{/tex}\xa0........... (2)Again using formula to find area of triangle:Area of {tex}\\triangle{/tex}BCD\xa0{tex} = \\frac{1}{2}\\left[ {{x_1}({y_2} - {y_3}) + {x_2}({y_3} - {y_1}) + {x_3}({y_1} - {y_2})} \\right]{/tex}{tex} = \\frac{1}{2}\\left[ { - 3( - 2 - 3) + 3\\left\\{ {3 - ( - 5)} \\right\\} + 2\\left\\{ { - 5 - ( - 2)} \\right\\}} \\right]{/tex}{tex} = \\frac{1}{2}(15 + 24 - 6) = \\frac{1}{2}{/tex}\xa0(33) = 16.5 sq units ........ (3)Putting (2) and (3) in (1), we getArea of Quadrilateral ABCD = 11.5 + 16.5 = 28 sq units | |
| 4271. |
In an ap if the pth term is q and 8th term is P then find its (P + Q )th term |
| Answer» q = a + (p - 1)d ..... (i)p = a + (q - 1)d ...... (ii)q - p = (p - 1 - q + 1)d{tex}\\frac{{q - p}}{{p - q}} = d{/tex}{tex} \\Rightarrow d = - 1{/tex}Put the value of d in eq (i)q = a + (p - 1) (-1){tex} \\Rightarrow {/tex}\xa0q = a - p + 1{tex} \\Rightarrow {/tex}\xa0a = q + p - 1ap+q = a +(p + q - 1)d= (q + p - 1) + (p + q - 1) (-1)= q + p - 1 - p - q + 1= 0 | |
| 4272. |
find all the zeros of X the power 4 minus 3 x cube minus 3 X square + 6 x minus 2 of 2 of 8002 - 2 |
| Answer» | |
| 4273. |
prove that (sinθ-2sin cube θ) =( 2 cos cube -cosθ)tanθ |
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Answer» shi baat h... Yha kese kre...?? Refer rd sharma.. |
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| 4274. |
Connection between volume of two objects |
| Answer» | |
| 4275. |
If cos =2/5 find the 4+4 tan2A |
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Answer» b/h =2/5. .•. p=√5² -2² =√21. Therefore, 4+4tan² =4+4(√21/2)² = 4+21 = 25. If cos =2/5 find the 4+4 tan2A |
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| 4276. |
2x is equal to b |
| Answer» | |
| 4277. |
An equil |
| Answer» | |
| 4278. |
What is the identity use for cube In real no. |
| Answer» | |
| 4279. |
The ratio of the sum of n terms of two AP\'s is 7n+1:4n+27. Find the ratio of their 9th terms. |
| Answer» 64:63 | |
| 4280. |
Question papers for 2018 |
| Answer» Maene thode hi banaye hau | |
| 4281. |
What is the value of log10? |
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Answer» 1 1 1 1 It is 1. |
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| 4282. |
29374837+648463836391-738462836392+83847392947-= -638363846394639373936384628 |
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Answer» Correct don\'t waste our time Do on calculator.???? |
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| 4283. |
Check whether (5,-2) (6,4) & (7,-2) are the vertices of an isoceles triangle |
| Answer» Let A {tex}\\rightarrow{/tex} (5, -2), B {tex}\\rightarrow{/tex} (6, 4) and C {tex}\\rightarrow{/tex} (7, -2)Then,{tex}AB = \\sqrt {{{(6 - 5)}^2} + {{(4 - ( - 2))}^2}} = \\sqrt {{{(1)}^2} + {{(6)}^2}}{/tex}{tex} = \\sqrt {1 + 36} = \\sqrt {37}{/tex}{tex}BC = \\sqrt {{{(7 - 6)}^2} + {{( - 2 - 4)}^2}} = \\sqrt {{{(1)}^2} + {{( - 6)}^2}}{/tex}{tex}= \\sqrt {1 + 36} = \\sqrt {37}{/tex}We see that AB = BCtherefore, ABC is an isosceles triangle.Let D be the mid-point of BC. Then, coordinates of D are {tex}\\left( \\frac { 5 + 7 } { 2 } , \\frac { - 2 - 2 } { 2 } \\right){/tex}\xa0i.e, (6, -2)Therefore, AD =\xa0{tex}\\sqrt { ( 6 - 6 ) ^ { 2 } + ( 4 + 2 ) ^ { 2 } } = \\sqrt { 36 } = 6{/tex} | |
| 4284. |
what is the formula of trignomerty |
| Answer» It is in Ncert book | |
| 4285. |
A man goes 15m due west and then 8m due north how far is he from the starting point |
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Answer» 17m 17cm Using Pythagoras theorem we can say that 15*15+8*8=root of 289=17m |
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| 4286. |
How to do completing the square method |
| Answer» I think u understand from any help book ? ? | |
| 4287. |
What type of central tendency we have to find when average of given data is asked? |
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Answer» I checked out this que recently on internet and the answer was mode. Mode is the right answer. Arithmetic mean Mean |
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| 4288. |
1/cosec A- cot A-1/sin A=1/sinA-1/cosecA+cot A |
| Answer» We have,\xa0{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 1 } { \\text{cosec} A - \\cot A } - \\frac { 1 } { \\sin A }{/tex}=\xa0{tex}\\frac { 1 } { \\sin A } - \\frac { 1 } { \\text{cosec} A + \\cot A }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 1 } { \\text{cosec} A - \\cot A } + \\frac { 1 } { \\text{cosec} A + \\cot A }{/tex}=\xa0{tex}\\frac { 1 } { \\sin A } + \\frac { 1 } { \\sin A }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 1 } { \\text{cosec} A - \\cot A } + \\frac { 1 } { \\text{cosec} A + \\cot A }{/tex}=\xa0{tex}\\frac { 2 } { \\sin A }{/tex}LHS =\xa0{tex}\\frac { 1 } { \\text{cosec} A - \\cot A } + \\frac { 1 } { \\text{cosec} A + \\cot A }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { \\text{cosec} A + \\cot A + \\text{cosec} A - \\cot A } { ( \\text{cosec} A - \\cot A ) ( \\text{cosec} A + \\cot A ) }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { 2 \\text{cosec} A } { \\text{cosec} ^ { 2 } A - \\cot ^ { 2 } A }{/tex}{tex}\\Rightarrow{/tex}\xa0{tex}\\frac { \\frac { 2 } { \\sin A } } { 1 } = \\frac { 2 } { \\sin A }{/tex}= RHS.Hence Proved. | |
| 4289. |
Divide and show 1390÷35 |
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Answer» Show is not there 39.71428571 |
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| 4290. |
A set of numbers consists of eight 4\'s,four 5\'s,five 7\'s and six 9\'s. Then find the mode. |
| Answer» 4 is a mode | |
| 4291. |
Paythagoras theoram |
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Answer» Hypotenuse square =square of base+square of the third side H2=p2+b2 (H)²=(B)²+(P)² |
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| 4292. |
How can we prepare for 10 board |
| Answer» Study karke | |
| 4293. |
Angle of elevation |
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Answer» Angle formed by our eyes above horizontal level. The angle of elevation of an object as seen by an observer is the angle between the horizontal and the line from the object to the observer\'s eye (the line of sight). |
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| 4294. |
10/x+y + 2/x-y =4 15/x+y -5/x-y=3 |
| Answer» Let 1/x+y=p and 1/x-y=q then two eqns are formed....10p+2q=4 and 15p-5q=3 then solve it then u will find value of p and q then u can easily find x&y.... | |
| 4295. |
In triangle ABC,DE parallel to BC, Find the value of x,if AD = 3x+19,BE = 3x+4,CD = x+3,CE =x |
| Answer» | |
| 4296. |
What is Eculid division lemma |
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Answer» O<=r<=b Given positive integer a and b there exist a unique integer q and r satisfying a= bq+r |
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| 4297. |
Prove that frustm of cone ? Please describe this question ? ???? |
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Answer» Ye kesa question hai???? Go see in ncert |
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| 4298. |
Prove th |
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Answer» Achaa .....and yes vese i guess we are friends now??? Nhi aaj pehle pee aya... 8 bje gya tha... Pal jaega chai peene? Ajj Busy haii kya??? Achaaa ???? Anushka me 109 haribhau upadhyaya nagar main... me rehta hu... Aisa kuch nhi hai... aapse baat hi nhi hui aaj.. Aaj toh shayad Ishan mujhse naraz haii tabhi toh reply nahi de raha? F9... Hii how are you ?? Whatt?? Madam |
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| 4299. |
Triangles chapter apko easy lagta hai ya hard?? Mujje bada bakwas lagta hai |
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Answer» From which skol and state are uh nishu Hii Hii Normal Same here |
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| 4300. |
If 3p+1,7and 6 p +1 are three consective term of ap then find p |
| Answer» Since common difference between two consecutive terms in an AP is same.So,d=(7-3p-1)=(6p+1-7)therefore,6-3p=6p-69p=12p=4/3 | |