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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.
| 35251. |
Surface area |
| Answer» Kiska | |
| 35252. |
What do you think about these questions |
| Answer» What questions????????????? | |
| 35253. |
If cosec theta is equal to root 10 find other trigonometric ratios |
| Answer» We have,\xa0{tex}cosec \\;A = \\frac { \\text { Hypotenuse } } { \\text { Perpendicular } } = \\frac { \\sqrt { 10 } } { 1 }{/tex}So, we draw a right triangle ABC, right-angled at B such thatPerpendicular = BC = 1 unit and, Hypotenuse{tex}= A C = \\sqrt { 10 }{/tex}.By Pythagoras theorem, we have{tex}A C ^ { 2 } = A B ^ { 2 } + B C ^ { 2 }{/tex}{tex}\\Rightarrow \\quad ( \\sqrt { 10 } ) ^ { 2 } = A B ^ { 2 } + 1 ^ { 2 }{/tex}{tex}\\Rightarrow \\quad A B ^ { 2 } = 10 - 1 = 9{/tex}{tex}\\Rightarrow \\quad A B = \\sqrt { 9 } = 3{/tex}When we consider the trigonometric ratios of {tex}\\angle A{/tex}, we haveBase = AB = 3 units, Perpendicular = BC = 1 units and, Hypotenuse{tex}= A C = \\sqrt { 10 }{/tex}\xa0units{tex}\\therefore \\quad \\sin A = \\frac { \\text { Perpendicular } } { \\text { Hypotenuse } } = \\frac { 1 } { \\sqrt { 10 } } , \\cos A = \\frac { \\text { Base } } { \\text { Hypotenuse } } = \\frac { 3 } { \\sqrt { 10 } }{/tex}\xa0{tex}\\tan A = \\frac { \\text { Perpendicular } } { \\text { Base } } = \\frac { 1 } { 3 } , \\quad \\sec A = \\frac { \\text { Hypotenuse } } { \\text { Base } } = \\frac { \\sqrt { 10 } } { 3 }{/tex}.and,\xa0{tex}\\cot A = \\frac { \\text { Base } } { \\text { Perpendicular } } = \\frac { 3 } { 1 } = 3{/tex} | |
| 35254. |
TanA = √3-1 then find sinA.cosA =? |
| Answer» | |
| 35255. |
Expres An of an AP in term of Sn and Sn-1 |
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Answer» Any one Short |
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| 35256. |
If one zero of the polynomial (a^2+ a)X^2+ 13 x + 6 x A reciprocal is other •find the value of a |
| Answer» Let {tex} \\alpha{/tex}\xa0and\xa0{tex} \\frac { 1 } { \\alpha }{/tex} be the zeros of\xa0(a2\xa0+ 9)x2\xa0+ 13x\xa0+ 6a.Then, we have{tex} \\alpha \\times \\frac { 1 } { \\alpha } = \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa01 =\xa0{tex} \\frac { 6 a } { a ^ { 2 } + 9 }{/tex}⇒\xa0a2\xa0+ 9 = 6a⇒ a2 - 6a + 9 = 0⇒\xa0a2\xa0- 3a - 3a + 9 = 0⇒\xa0a(a - 3) - 3(a - 3) = 0⇒\xa0(a - 3) (a - 3) = 0⇒\xa0(a - 3)2\xa0= 0⇒\xa0a - 3 = 0⇒ a = 3So, the value of a in given polynomial is 3. | |
| 35257. |
can anyone.give me an innovative idea for maths exhibition class 10 |
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Answer» Trigonometric ratios Theorems |
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| 35258. |
secQ + tanQ =mSecQ +tanQ =n Show that mn=1 |
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Answer» (secQ+tanQ)(tanQ-secQ)-Sec²Q+secQtanQ-secQtanQ+tan²Q=Tan²Q-sec²Q=1 The question is not correctIt is secQ+tanQ=m and secQ-tanQ=n |
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| 35259. |
What is concentric |
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Answer» the circles which drawn from same centre Circles with same centre |
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| 35260. |
What is modal class? |
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Answer» So...if we are given a question in which we have a table with wages like less than 10 ,less than 20 ,less than 30 ,less than 40 ,less than 50,and less than 60 with their respective no. of students like 2 ,11,25,45,57,75.....so here what is the modal class ..."75" or "less than 60"??? Modal is highest no. Of frequency Mode |
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| 35261. |
Prove that 2 equilateral triangles are similar |
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Answer» angles of equilateral triangle are equal than ABC~PQR by sss rule Equilateral triangles have equal sides so that all angles are equal let one triangle is ABC And another is PQR |
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| 35262. |
The angle of depression of the top and the bottom of 7m tall building |
| Answer» Incomplete question | |
| 35263. |
What type of paper will cbse choose for 2019 board exams for class 10 ?esay or difficult? |
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Answer» i think level ka hi hoga Ofcorse difficult hoga lekin padhne walo ke liye nahi Easy |
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| 35264. |
(2x-x)=(3x+1) |
| Answer» -1/2 | |
| 35265. |
Find the value of x the distance between the points (1,x) and (-2,-2) is 3 root 2 units |
| Answer» | |
| 35266. |
The number lies between 100000 and 110000 which is exactly divisible by 8,15 and 21 is |
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Answer» How 109200 |
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| 35267. |
Write all the formulas from chapter 1 to chapter 15 |
| Answer» There are many. U can get all by noting from the cbse revision notes of all mathematics chapters from my cbseguide app. | |
| 35268. |
1.The first three terms of an AP are respectively (3y_ 1),(3y+5),and (5y+1),find the value of y |
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Answer» Y = 5 The common difference between a1 and a2Common difference = a2 - a1=(3y + 5) - (3y - 1)=3y + 5 - 3y + 1=6. ――1Now the common difference between a3 and a2Common difference =a3 - a2=(5y + 1) - (3y + 5)=5y + 1 - 3y - 5= 2y - 4 ――2On equating 1 and 2 we get2y - 4 = 62y = 10y = 5So the value of y is 5 |
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| 35269. |
Sin360 is what |
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Answer» Correct 0 |
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| 35270. |
If sinQ =cosQ ,find the value |
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Answer» kiski value nikale 45 |
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| 35271. |
Find the value of k, if the quadratic equations 3x square -k under root +4 has eual root |
| Answer» 4under root 3 | |
| 35272. |
Vvv |
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Answer» What you say , I don\'t understand You need xxx Vvvv |
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| 35273. |
Given that HCF (253,440) = 11 and LCM (253,440)=253 x R, find the value of R |
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Answer» HCF×LCM= 253×44011×253R=111320. (÷11)253R= 10120. (÷253)R= 40 55555 |
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| 35274. |
Find the condition that the zeros of the polynomial f(x)=x3+3px2+3qx+r may be in A.P. |
| Answer» The given quadratic polynomial is:f(x) = x3 + 3px2 + 3qx + rwe have to show that the zeroes of given polynomial are in the form of AP.Let, a - d, a, a + d be the zeroes of the polynomial, thenThe sum of zeroes = {tex}\\frac{{ - b}}{a}{/tex}a + a - d + a + d = -3p3a = - 3pa = - pSince, a is the zero of the polynomial f(x),Therefore, f(a) = 0f(a) = a3 + 3pa2 + 3qa + r = 0{tex}\\Rightarrow{/tex}\xa0a3 + 3pa2 + 3qa + r = 0{tex}\\Rightarrow{/tex}\xa0(-p)3 + 3p(-p)2 + 3q(-p) + r = 0{tex}\\Rightarrow{/tex}\xa0-p3 + 3p3 - 3pq + r = 0{tex}\\Rightarrow{/tex} 2p3 - 3pq + r = 0Which is the required condition. | |
| 35275. |
Solve the value of sin A |
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Answer» P/H B/H Sorry B/H |
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| 35276. |
Simplify DMAS (12+7)1 |
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Answer» 19 19 19 19 |
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| 35277. |
Evalute - cosec square 63°+ten square 27° / cot square 66°+sec square 24° |
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Answer» Plese evalute this question and explain I know answer is1 but not write how to explain 1 |
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| 35278. |
Determine the Ap whose 3rd term is 5 and the 7th term is 9? |
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Answer» X+2d=5X+6d=9Solution-4d=-4d=1Now the value of x isX+2=5X=3So first term is 3,The Ap is= 3,4,5,6...... Ap :3,4,5,6,7,8,9,... |
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| 35279. |
Prove that n2 - n is divisible by 2 for any positive integer n.?? |
| Answer» Any positive integer is of the form 2q or 2q + 1, for some integer q.{tex}\\therefore{/tex} When n = 2q{tex}\\style{font-family:Arial}{n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;2q(2q\\;-\\;1)=\\;2m,}{/tex}where m = q(2q - 1) ( m is any integer)This is divisible by 2When n = 2q + 1{tex}\\style{font-family:Arial}{\\begin{array}{l}n^2\\;-\\;n\\;=\\;n(n\\;-\\;1)\\;=\\;(2q\\;+\\;1)(2q+1-1)\\\\=2q(2q+1)\\end{array}}{/tex}= 2m, when m = q(2q + 1) ( m is any integer)which is divisible by 2.Hence, n2 - n is divisible by 2 for every positive integer n. | |
| 35280. |
Pove √3 |
| Answer» Plz write full question ....? | |
| 35281. |
Hcf of 405 and 2520 |
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Answer» We know that 2520 greater than 405 so we apply division algorithm to 2520 and 405a=bq+r (r=0)2520=405×6+900Here r is not equal to 0 so aplly division lemma on 900 and 405900=405×2+90Here r is not equal to 0So aplly division lemma on 405 and 90405=90×5+5Here r is not equal to 0 so aplly division lemma on 5 and 5 5=5×1+0 here r = 0 So the HCF of 2520 and 405 is 5 8 |
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| 35282. |
Find the point on x-axis which is equidistant fron (2,-5)and (-2,9) |
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Answer» Point is (-7,0) Draw a pair up to down and left to right centre point origin hoga or up or down line ko y bolengay or left to right ko x abb number likh loo 12345678 both side and fir +- mai karna fir point show kardooo |
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| 35283. |
Find n term of A.P n+7:3n+1 find ratio of 7^th term |
| Answer» | |
| 35284. |
How many Formula in class 10th |
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Answer» 000000 About 200 In which subject or in whicb chapter ?? How many Formula in class 10th Math subject |
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| 35285. |
How many theorems in circle |
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Answer» Two theorems. 2 in ncert and 2 are in reference books |
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| 35286. |
Sum of two degit is 31. If 27 added to this number that number is interchanged. What is the number |
| Answer» | |
| 35287. |
(xy+yz)-(xy-yz)=? |
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Answer» 2yz (xy)square -(yz)square Thank you very much!☺ (xy + yz) - (xy - yz) = xy + yz - xy + yz = yz + yz = 2yz 2yz 2yz |
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| 35288. |
Prove that 1÷a+b+x=1÷a+1÷b+1÷x |
| Answer» Given,{tex}\\frac { 1 } { ( a + b + x ) } = \\frac { 1 } { a } + \\frac { 1 } { b } + \\frac { 1 } { x }{/tex}{tex}\\Rightarrow \\quad \\frac { 1 } { ( a + b + x ) } - \\frac { 1 } { x } = \\frac { 1 } { a } + \\frac { 1 } { b } \\Rightarrow \\frac { x - ( a + b + x ) } { x ( a + b + x ) } = \\frac { b + a } { a b }{/tex}{tex}\\Rightarrow \\quad \\frac { - ( a + b ) } { x ( a + b + x ) } = \\frac { ( a + b ) } { a b }{/tex}On dividing both sides by (a+b){tex}\\Rightarrow \\quad \\frac { - 1 } { x ( a + b + x ) } = \\frac { 1 } { a b }{/tex}Now cross multiply{tex}\\Rightarrow{/tex}\xa0x(a + b + x) = -ab\xa0{tex}\\Rightarrow{/tex}\xa0x2 + ax + bx + ab = 0{tex}\\Rightarrow{/tex}\xa0x(x +a) + b(x +a) = 0{tex}\\Rightarrow{/tex}\xa0(x\xa0+ a) (x + b) = 0{tex}\\Rightarrow{/tex}\xa0x + a = 0 or x + b = 0{tex}\\Rightarrow{/tex}\xa0x = -a or x = -b.Therefore, -a and -b\xa0are the roots of the equation. | |
| 35289. |
Class 10 ncert important ques |
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Answer» Thanks Also, You can find last year questions on this cbse guide app:)? https://www.cbsetuts.com/chapter-wise-important-questions-class-10-social-science/Visit the site you can get all subject important questions Which chapter? All chapters |
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| 35290. |
Compound interest |
| Answer» Compound interest is the addition of interest to the principal sum of a loan or deposit, or in other words, interest on interest. It is the result of reinvesting interest, rather than paying it out, so that interest in the next period is then earned on the principal sum plus previously accumulated interest. | |
| 35291. |
Is any sample question paper is sended |
| Answer» Cbseacademic.nic.in | |
| 35292. |
If 21 term of AP is 25.find sum of AP whose 41 term? |
| Answer» | |
| 35293. |
Guys jo CBSE ne sample paper upload kiye h usme se kuch aa sakta h Kya plz reply. |
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Answer» Cbseacademic.nic.in Cbse academics se download kar lo jo cbse official website h Sample paper ki link kya ha Yes Haan usse related questions aa sakte hai |
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| 35294. |
What is Maths?? |
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Answer» Nice que Mathematics is a old, board, deep discipline (field of study).people wrking to improve math education need to understand "what is math?"✔ Maths is a field of study ?? Full form of maths mere aatma tuse hamesa satayegi . It is a book by which we can feel bore. |
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| 35295. |
Rr |
| Answer» ? | |
| 35296. |
Evaluate this:Cos58/sin32 + sin22/cos68 - cis38 cosec52 / tan18 tan35 tan60 tan72 tan55 |
| Answer» We have,{tex}2\\left(\\frac{\\cos58^\\circ}{\\sin32^\\circ}\\right)-\\sqrt3\\left(\\frac{\\cos38^\\circ cosec52^\\circ}{\\tan15^\\circ\\tan60^\\circ\\tan75^\\circ}\\right){/tex}{tex}=2\\left\\{\\frac{\\cos\\left(90^\\circ-32^\\circ\\right)}{\\sin32^\\circ}\\right\\}-\\sqrt3\\left\\{\\frac{\\cos38^\\circ cosec\\left(90^\\circ-38^\\circ\\right)}{\\tan15^\\circ\\tan60^\\circ\\tan\\left(90^\\circ-15^\\circ\\right)}\\right\\}{/tex}{tex}= 2 \\left( \\frac { \\sin 32 ^ { \\circ } } { \\sin 32 ^ { \\circ } } \\right) - \\sqrt { 3 } \\left\\{ \\frac { \\cos 38 ^ { \\circ } \\sec 38 ^ { \\circ } } { \\tan 15 ^ { \\circ } \\times \\sqrt { 3 } \\times \\cot 15 ^ { \\circ } } \\right\\}{/tex}{tex}\\left[\\because\\;\\cos\\left(90-\\theta\\right)=\\sin\\theta\\;,\\;\\cos ec\\left(90-\\theta\\right)=sec\\theta,\\;\\tan\\left(90-\\theta\\right)=cot\\theta\\;\\right]{/tex}{tex}= 2 - \\sqrt { 3 } \\left\\{ \\frac { \\cos 38 ^ { \\circ } \\times \\frac { 1 } { \\cos 38 ^ { \\circ } } } { \\tan 15 ^ { \\circ } \\times \\sqrt { 3 } \\times \\frac { 1 } { \\tan 15 ^ { \\circ } } } \\right\\} = 2 - \\frac { \\sqrt { 3 } } { \\sqrt { 3 } } = 2 - 1 = 1{/tex}{tex}\\left[sec\\theta=\\frac1{\\cos\\theta},\\;cot\\theta=\\frac1{\\tan\\theta}\\right]{/tex}therefore,\xa0{tex}2\\left(\\frac{\\cos58^\\circ}{\\sin32^\\circ}\\right)-\\sqrt3\\left(\\frac{\\cos38^\\circ cosec52^\\circ}{\\tan15^\\circ\\tan60^\\circ\\tan75^\\circ}\\right)=1{/tex} | |
| 35297. |
Find the ap of 1st 8 multiple of 3 |
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Answer» 3,6...a=3 ,d=3,n=8=a+(n-1)×d=3+(8-1)×3=3+7×3=3+21=24T8=243,6,9,12,15,18,21,24 1st 8 multiples of 3 are- 3,6,9,12,15,18,21,24.Here, a=3 d=3 an=24 n=8(using an formula) Sn=112(using Sn formula). |
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| 35298. |
Sin square +cos square |
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Answer» 1 1 1 2 1 |
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| 35299. |
Agahs |
| Answer» What is this | |
| 35300. |
Alpha + beta=? |
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Answer» -b/a -b/a Alpha +beta=-b/a In an quadratic equations -b/a -b/a -b/a |
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