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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.
| 28001. |
Important question of ch 6 |
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Answer» These question is most important that is Pythagoras theorem , similar triangles and exercise 6.4 ka 7 question All theorems and last exercise questions.. Example and theorem are important All theorem and 6.5. 4 last question 6.5 que 14,15,16 |
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| 28002. |
(sinA +cosecA) ²(cosA+secA)²=7+tan²A+cot²A |
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Answer» Ncert ka question hy Check it on this appp |
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| 28003. |
Find value of k for which root of tha following quadratic equation is kx^-14x+8=0 is 2 |
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Answer» Ncert ko hai K = 5 |
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| 28004. |
Sintheta ÷1-cos theta +tantheta ÷1 +costheta =costheta×cosectheta+cottheta |
| Answer» Abe vedantu ka question yaha kyo kr rha hy | |
| 28005. |
If two zeroes of the polynomial x4_6x3+16x2+138x_35 are 2+_root3 find other zeroes |
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Answer» Ab x- 4x+1 se diye polynomial se divide kr do Let the other roots be ( x-2-√3) ( x-2+√3) = (X-2)² - (√3)² = X² -4x +1 Poly. ko x^2-4x+1 se divide kar do ur quotient ko factorise Ncert ka question hai.....optional exercise check kijiye I think ur que is wrong. Check it again |
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| 28006. |
Ap.3;15;27;39 whos term ap;s terms54 high from132 |
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Answer» a=3 , d= 12 , n = 54 , an=? ..... a54= 3+53x12 ..... a54= 639 , now 54th+ 132 , 639+132= 771 ... an=771 , 771 = 3+(n-1)x12 .... Then yha se n ki value 65 aayegi and whi answer hoga Or iska answer hai 65term 65th term |
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| 28007. |
Converse of BPT and Converse of Pythagoras Theorem |
| Answer» Refer ncert book | |
| 28008. |
Does optional questions come in exam |
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Answer» Yes optional questions r most important. Please answer ASAP. So, do we need to prepare optional exercise of chapter 6 and chapter 13 also? Comes Yes optional question comes in all the sections of the paper.Even in section -A and section -B. |
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| 28009. |
x-1,x+2,x-3 are in the form of AP.Find the value of x? |
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Answer» Because common difference is not same They are not in Ap Question is wrong |
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| 28010. |
Using the section formula show that the points (-3,-2),(5,2),(1,0) are collinear. |
| Answer» Bhai sun :Let A(-3,-2) , B(5,2) ,C(1,0) be the required points For collinear points We know that,Distance AB+Distance BC=DISTANCE AC | |
| 28011. |
Clash royale |
| Answer» Namaste | |
| 28012. |
x(sin)3A+y(cos)3A=sinA×cosA and xsinA=ycosA , prove that x2+y2= 1 |
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Answer» x sinA×sin^2A + y cosA×cos^2A = sinA ×cosA.... x sinA × sin^2A + x sinA × cos^2A= sinA×cosA... as xsinA=ycosA..... ab common le jao xsinA ko yu will get x=cosA and y= sinA.... and we know that sin^2A+ cos^2A=1... so, x^2+y^2=1... proved x sin^3A me se xsinA baahar lelo.. Then xsinA ki jagah ycosA put karo.. Then if u solve u will get y=sin A nd x=cosA... Phir ye value vahan daal do jahan aapko proof karna hai.. |
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| 28013. |
Theorem 1.4 that root 2is irrational proof of real number chapter 1 of class 10 |
| Answer» Let us assume that {tex}\\sqrt 2{/tex} is rationalso let {tex}\\sqrt 2{/tex} ={tex}\\frac pq{/tex}where p and q are integers and also let p and q are not having any common factor\xa0{tex}\\begin{array}{l}2=\\frac{\\mathrm p^2}{\\mathrm q^2}\\\\\\mathrm p^2=2\\mathrm q^2---(1)\\\\\\mathrm{Hence}\\;\\mathrm p^{\\;2}\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2\\\\\\mathrm{so}\\;\\mathrm p^\\;\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2---(2)\\\\\\mathrm{let}\\;\\mathrm p=2\\mathrm t\\;(\\;\\mathrm t\\;\\;\\mathrm{is}\\;\\mathrm a\\;\\mathrm{positive}\\;\\mathrm{integer})\\\\\\mathrm{So}\\;\\mathrm{from}(1)\\\\2\\mathrm q^2=4\\mathrm t^2\\\\\\mathrm q^2=2\\mathrm t^2\\\\\\mathrm{Hence}\\;\\mathrm q^{\\;2}\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2\\\\\\mathrm{so}\\;\\mathrm q^\\;\\;\\mathrm{is}\\;\\mathrm{divisble}\\;\\mathrm{by}\\;2---(3)\\end{array}{/tex}From (2) and (3) it follows thatp and q both having 2 as the common factorBut this makes wrong our assumption that p and q are not having any common factorThis has come as we assumed that {tex}\\sqrt 2{/tex} is rational.Hence {tex}\\sqrt 2{/tex} is a irrational number | |
| 28014. |
If the radius of the circle is 6cm and length of an arc is 12 cm find area of sector |
| Answer» Let angle of arc ={tex}\\theta{/tex}{tex}\\begin{array}{l}\\mathrm{Length}\\;\\mathrm{of}\\;\\mathrm{arc}=\\frac{\\mathrm\\theta}{360}\\times2\\mathrm{πr}\\\\\\mathrm{so}\\;\\frac{\\mathrm\\theta}{360}\\times2\\mathrm\\pi\\times6=12\\\\\\frac{\\mathrm{θπ}}{360}=1.....(1)\\\\\\mathrm{now}\\;\\mathrm{area}\\;\\mathrm{of}\\;\\mathrm{sector}\\\\=\\frac{\\mathrm\\theta}{360}\\times\\mathrm{πr}^2=1\\times\\mathrm r^2=\\mathrm r^2\\;(\\;\\mathrm{from}\\;(1))\\\\=6^2=36\\;\\mathrm{cm}^2\\end{array}{/tex} | |
| 28015. |
6 & 13 chapter ki optional krni zaroori hai kya? ? |
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Answer» Thanks to answer. ? No it depends upon you that you want to solve miscellaneous questions. Its not compulsory |
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| 28016. |
Find the ratio in which the line segment p(x1,y1) and (x2,y2) is divided by x-axis. |
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| 28017. |
Cosec theta + cot theta= p Then find the value of=P2+1---------- = cos thetaP2-1 |
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Answer» 2016 ke paper me hai ja ke phle dekh Ye question pichle 10 saalon m na puccha gya aur n hi puccha jaega |
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| 28018. |
Express 32760 into product of prime factors |
| Answer» 32760 = 23 X 32 X 5 X 7 X 13 | |
| 28019. |
Prove that :(Tan A + Sec A ) - 1(Tan A - Sec A) +1 =1+Sin A Cos A |
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| 28020. |
Please prove this no equal3=-3??If you have guds solve this above questions |
| Answer» 3=-3Squaring both side (3)power 2=(-3)power 29=9??Hence proved I have a guds to solve these logical thinking questions | |
| 28021. |
Prove:- (tan A + cosec B) square - (cot B - sec A)square = 2tan A cot B (cosec A + sec B) |
| Answer» tanA + cosecB)² - (cotB - secA)²= tan²A + cosec²B + 2 tanA cosecB -(cot²B + sec²A - 2 cotB secA) ( using (a±b)² = a² + b² ± 2ab)= tan²A + cosec²B + 2tanA cosecB - cot²B - sec²A + 2cotB secARearranging above equation we getcosec²B - cot²B - sec²A + tan²A + 2tanA cosecB + 2cotB secA= (cosec²B - cot²B) - (sec²A - tan²A) + 2tanA cotB ( cosecB/cotB + secA/tanA )= 1 - 1 + 2tanA cotB { (1/sinB)/(cosB/sinB) + (1/cosA)/(sinA/cosA) } ( As cosec²ø - cot²ø = 1 sec²ø - tan²ø = 1)= 0 + 2tanA cotB { (1/sinB) x (sinB/cosB) + (1/cos A) x (\xa0cosA/sinA) }= 2tanA cotB ( 1/cosB + 1/sinA) = 2tanA cotB (secB + cosecA)= 2tanA cotB (cosecA + secB)Read more on Brainly.in - https://brainly.in/question/2314597#readmore | |
| 28022. |
What is the trick to solve the any trigonometry question |
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Answer» Try your level best.... If u still won\'t get, multiply both LHS and RHS by zero. Therefore, LHS=RHS..U WILL SURELY GET 2MARKS FOR 4???? U just need to learn all the identities properly.. Learn the trigonometric ratios.. Practice as many questions of this chapter as possible.. Soon u will find it easy.. |
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| 28023. |
Do compound interest and simple interest form AP???? |
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| 28024. |
Why we are taking a-3d,a-d,a+d,a+3d in ap as consecutive no.? |
| Answer» Because their differences are equal | |
| 28025. |
Find two no.whose sum is 27and product is 182 |
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Answer» Let the two numbers be x &yx+y=27 .... (I) xy=182(x+y) ²=x²+2xy+y²27²=x²+y²+2×182729=x²+y²+364x²+y²=729-364x²+y²=365(x-y) ²=x²+y²-2xy(x-y) ²=365-364(x-y) ²=1x-y=1 .... (II) adding (I) &(II) 2x=28x=14y=13Read more on Brainly.in - https://brainly.in/question/1858456#readmore Its ncert question |
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| 28026. |
Maths important questions |
| Answer» Check out in this app | |
| 28027. |
3x + y =1(2k-1) x+(k-1)y = 2k + 1 |
| Answer» What\'s ur question | |
| 28028. |
If the points A (-1,2) B 2,3 Ca,2, |
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Answer» a=-1??? Complete your question |
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| 28029. |
If the points A ,92034#3 |
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| 28030. |
If the HCF of 408 and 1032 is expressible in the 1032m - 408×5,find m. |
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Answer» Ok 8 is the HCF of both the nos.HCF is here expressible in the form 1032m- 408×5. Now,8= 1032m - 408×58=1032m- 20408+2040= 1032m2048= 1032m2048÷1032= m128/43=m (ans.) That\'s what my answer is coming ..If any mistake is done by me please rectify it. |
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| 28031. |
Find all zeros of polynomial p (x)=2xpower3 +xsquare -6x-3 if -root under 3 +root under 3 |
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Answer» Cubic polynomial mein maximum 3 zeroes hi hote hain 4th zero √3 and -√3 are zeroes ,then 3rd zero is -1/2 |
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| 28032. |
If 1and -2 are two zeroes of the quadratic polynomial p(x)= 2x^3-x^2-5x-2 find its third zero |
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Answer» Its 1/2 Sorry wrong ans 1 |
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| 28033. |
Find the quadratic polynomial where sum and product of the zeroes are a and 1÷a |
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Answer» A+b=aA.b=1\\aX^2-ax+1\\a x^2 - ax + 1/a x^2-ax+1/a X sq. -ax+1/a |
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| 28034. |
In a triangle ABC, DE is parallel to BC. If AD= 2x+1, BD= x-3, AE=2x+5, EC=x-1. Find the value of x. |
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Answer» Mathematics ➡user submitted ➡paper 03 septeaber2008-2009 Check diagram from Use bpt For no value or for infinity 7/2 |
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| 28035. |
The sum of number and its positive square root is 6/25 Find the number |
| Answer» We get no. As -3/5 or -2/5 | |
| 28036. |
Find the value of a and b for which x= 3/4 and x=-2 are the roots of equation axpower square+bx-6=0 |
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Answer» Right answer is a= 4 and b= 5 My ans is a=+-√6 and b=0 a=4 and b=5 |
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| 28037. |
All formulas of chapter surface areas and volume |
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Answer» Go to chap. 13 and click on cbse revision notes It is given on this appp |
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| 28038. |
the sum of all odd num under the 0 to 100 |
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Answer» 2500 The square of the numbers of odd no. |
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| 28039. |
Is Ncrt and RD sharma enough to score 90+ marks |
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Answer» Yes Offcourse Yes |
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| 28040. |
Circles sector area formula |
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Answer» Theta/360°×πr^2 Theta/ 360°×πr² Tita / 360 × pie raduis sq |
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| 28041. |
Is ncert is enough to get 90 + marks |
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Answer» Yes Yes ncert is enough ,if you focus on concepts of questions No Only 60% I think yes |
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| 28042. |
Tan 60th+4cos²45+3sec²30+5cos²90/cossec30+sec60-cot²30= me |
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| 28043. |
Is it right that cbse Will ask 25%objective questions |
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Answer» May be i am not confirm but there is no any objectice in sylabllus and also no pattern I am not confirm but i heared Really |
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| 28044. |
What in short we can write theorem 6.7?? |
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Answer» Sorry (1)in tri, |
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| 28045. |
Find two no whose sum is 27 and product is182 |
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Answer» Two no.s x and y Where x+y=27and xy=182X=27-y (27-y)y=182y2 -27y+182=0 12 and 13 |
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| 28046. |
Find the 30th term of an AP 10,7,,4 |
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Answer» -77 -77 -77 |
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| 28047. |
Plz supply ncert example questions too |
| Answer» | |
| 28048. |
Prove that the parallekogram circumscribing a circle is a rhombus |
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Answer» Prove by equalling adjacent sides of a parallelogram Given ABCD is a ||gm such that its sides touch a circle with centre O.∴ AB = CD and AB || CD,AD = BC and AD || BCNow, P, Q, R and S are the touching point of both the circle and the ||gmWe know that, tangents to a circle from an exterior point are equal in length.∴ AP = AS [Tangents from point A] ... (1) BP = BQ [Tangents from point B] ... (2) CR = CQ [Tangents from point C] ... (3) DR = DS [Tangents from point D] ... (4)On adding (1), (2), (3) and (4), we getAP + BP + CR + DR = AS + BQ + CQ + DS⇒ (AP + BP) + (CR + DR) = (AS + DS) + (BQ + CQ)⇒ AB + CD = AD + BC⇒ AB + AB = BC + BC [∵ ABCD is a ||gm . ∴ AB = CD and AD = BC]⇒ 2AB = 2BC⇒ AB = BCTherefore, AB = BC impliesAB = BC = CD = ADHence, ABCD is a rhombus |
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| 28049. |
find the HCF of 9 and 12 |
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Answer» I know this answer but I think your mind 3 HCF of 9 and 12 is 3 3 3 |
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| 28050. |
I need proof of theorem 1 of circle |
| Answer» How can I proof here | |