No products in the cart.

Ask questions which are clear, concise and easy to understand.

Ask Question

    • Posted by Ayush Singh 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Khatik 5 years, 6 months ago

    Shuru prove that the product of two consecutive positive integer is divisible by 2

    1Thank You

    ANSWER

    • Posted by Naitik Kumar. 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 4 months ago

    If AOBC is a rectangle whose three vertices are A(0, 3), O(0, 0) and B(5, 0), then the length of its diagonal is

    Solution:

    Now, length of the diagonal AB = Distance between the points A(0, 3) and B(5, 0).

     

    ∴ Distance between the points (x,, y,) and (x , y ),

    Hence, the required length of its diagonal is √34.

    0Thank You

    ANSWER

    • Posted by Abhay Singh 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 6 months ago

    Steps to Create Trigonometric Table:

    Step 1: Draw a tabular column with the required angles such as 0, 30o, 45o, 60o, 90o, 180o, 270o, 360o in the top row and all 6 trigonometric functions such as sine, cosine, tangent, cosecant, secant, and cotangent in first column.

    Step 2:

    Step 3:

    Step 4:

    Step 5:

    Step 6:

    Step 7:

     

    0Thank You

    ANSWER

    • Posted by आदित्य बड़गुर्जर 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Atoria Atoria Lyngthong 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yogita Ingle 5 years, 6 months ago

    0Thank You

    Atoria Atoria Lyngthong 5 years, 6 months ago

    Circumference of the circle exceed the diameterby 16.8. 2pir

    1Thank You

    ANSWER

    • Posted by Preeti Kumari Preeti Kumari 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Akásh Pànchàl 5 years, 6 months ago

    a=2 d=4 n=10 a^n=a+(n-1)d

    1Thank You

    Preeti A 5 years, 6 months ago

    a=2 d=6-2 d=4 an=a+(n-1)*d a10=2+(10-1)*4 a10=2+9*4 a10=2+36 a10=38

    1Thank You

    D.R.K Mishra 5 years, 6 months ago

    A.P:- 2,6,10,14,18,22,26,30,34,38.

    2Thank You

    ANSWER

    • Posted by Preeti Kumari Preeti Kumari 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Anuj Kumar 5 years, 6 months ago

    ???

    0Thank You

    ANSWER

    • Posted by Preeti Kumari Preeti Kumari 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 4 answers

    Rohan £#@#÷׶¥€~ 5 years, 6 months ago

    l^2 =h^2 +r^2

    0Thank You

    Khushi Shama 5 years, 6 months ago

    L=√r2+h2

    0Thank You

    Adarsh Jaiswal 5 years, 6 months ago

    Answer - under root h square minus r square

    0Thank You

    Himanshu Sharma 5 years, 6 months ago

    L ka square is equal to under root r ka square plus h ka square

    0Thank You

    ANSWER

    • Posted by Avdhi Kasliwal 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 4 months ago

    The value of  is 

    Step-by-step explanation:

    Given,

     is trapezium where diagonals intersect at ,,,,  and 

    From figure,

     and 

    •  (vertically opposite angle)
    •  ( and alternate angle)
    •   ( and alternate angle)  

    ∴ ≅

    So, We can write,

     

    Plug the value in above equation,

    ⇒ 

    ⇒

    ⇒

    ⇒

    0Thank You

    ANSWER

    • Posted by Aalok Yadav 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Honey Jain 5 years, 6 months ago

    ax2+bx+c=0, a is not=to 0

    0Thank You

    Devansh Goal 5 years, 6 months ago

    ax^2+ bx + c = 0

    0Thank You

    Yogita Ingle 5 years, 6 months ago

    The standard form of a quadratic equation is ax2+bx+c=0, where a,b and c are real numbers and a≠0.
    ‘a’ is the coefficient of x2. It is called the quadratic coefficient. ‘b’ is the coefficient of x. It is called the linear coefficient. ‘c’ is the constant term.

    1Thank You

    ANSWER

    • Posted by Jaspreet Kaur Kalra 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Honey Jain 5 years, 6 months ago

    2 × 2 × 3 × 13 are the factor of 156.

    1Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 6 months ago

    2^2*3*13

    3Thank You

    ANSWER

    • Posted by Mohd Zubair 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Tanu Man 5 years, 6 months ago

    Write complete question ?? plz??

    0Thank You

    ANSWER

    • Posted by Rina Goswami 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    S. Yarthishaa 5 years, 6 months ago

    Tan 60degree is equal to root3

    0Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 6 months ago

    1.73205080757

    1Thank You

    ANSWER

    • Posted by Nischita Nishi 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Prakash Samrat 5 years, 6 months ago

    X=3,Y=0

    2Thank You

    ANSWER

    • Posted by Anushka Sisodiya 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Riddhima Joshi 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 5 answers

    Tanu Man 5 years, 6 months ago

    NCERT Exemplar

    0Thank You

    आदित्य बड़गुर्जर 5 years, 6 months ago

    Together with

    1Thank You

    Honey Jain 5 years, 6 months ago

    U can download ncert class 10 book offine all books are which u want

    1Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 6 months ago

    S Chand.. Pearson etc...

    2Thank You

    Rahul Yadav 5 years, 6 months ago

    Rd Sharma ,Rs Aggrawal,

    2Thank You

    ANSWER

    • Posted by Bijisreejesh Bijisreejesh 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Vinay Pandey 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Ambika Ambika 5 years, 6 months ago

    Solution :- (i) given that a=5 , d=3, an=50 we know nth term of an AP, an= a+(n-1)d =50=5+(n-1)3 =50-5=(n-1)3 =45=(n-1)3 =45/3=n-1 =15=n-1 =n=15+1 =n=16 Now,sum of n terms, Sn=n/2 [a+an] Sn=16/2 [5+50] sn=16/2 ×55 Sn=8×55=440 . (ii) given that,a=7,a13 =35 we know nth term of an AP, an =a+n-1 =a13=7+(13-1)d =35=7=12d =35-7=12d =28=12d =d=28/12=7/3 Now,sum of n terms Sn =n/2 [a+an] S13=13/2 [7+25] S13=13/2×42 S13=273. (iv) given that a3=15, S10 =125 we know, nth term of an AP, An=a+(n-1) = a3=a+(3-1)d--------------> (1) Sum of n terms Sn=n/2 [2a+(n-1)d] =S10=10/2 [2a+(10-1)d] =S10=5(2a+9d)-------------> (2) multiplying equ.(1) by (2) we get 30=2a+4d ----------> (3) Subtracting equ.(2) from equ (3) we get 2a+4d=30 2a+9d=25 (-) (-) _____________ -5d =5 d=___5__= -1 -5 subtracting the value of d in the equ (1)we get 15=a+2(-1) =15=a+2=a =15+2=a =a=17 Therefore a10 =a+(10-1)d a10=17+(10-1)(-1) a10=17+(10-1)(-1) a10=17+9×(-1) a10=17-9 a10=8 thus , d= -1 and a10=8 (v) given that, d=5, Sq=75 we know sum of n terms, Sn=n/2 [2a+(n-1)d] = Sq=9/2 [2a+(9-1)] =75=9/2 [2a+40] =75×2=9[2a+40] =150×18a+360 =18a=150-360 =18a= -210 =a= -210/18=-35/3 we know, an =a+(9-1)d a9=-35/3 +(9-1)5 a9= -35/3 +40 a9= -35+120/3 =85/3 thus a=-35/3 and a9=85/3 (vi) given a=2, d=8, Sn=90 we know,sum of n terms,Sn =n/2 [2a+(n-1)] =90=n/2 [2 (2)+(n-1)4] =90= n/2×2 [2+(n-1)4] =90=n [2+(4n-4] =90= n[4n-2] =90=4n^ - 2n =4n^-2n-90=0 =2n^-n-45=0 =2n^-10n+9n-45 (splitting the middle term) =2n(n-5)+9(n-5)=0 =(n-5)(Zn+9)=0 =n-5=0 we know nth term of an AP = an =an=2+(5-1)8 =an=2+2 4×8 =an= 2+32 =an=34 thus, n=5 and an=34 (viii) given that an=4,d=2, Sn=-14 we know nth term of an AP,an =an=a+(n-1)2 =4=a+2n-2 =6=a+2n =a=6-2n ------------>(1) we know sum of n terms Sn=n/2[2a+(n-1)] = -14=n/2×2[a+(n-1)] = -14=n[a+(n-1)] --------------> subtracting equ (1) in equ.(2) we get -14= [6-2n+n-1] = -14=n [5-n] = -14=5n-n2 =n2-5n-14=0 =n2-7n +2n-14=0 =n(n-7)+2 (n-7)=0 =(n-7)(n+2)=0 =n-7=0 =n=7 subtracting the value of'n' in the equ (1)we get =a=6-2(7) =a=6-14 =a= -8 thus n=7 and a= -8 (ix) given that a=3, n=8 and s=192 we know the sum of n terms of an AP Sn= n/2[a+an] =192=8/2 [3+an] =192=8/2[3+an] =192/4=[3+an] =192/4=3+an =48=3+an =an=48-3 =an=45 we know nth term of an AP an=a+(n-1)d =45=3+(8-1)d =45=7d =d=6 thus d=6

    0Thank You

    ANSWER

    • Posted by Krishan Aanand Patel 4 years, 11 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 4 years, 11 months ago

    1. One point
      Circle is the locus of points equidistant from a given point, the center of the circle, and nd Tangent is the line which intersect circle at one point only
    2. Secant
      A line which intersect circle in two points is called a chord if this line passes through center then it is called secant. (Points are A and B)
    3. Two
      As circle has infinite points, so there will be infinite tangents can be drawn on these points which touches at only one point. (P and Q)
      So there will be infinite pairs of tangents which are parallel.
    4. Point of Contact
      The common point of a tangent to a circle and the circle is called point of contact. (P or Q in the given figure)

    0Thank You

    ANSWER

    • Posted by Noobi Master 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Rahul Yadav 5 years, 6 months ago

    It is removed from shllybus

    0Thank You

    ANSWER

    • Posted by Tannu Raj 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Siddharth Avhad 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Siddharth Avhad 5 years, 6 months ago

    #ß$%&--JJ hu v C

    1Thank You

    Siddharth Avhad 5 years, 6 months ago

    #ß$%&--JJ hu v C

    2Thank You

    ANSWER

    • Posted by Rishi Dixit 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Siddharth Avhad 5 years, 6 months ago

    Giorgi

    0Thank You

    ANSWER

    • Posted by Sumona Mukherjee 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 6 months ago

    p[a+(p−1)a]=q[a+(q−1)d]
    ora(p−q)+(p2−q2)d+(q−p)d=0
    ora(p−a)+(p+a)(p−q)−(p−q)d=0
    ora+(p+q−1)d=0
    This is the (p+q) th term.

    0Thank You

    ANSWER

    • Posted by Khushi Gangwani 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Khushi Gangwani 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Anil Tharkar 5 years, 6 months ago

    Let 'a' be the first term and 'd' be common difference of the A.P ⇒Tm​=a+(m−1)d=n⇒a−d=n−md...(1) and Tn​=a+(n−1)d=m⇒a−d=m−nd..(2) Using (1) and (2) we get, d=−1 and a=n+m−1 ∴Tp​=a+(p−1)=n+m−1+(p−1)(−1)=n+m−p Hence, option 'A' is correct.

    2Thank You

    ANSWER

    • Posted by Flickzi Akshat 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 6 months ago

    Given zeroes,
    5+√2 and 5-√2
    Sum of the zeroes=5+√2+5-√2
    Sum of the zeroes=10
    Product of zeroes=(5+√2)(5-√2)
    Product of zeroes=25-2=23
    We know that quadratic polynomial is in the form of,
    =k{x²-(sum of zeroes)x+product of zeroes}
    By putting required values we get,
    =k{x²-10x+23}

    =x²-10x+23

    Hence x²-10x+23 is the required polynomial.

     

    Or

    So, f(x) = k {x2 - 10x + 23}, where, k is any non-zero real number.

    1Thank You

    ANSWER

    • Posted by Ritu Raj 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Tapo Kena 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Prachi Jain 5 years, 6 months ago

    510 = 92 x 5 + 50 92 = 50 x 1 + 42 50 = 42 x 1 + 8 42 = 8 x 5 + 2 8 = 2 x 4 + 0 ∴ HCF of 510 and 92 = 2 Product of two numbers = Product of their LCM and HCF 510 x 92 = 2 x LCM LCM = (510 x 92) / 2         = 23460 ∴ LCM of 510 and 92 = 23460.

    0Thank You

    Yogita Ingle 5 years, 6 months ago

    510 = 92 x 5 + 50
    92 = 50 x 1 + 42
    50 = 42 x 1 + 8
    42 = 8 x 5 + 2
    8 = 2 x 4 + 0

    ∴ HCF of 510 and 92 = 2

    Product of two numbers = Product of their LCM and HCF

    510 x 92 = 2 x LCM

    LCM = (510 x 92) / 2

            = 23460

    ∴ LCM of 510 and 92 = 23460.

    1Thank You

    ANSWER

    • Posted by Tapo Kena 5 years, 6 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Yogita Ingle 5 years, 6 months ago

    510 = 92 x 5 + 50
    92 = 50 x 1 + 42
    50 = 42 x 1 + 8
    42 = 8 x 5 + 2
    8 = 2 x 4 + 0

    ∴ HCF of 510 and 92 = 2

    Product of two numbers = Product of their LCM and HCF

    510 x 92 = 2 x LCM

    LCM = (510 x 92) / 2

            = 23460

    ∴ LCM of 510 and 92 = 23460.

    0Thank You

    ANSWER

    myCBSEguide App

    myCBSEguide

    Trusted by 1 Crore+ Students

    Install Now

    Test Generator

    Test Generator

    Create papers online. It's FREE.

    Try Now

    CUET Mock Tests

    CUET Mock Tests

    75,000+ questions to practice only on myCBSEguide app

    Install Now
    Google Play

    Download myCBSEguide App

    All Courses

    Other Websites

    CBSE Courses

    Resources

    Corporate

    NCERT Solutions

    CBSE Sample Papers

    All Content and Intellectual Property is under Copyright Protection | myCBSEguide.com ©2007-2026