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    • Posted by Ayush Singh 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Khushi Khatik 5 years, 1 month ago

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

    1Thank You

    ANSWER

    • Posted by Naitik Kumar. 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years 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.

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    • Posted by Abhay Singh 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 1 month 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:

     

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    • Posted by आदित्य बड़गुर्जर 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Atoria Atoria Lyngthong 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Yogita Ingle 5 years, 1 month ago

    0Thank You

    Atoria Atoria Lyngthong 5 years, 1 month ago

    Circumference of the circle exceed the diameterby 16.8. 2pir

    1Thank You

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    • Posted by Preeti Kumari Preeti Kumari 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Akásh Pànchàl 5 years, 1 month ago

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

    1Thank You

    Preeti A 5 years, 1 month 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, 1 month ago

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

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    • Posted by Preeti Kumari Preeti Kumari 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Anuj Kumar 5 years, 1 month ago

    ???

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    • Posted by Preeti Kumari Preeti Kumari 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 4 answers

    Rohan £#@#÷׶¥€~ 5 years, 1 month ago

    l^2 =h^2 +r^2

    0Thank You

    Khushi Shama 5 years, 1 month ago

    L=√r2+h2

    0Thank You

    Adarsh Jaiswal 5 years, 1 month ago

    Answer - under root h square minus r square

    0Thank You

    Himanshu Sharma 5 years, 1 month ago

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

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    • Posted by Avdhi Kasliwal 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years 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,

    ⇒ 

    ⇒

    ⇒

    ⇒

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    • Posted by Aalok Yadav 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 3 answers

    Honey Jain 5 years, 1 month ago

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

    0Thank You

    Devansh Goal 5 years, 1 month ago

    ax^2+ bx + c = 0

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    Yogita Ingle 5 years, 1 month 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

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    • Posted by Jaspreet Kaur Kalra 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Honey Jain 5 years, 1 month ago

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

    1Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

    2^2*3*13

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    • Posted by Mohd Zubair 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Tanu Man 5 years, 1 month ago

    Write complete question ?? plz??

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    • Posted by Rina Goswami 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    S. Yarthishaa 5 years, 1 month ago

    Tan 60degree is equal to root3

    0Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

    1.73205080757

    1Thank You

    ANSWER

    • Posted by Nischita Nishi 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Prakash Samrat 5 years, 1 month ago

    X=3,Y=0

    2Thank You

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    • Posted by Anushka Sisodiya 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Riddhima Joshi 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 5 answers

    Tanu Man 5 years, 1 month ago

    NCERT Exemplar

    0Thank You

    आदित्य बड़गुर्जर 5 years, 1 month ago

    Together with

    1Thank You

    Honey Jain 5 years, 1 month ago

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

    1Thank You

    Priya ✧*｡٩(๑˙╰╯˙๑)و✧*｡ 5 years, 1 month ago

    S Chand.. Pearson etc...

    2Thank You

    Rahul Yadav 5 years, 1 month ago

    Rd Sharma ,Rs Aggrawal,

    2Thank You

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    • Posted by Bijisreejesh Bijisreejesh 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Vinay Pandey 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Ambika Ambika 5 years, 1 month 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

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    • Posted by Krishan Aanand Patel 4 years, 7 months ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Sia ? 4 years, 7 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)

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    • Posted by Noobi Master 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Rahul Yadav 5 years, 1 month ago

    It is removed from shllybus

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    • Posted by Tannu Raj 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Siddharth Avhad 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Siddharth Avhad 5 years, 1 month ago

    #ß$%&--JJ hu v C

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    Siddharth Avhad 5 years, 1 month ago

    #ß$%&--JJ hu v C

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    • Posted by Rishi Dixit 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Siddharth Avhad 5 years, 1 month ago

    Giorgi

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    • Posted by Sumona Mukherjee 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 1 month 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, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Khushi Gangwani 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Anil Tharkar 5 years, 1 month 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

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    • Posted by Flickzi Akshat 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Gaurav Seth 5 years, 1 month 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

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    • Posted by Ritu Raj 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 0 answers
    ANSWER

    • Posted by Tapo Kena 5 years, 1 month ago

    CBSE > Class 10 > Mathematics
    • 2 answers

    Prachi Jain 5 years, 1 month 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, 1 month 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, 1 month ago

    CBSE > Class 10 > Mathematics
    • 1 answers

    Yogita Ingle 5 years, 1 month 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

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