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find the prime factorisation of the denominator of the rational number equivalent to 8.39

Posted by Aan Anna Philip (May 26, 2017 10:50 p.m.) (Question ID: 5511)

Name shape of the curve of a quadratic polinomial 

Posted by Gayaprasad Yadav (May 26, 2017 10:53 a.m.) (Question ID: 5498)

  • Answers:
  • The graph of a quadratic function is called a Parabola.

    A parabola is roughly shaped like the letter "U".

    Answered by Payal Singh (May 26, 2017 11:49 a.m.)
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A Boat takes 3 hours to go 45 km downstream and IT returns in 9 hours find the speed of stream and that a boat in still water

Posted by Binod Sharma (May 25, 2017 1:43 p.m.) (Question ID: 5483)

  • Answers:
  • Ans. Let Speed of boat in still water = x km/h

    Speed of stream = y km/h

    According to question, 

    Case 1 Downstream :

    Time taken = 3 hours

    total distance = 45 km 

    Speed of  boat downstream = x + y 

    So, {tex}{45\over 3} = x+y{/tex}

    => x+y = 15    ............. (1)

    Case 2 Upstream : 

    Time taken = 9 hours

    total distance = 45 km 

    Speed of  boat upstream = x - y 

    So, {tex}{45\over 9} = x-y{/tex}

    => x -  y = 5   ..............(2)

    Adding (1) and (2) we get

    => 2x = 20 

    x = 10 

    From (2). 10 - y =  5 

    => y = 5 

    So speed of boat in still water = 10 Km/h

    Speed of stream = 5 Km/h

    Answered by Naveen Sharma (May 25, 2017 2:26 p.m.)
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GIVE EXAMPLE OF POLYNOMIALS p(x),g(x)and r(x), which satisfy the division algorithm and:

  1. deg p(x)=deg q(x)
  2. deg q(x)=deg r(x)
  3. deg r(x)=0

 

PLS GIVE THE ANSWER........

Posted by Archith jayalal (May 25, 2017 9:15 a.m.) (Question ID: 5472)

  • THANK YOU VERY MUCH.........

     

    Posted by Archith jayalal (May 24, 2017 7:27 p.m.)
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  • Answers:
  • We can write many such examples

    (i) deg p(x) = deg q(x)

    P(x) = x2, q(x) = x2, g(x) = 1, R(x) = 0

    (ii) deg q(x) = deg r(x)

    q(x) = x , R(x) = x , p(x) = x2 + x, g(x) = x2

    (iii) deg r(x) = 0

    q(x) = 1 , R(x) = 1 , p(x) = x+1, g(x) = x

    Answered by Payal Singh (May 25, 2017 9:15 a.m.)
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Let a b c p be rational numbers such that p is not perfect cube . if a +bp^1/3+cp^2/3=0 then prove a=b=c=0

Posted by Sidharth Thakur (May 25, 2017 5:31 p.m.) (Question ID: 5471)

  • Answers:
  • As p is not a perfect cube. Hence p1/3 and  p1/3 , p2/3  are irrartional numbers. Then we have b p1/3 +c p2/3   as irrational and a as rational. As sum of a rational number and an irrational number cannot give us a zero as answer so all the coefficiants (a,b and c) must be zero in general.

    Answered by Shivam .G (May 25, 2017 5:19 p.m.)
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If a+b=5 and 3a+2b=20,  Then value of a+b is:

1) 10

2) 15

3) 20

4) 25

Posted by Sudhanshu Baranwal (May 24, 2017 10:39 a.m.) (Question ID: 5460)

  • you have already given value of a+b = 5 then what is the use of asking.

    it should be a printing mistake. you have to find value of 3a+b 

    Posted by Naveen Sharma (May 24, 2017 3:33 p.m.)
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  • Answers:
  • Ooo.

    But this ques. had come in chs exam

    Answered by Sudhanshu Baranwal (May 24, 2017 2:56 p.m.)
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  • Question Must be : If a + b = 5 and 3a + 2b = 20, then 3a + b will be

    Ans. 4) 25

    a + b = 5  .............(1)

    3a +  2b =  20    .........(2)

    From (1), a = 5 - b 

    put this value in (2), we get 

    => 3( 5 - b ) + 2b = 20

    => 15 - 3b + 2b = 20 

    => - b = 5 

    => b = -5 

    put values of b in (1), we get

    => a - 5 = 5 

    => a = 10 

    Now, 3a + b = 3(10 ) - 5

    = 30 - 5 = 25

    Answered by Naveen Sharma (May 24, 2017 11:55 a.m.)
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If cos (a+b)=0, then sin (a-b)=

1) cos 2b

2) cos b

3) sin 2a

4) sin a

Posted by Sudhanshu Baranwal (May 23, 2017 8:25 a.m.) (Question ID: 5442)

  • Answers:
  • 1) cos 2b

    Explanation:

    cos (a+b) =0

    => cos (a+b) = cos 90

    => a+b = 90

    => a = 90- b

    Now

    sin(a+b)

    = sin(90-b-b)

    = sin(90-2b)

    = cos 2b     [as sin (90 - x) = cos  x]

    Answered by Payal Singh (May 23, 2017 8:34 a.m.)
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According to trigonometric ratios of 30 degree and 60 degree, in an isosceles triangle ABC in which D is mid point of side BC, the values of side AB and AC is 2a and BD and CD is why??

Posted by Anurag Patel (May 21, 2017 6:56 p.m.) (Question ID: 5412)

The value of 

      cot x

<hr/>

cot x-cot 3x

+      

tan x

<hr/>

tan x -tan 3x 

Will be??

Posted by Sudhanshu Baranwal (May 21, 2017 2:16 p.m.) (Question ID: 5410)

In an isosceles triangle ABC, if AB=BC and AB2=2AC2 ,then angle C will be???

 

Posted by Sudhanshu Baranwal (May 21, 2017 2:11 p.m.) (Question ID: 5409)

  • Answers:
  • Ans.

    Given: AC = BC
    ABC is an Isosceles triangle

    AB2 = 2 AC2

    To Find = {tex}\angle C{/tex}

    Proof : 

    As AB2 = 2AC2

    => AB2 = AC + AC2

    =>  AB2 = AC + BC2     [Given AC =  BC]

    So By Converse of Pythagoras theorem 

    {tex}\angle C = 90^o{/tex}

     

    Answered by Naveen Sharma (May 23, 2017 10:24 a.m.)
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SecA (1+SecA)(SecA-tanA)=1

 

Posted by hrusikesh sahu (May 21, 2017 10:14 a.m.) (Question ID: 5403)

  • Answers:
  • secA(1-sinA)(secA+tanA)

     

    =(secA-tanA)(secA+tanA)

     

    =1... using identity

    Answered by Shivam .G (May 25, 2017 5:22 p.m.)
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Find the zero of the quadratic polynomials and verify the relationship between the zero and the coefficient :-

x² -2x -8

Posted by Anurag Patel (May 20, 2017 10 a.m.) (Question ID: 5389)

  • Answers:
  • Ans. To find zeros, we ll equate it with 0

    => x- 2x - 8  = 0

    => x2 - 4x + 2x - 8 = 0 

    => x(x-4) + 2( x-4) = 0 

    => (x - 4) ( x + 2 ) = 0

    => x - 4 = 0  Or   x + 2 = 0 

    => x = 4, -2 

    Zeros of the polynomial are 4 , -2

    On comparing with ax2 +  bx + c, we get a = 1, b = -2 and c = -8

    We know Sum of zeroes = {tex}-{b\over a}{/tex}

    => 4 + (-2) = {tex}- {-2\over 1}{/tex}

    => 2 = 2 

    Again Product of zeroes = {tex}c\over a{/tex}

    => {tex}4 \times (-2) = {-8\over 1}{/tex}

    => -8 = -8 

    Hence verified 

    Answered by Naveen Sharma (May 20, 2017 10:06 a.m.)
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How to get full marks in maths exam 2017-18

Posted by Himanshi Sharma (May 19, 2017 11 p.m.) (Question ID: 5386)

Unit tests of class 10 pattern 2017

 

Posted by Himanshi Sharma (May 17, 2017 11:56 p.m.) (Question ID: 5360)

  • Answers:
  • Every school has to take three periodic tests for every subject and the best two will be considered. 

    Periodic tests has a weightage of 10 marks. 

    2-3 chapters will be assessed in every subject. 

    All the best. Make the most of your summer vacations! 

    Answered by Shweta Gulati (May 18, 2017 10:27 p.m.)
    Thanks (1)
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What is best way of solving out an root term?

Posted by Prashant Gautam (May 17, 2017 9:18 p.m.) (Question ID: 5357)

what are properties of integers

 

Posted by swapnil pawar (May 17, 2017 11:56 a.m.) (Question ID: 5347)

  • Answers:
  • Ans. 

    Answered by Naveen Sharma (May 17, 2017 2:14 p.m.)
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What is procedure of using elimination method?

Posted by Prashant Gautam (May 16, 2017 8:21 p.m.) (Question ID: 5336)

  • Answers:
  • Ans. The elimination method of solving systems of equations is also called the addition method. To solve a system of equations by elimination we transform the system such that one variable "cancels out".

    Here are the steps to follow:

    Step 1 : Try to eliminate a variable as you add the left sides and the right sides of the two equations

    Step 2 : Set the sum resulting from adding the left sides equal to the sum resulting from adding the right sides

    Step 3 : Solve for the variable that was not cancelled or eliminated

    Step 4 : Use the answer found in step 3 to solve for the other variable by substituting this value in one of the two equations

    Answered by Naveen Sharma (May 17, 2017 2:18 p.m.)
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tan²A+cot²A+2=sec²A cosec²A

Posted by Davidson Louriyam (May 14, 2017 4:43 p.m.) (Question ID: 5309)

  • i forgot to add 2

     

    Posted by sukdeb mukherjee (May 14, 2017 8:38 p.m.)
  • question is wrong . it should be tan2a+cot2a = sec2a+cosec2a

    Posted by sukdeb mukherjee (May 14, 2017 8:34 p.m.)
  • Add Comment
  • Answers:
  • tan2a + cot2a + 2

    = sec2a - 1 + cosec2a - 1 + 2

    = sec2a - 2+2 + cosec2a

    =sec2a + cosec2a

    Answered by sukdeb mukherjee (May 14, 2017 8:37 p.m.)
    Thanks (1)
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When to do LCM in problem sums?

Posted by anand yadav (May 13, 2017 3:09 p.m.) (Question ID: 5271)

  • Answers:
  • When you have to find smallest least minimum number 

    Answered by Payal Singh (May 13, 2017 4:28 p.m.)
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When to do HCF in problem sums?

Posted by anand yadav (May 13, 2017 3:08 p.m.) (Question ID: 5270)

  • Answers:
  • When you need to find out maximum or largest value 

    Answered by Payal Singh (May 16, 2017 9:31 a.m.)
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What are the zeroes of quadratic polynomial x2+99x+127?

Posted by anand yadav (May 13, 2017 2:08 p.m.) (Question ID: 5268)

  • Answers:
  • We need to find zeroes of x^2+99*x+127=0
    We have certainly got a formula for determining the zeroes of a 2 degree equation.That is called Discriminant formula or D- formula or  Shri Dharacharya Method.

    Roots of the 2 degree equation are given by=(-b+(b^2-4*a*c)^(1/2))/2*a and (-b-(b^2-4*a*c)^(1/2))/2*a 

    So, Zeroes of this equation is given by 
    (-99+(99^2-4*1*127)^(1/2))/2*1 and (-99-(99^2-4*1*127)^(1/2))/2*1

    Hence,zeroes are -1.3 and -97.7 respectively.

    Answered by Shivam .G (May 25, 2017 5:26 p.m.)
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If the product of zeroes of the quadratic polynomial f (x) =x square -4x+k is 3 . Find the value of k

Posted by Shweta Choudhary (May 09, 2017 6:52 p.m.) (Question ID: 5177)

  • Please answer me fast

    Posted by Shweta Choudhary (May 09, 2017 8:36 p.m.)
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  • Answers:
  • Product of the zeroes =c/a

                                             =k/1=3

                                              =k=3

    Answered by anand yadav (May 13, 2017 3:14 p.m.)
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  • Ans. x2 -4x +  k 

    on comparing with ax + bx + c we get 

    a= 1 b = -4 and c = k 

    we know that 

    product of roots = {tex}c\over a{/tex}

    => {tex}3 = {k\over 1}{/tex}

    => k = 3

    Answered by Naveen Sharma (May 10, 2017 10:10 a.m.)
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Write the quadratic polynomial having -1/4,1 as it's zeroes 

Posted by Shweta Choudhary (May 09, 2017 4:52 p.m.) (Question ID: 5173)

  • Answers:
  • Ans. Let equation is ax2 + bx + c = 0, roots are {tex}-{1\over 4},1{/tex}.

    We know that,

    Sum of roots = {tex}-{b\over a}{/tex}

    => {tex}-{1\over 4} + 1 = -{b\over a}{/tex}

    {tex}=> {3\over 4} = -{b\over a}{/tex}

    => b = -3 and a = 4

    Also

    Product of roots = {tex}c\over a{/tex}

    => {tex}-{1\over 4}\times 1 = {c\over a}{/tex}

    {tex}=> -{1\over 4}= {c\over a}{/tex}

    c = -1 and a = 4

    Put Values of a,b,c in equation we get,

    4x2 - 3x -1 = 0    Requried equation

    Answered by Naveen Sharma (May 09, 2017 5:11 p.m.)
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Ashish

Posted by Deepak Yadav (May 05, 2017 12:23 a.m.) (Question ID: 5122)

the area of isoceles triangle EFG is 60 sq .m and  EF=EG=13 m find the FG

Posted by chethan kb (May 04, 2017 3:03 p.m.) (Question ID: 5107)

  • Answers:
  • Let the length of FG be x m.

    Using Heron's formula,

    s = {tex}{{13 + 13 + x} \over 2} = {{26 + x} \over 2}{/tex}

    Area of triangle EFG = 60 sq.m

    =>          {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x} \over 2} - 13} \right)\left( {{{26 + x} \over 2} - 13} \right)\left( {{{26 + x} \over 2} - x} \right)} {/tex} = 60

    =>          {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{{26 + x - 26} \over 2}} \right)\left( {{{26 + x - 26} \over 2}} \right)\left( {{{26 + x - 2x} \over 2}} \right)} {/tex} = 60

    =>         {tex}\sqrt {\left( {{{26 + x} \over 2}} \right)\left( {{x \over 2}} \right)\left( {{x \over 2}} \right)\left( {{{26 - x} \over 2}} \right)} {/tex} = 60

    =>          {tex}{x \over 2}\sqrt {\left( {{{676 - {x^2}} \over 4}} \right)} = 60{/tex}

    =>          {tex}{x \over 4}\sqrt {676 - {x^2}} = 60{/tex}

    =>          {tex}x\sqrt {676 - {x^2}} = 240{/tex}

    Squaring both sides,

    =>          {tex}{x^2}\left( {676 - {x^2}} \right) = 57600{/tex}

    =>          {tex}{x^4} - 676{x^2} + 57600 = 0{/tex}

    =>          {tex}{x^4} - 576{x^2} - 100{x^2} + 57600 = 0{/tex}

    =>          {tex}{x^2}\left( {{x^2} - 576} \right) - 100\left( {{x^2} - 576} \right) = 0{/tex}

    =>          {tex}\left( {{x^2} - 576} \right)\left( {{x^2} - 100} \right) = 0{/tex}

    =>          {tex}{x^2} - 576 = 0{/tex}     and     {tex}{x^2} - 100 = 0{/tex}

    =>          {tex}{x^2} = 576{/tex}     and     {tex}{x^2} = 100{/tex}

    =>          {tex}x=24{/tex} m      and      {tex}x = 10{/tex} m

     

    Answered by Rashmi Bajpayee (May 06, 2017 1:04 p.m.)
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ax + by = 1 

bx + ay = (a+b)^2 / a2 + b2

Posted by purva jhingan (May 04, 2017 7:49 a.m.) (Question ID: 5104)

the age of the father is equal to sum of the ages of his 6 children. after 15 years twice the age of the father will be the sum of ages of his children. find age of father

Posted by Khushbu B (May 03, 2017 9:24 p.m.) (Question ID: 5103)

  • Answers:
  • Let the age of father be x. 

    and the average age of one child be y. 

    Present age of father = sum of ages of 6 children

                              x = 6y  .................................  (i) 

    After 15 years

    Father's age = x + 15

    Age of one child = y + 15

    According to question

    2(x + 15) = 6 (y + 15)

    2x + 30 = 6y + 90

    2 × 6y + 30 = 6y + 90 ....... [substitude x = 6y from (i)]

    12y + 30 = 6y + 90

    12y - 6y = 90 - 30 

    6y = 60

    {tex}\eqalign{ & \,\,\,\,\,y\,\, = \,\,{{60} \over 6}\,\, = \,\,10 \cr & \,\,x\,\, = \,\,6y\,\, = \,\,6\,\, \times \,\,10\,\, = \,\,60 \cr} {/tex}

    Hence the age of father be = x = 60 years.

    Answered by Rupender singh (May 04, 2017 8:49 a.m.)
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From where can I get free Cbse sample papers class 10 (2017-18) for downloading, for all subjects. 

Posted by Rachna gupta (Apr 29, 2017 3:29 p.m.) (Question ID: 5066)

  • Answers:
  • Log into mycbseguide.com and go in the Downloads section. 

    You can easily download the papers of all subjects according to your need. You may also download solutions. 

    Answered by Shweta Gulati (May 01, 2017 2:08 p.m.)
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If one zero of the quadratic polynomial x2+3x+k is 7,then find the value of k

Posted by Singlaparvesh Singla (Apr 29, 2017 9:16 a.m.) (Question ID: 5061)

  • Answers:
  • Let p(x) = x<font size="2">2 </font>+ 3x + k

    If one zero of the p(x) is 7, then

    p(7) = 0

    =>          (7)2 + 3 x 7 + k = 0

    =>          49 + 21 + k = 0

    =>          70 + k = 0

    =>          k = -70

    Answered by Rashmi Bajpayee (Apr 29, 2017 12:02 p.m.)
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Find the zeroes and verify the relationship between zeroes and coefficients:

4u2 + 8u

Please give step by step instructions and explain every step.

Please.

Posted by Rachna gupta (Apr 28, 2017 5:52 p.m.) (Question ID: 5057)

  • Answers:
  • 4u^2 + 8 u +0 = 0

    D=64

    x = -b + √ D÷2a,-b -√D÷2a

    x = 0,-2

    -2+0 = -8÷4

    -2= -2

    -2×0 = 0÷4

    0=0

    Hence,verified.

    Answered by Kartik Bhardwaj (Apr 28, 2017 9:50 p.m.)
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  • Let p(u)=4u2+0.u+8

    =>4(u2+2)

    Therefore the req zeroes are -4 and -√2

    Verify it then

    Answered by Niraj Agarwal (Apr 28, 2017 6:49 p.m.)
    Thanks (1)
  • Add Answer

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