Q1) 3/7 lies between___ fractions

(a) 4/9,5/9

(b)43/99,4/9

(c)42/99,4/9

(d)41/99,42/99

Q2) if a= bm, then _____ (m>0;a,b>0)

&nb

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sometimes when we get the area of a triangle or a quadrilateral, we get the answer in roots.

so should we take an approximate value of the root to get an approximate anwer.

about 2 years ago 1 Answer 297 views

Q1) 3/7 lies between___ fractions

(a) 4/9,5/9

(b)43/99,4/9

(c)42/99,4/9

(d)41/99,42/99

Q2) if a= bm, then _____ (m>0;a,b>0)

&nb

about 2 years ago 0 Answer 426 views

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rn

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How to answer the following question?

Ramesh is a cashier at Canara Bank. He has notes of denominations of Rs 100. 50 and 10 respectively. The ratio of the number of these notes are is  2: 3: 5 respectively.

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answer 7/5

 

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Good Evening teacher,

In Vedic Maths, adding time, at the last 1.35+3.55=490 and the last 2 digits greater than 60 we added 40 and we got the answer 530. My question how do we get "40" ?

Abhijith C J

about 1 year ago 1 Answer 189 views

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find two rationals between 0.5 and 0.55

 

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Sum of the area of two squares is 468centimetre square if the difference of their perimeter is 24meter find the side of the square

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Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

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Sum of the digits of a two-digit numberis9.When we interchange the digits,it is found that the resulting new number is greater than the original number by 27.What is the two digit number.

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how to identify root problems in simply?

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അടുത്തടുത്ത രണ്ട് അധി ഒറ്റ സംഖ്യകളുടെ ഗുണനഫലം 399 ആയാൽ സംഖ്യകൾ ഏവ

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Give Examples.....

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Reliable latest lines and angles question 6

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Question No - 20

 

In the below figure ABleft | CDleft | EF: : and : : : GHleft | KL. Find angle HKL ?

In this question from lines and angles exercise.in the solution it was said that:-

-----------------------------X----------------------------X------------------------X------------X-----

Now, alternate angles are equal

angle CHG=angle HGN=60^0

angle HGN=angle KNF=60^0           [ corresponding angles]

Hence, angle KNG=180^{o}-60^{o}=120^{o}

Rightarrow angle GNK=angle AKL=120^0      [ corresponding angles]

angle AKH=angle KHD=25^0           [ alternative angles]

Therefore, angle HKL=angle AKH+angle AKL=25+120=145^0

-------------X-------------------------X--------------------------X------------------------------X-----------------------X----------

In the above said solution where did the 60 degree come from?

 

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Mydkyrkxb.xmhr b bsym

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In figure a-b=80 and POQ is a straight line,then find a and b

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A,B,C are the three angles of triangle ABC. if A-B=25, and B-C=400, find angles A, B, C. also what type of triangle is it?

 

thank you

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A and B have a ceratin number of mangoes. A says to B If you give 30 of your mangoes I will have twice as many left with you. B replies, if you give me 10, I will have thrice as many as left with you. How many mangoes does each have ?

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Draw acircle of radius 3cm.Draw a triangle with angles 50° and 70° with all sides are tangents to circle?

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Explain why 13233343563715 is a composite number.

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Question No - 15

 

Draw these figures, Four equal rhombuses

Question No - 16

 

Draw these figures., Five equal rhombuses:

Question No - 17

 

Draw these figures, Four rhombuses around a square:

 

Question No - 18

 

Parallelograms on two sides of a square :

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Sir/Madam

Please explain to me this through audio.

 

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Common difference of an arithmetic sequence is 8 and its one term is 45 can the sum of any 15 terms of this sequence be 2018?

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In a triangle , the internal bisector of an angle bisects the opposite side.Find the nature of triangle.
(a) right angled 
(b) eqilateral 
(c) scalene
(d) isosceles 

 

 

about 25 days ago 0 Answer 20 views

Identify an irrational number among the following numbers: 0.13 ,0.13overline{ 15}0.overline{ 1315} , 0.3013001300013.....

0.13 is a terminating number. So, it is not an irrational number.

0.13overline{15} = 0.131515......, 15 is repeating continuously so it is not an irrational number.

0.overline{ 1315} = 0.13151315........is repeating continuously so it is not an irrational number.

0.3013001300013....., non terminating and non recurring decimal. Hence, it is an irrational number. So, 0.3013001300013 is an irrational number.

Write the sum of 0.overline{3} and 0.overline{4}.

0.overline{3}+0.overline{4}=left ( 0.333.... right )+left ( 0.444.... right )

                   = 0.777...

               x  = 0.777

           10x  =  7.777

    10 x – x  = ( 7.777.... ) – ( 0.777....)

           9 x  =  7

               x = frac{7}{9}

Find six rational numbers between 3 and 4.

let  a = 3 , and b = 4

Here, we find six rational numbers i,e n = 6

 d = frac{b - a}{n + 1} = frac{4 - 3}{6 + 1} = frac{1}{7}

1st rational number =   a + d = 3 + frac{1}{7} = frac{22}{7}

2nd rational number =  a + 2d = 3 + frac{2}{7} = frac{23}{7}

3 rd rational number =  a + 3d = 3 + frac{3}{7} = frac{24}{7}

4th rational number  =  a + 4d = 3 + frac{4}{7} = frac{25}{7}

5th rational number  =  a + 5d = 3 + frac{5}{7} = frac{26}{7}

6th rational number  =   a + 6d = 3 + frac{6}{7} = frac{27}{7}

So six rational numbers are frac{22}{7} , frac{23}{7} , frac{24}{7} , frac{25}{7} , frac{26}{7} , frac{27}{7}

simlify   left ( 5+sqrt{5} right )left ( 5+sqrt{5} right )

left ( 5+sqrt{5} right )left ( 5+sqrt{5} right )= {5^{2}-left (sqrt{5} right )^{2}}

                                           = 25-5

                                          = 20

 

if sqrt{2} = 1.414, then, find the value of  frac{1}{sqrt{2 +1}}

 frac{1}{sqrt{2 }+1} times (frac{sqrt{2 }-1}{sqrt{2}-1})=sqrt{2}-1

                                =1.414-1

                                = 0.414

 

Alternative method

:frac{1}{sqrt{2 }+1} =frac{1}{sqrt{2 }+1}times (frac{sqrt{2 }-1}{sqrt{2}-1})

 

               = frac{sqrt{2}-1}{sqrt{2}^{2}-1^{2}}

              = frac{sqrt{2}-1}{2-1} =frac{1.414-1}{1} = 0.414

 

Rationalize the denominator of   frac{30}{5sqrt{3}-3sqrt{5}}

 frac{30}{5 sqrt{3}-3sqrt{5}} times frac{5sqrt{3 }+3sqrt{5}}{5sqrt{3 }+3sqrt{5}} =frac{30(5(sqrt{3 }+3sqrt{5})}{(5sqrt{3 })^{2}-(3sqrt{5})^{2}}

                                                    = frac{30(5sqrt{3}+3sqrt{5})}{30}

                                                    = 5sqrt{3}+3sqrt{5

Alternative Method

frac{30}{5sqrt{3}-3sqrt{5}} = frac{30}{(5sqrt{3}-3sqrt{5})}times frac{(5sqrt{3}+3sqrt{5})}{(5sqrt{3}+3sqrt{5})}

                      = frac{30(5sqrt{3}+3sqrt{5})}{(5sqrt{}3)^{2}-(3sqrt{5})^{2}}

                      = frac{30(5sqrt{3}+3sqrt{5})}{75-45} =frac{30(5sqrt{3}+3sqrt{5})}{30}

                      = 5sqrt{3 }+3sqrt{5}

 

Simplify   frac{1}{1+sqrt{2}} +frac{1}{sqrt{2}+sqrt{3}} +frac{2}{sqrt{3}+sqrt{5}}

frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{2}{sqrt{3 }+ sqrt{5}}

                            =frac{1}{sqrt{2}+1}times frac{(sqrt{2}-1)}{(sqrt{2}-1)}+frac{1}{sqrt{3}+sqrt{2}} times frac{sqrt{3}-sqrt{2}}{sqrt{3}-sqrt{2}} + frac{2}{sqrt{5}+sqrt{3}}times frac{sqrt{5}-sqrt{3}}{sqrt{5}-sqrt{3}}

                           =frac{sqrt{2}-1}{2-1} +frac{sqrt{3}-sqrt{2}}{3-2} +frac{2(sqrt{5}-sqrt{3})}{5-3}

                          =sqrt{2}-1+sqrt{3}-sqrt{2}+sqrt{5}-sqrt{3} =sqrt{5}-1

Simplify  frac{sqrt{6}}{sqrt{2}+sqrt{3}} + frac{3sqrt{2}}{sqrt{6}+sqrt{3}} -frac{4sqrt{3}}{sqrt{6}+sqrt{2}}

 

 

frac{sqrt{6}}{sqrt{2}+sqrt{3}}=sqrt{18}-sqrt{12}=3sqrt{2}-2sqrt{3}

frac{3sqrt{2}}{sqrt{6}+sqrt{3}}=sqrt{12}-sqrt{6}=2sqrt{3}-sqrt{6}

frac{4sqrt{3}}{sqrt{6}+sqrt{3}}=sqrt{18}-sqrt{6}=3sqrt{2}-sqrt{6}

therefore Given expression  =  3sqrt{2}-2sqrt{3}+2sqrt{3} -sqrt{6}-3sqrt{2}+sqrt{6}

                               = 0

 

Alternative Method

frac{sqrt{6}}{sqrt{2}+sqrt{3}} + frac{3sqrt{2}}{sqrt{6}+sqrt{3}} -frac{4sqrt{3}}{sqrt{6}+sqrt{2}} 

              =frac{sqrt{6}}{sqrt{2}+sqrt{3}}times frac{(sqrt{3}-sqrt{2})}{(sqrt{3}-sqrt{2})}+frac{(3sqrt{2})}{(sqrt{6}+sqrt{3})}times frac{(sqrt{6}-sqrt{3})}{(sqrt{6}-sqrt{3)}}-frac{4sqrt{3}}{(sqrt{6}+sqrt{2})}times frac{(sqrt{6}-sqrt{2})}{(sqrt{6}-sqrt{2})}

             =frac{sqrt{18}-sqrt{12}}{3-2}+frac{3sqrt{12}-3sqrt{6}}{6-3}-frac{4sqrt{18}-4sqrt{6}}{6-2}

             =sqrt{18}-sqrt{12}+frac{3(sqrt{12}-sqrt{6})}{3}-frac{4(sqrt{18}-sqrt{6})}{4}

             =sqrt{18}-sqrt{12}+sqrt{12}-sqrt{6}-sqrt{18}+sqrt{6}=0.

 

Simplify  frac{5+sqrt{3}}{7-4sqrt{3}} times frac{7+4sqrt{3}}{7+4sqrt{3}}

 frac{5+sqrt{3}}{7-4sqrt{3}} times frac{7+4sqrt{3}}{7+4sqrt{3}} =frac{35+20sqrt{3}+7sqrt{3}+12}{49-48}   

                                = frac{47+27sqrt{3}}{1}

     and: : : :: : : frac{5+sqrt{3}}{4+4sqrt{3}} =frac{(5+sqrt{3})(7-4sqrt{3})}{(7+4sqrt{3})(7-4sqrt{3})}

                               =frac{23-13sqrt{3}}{1}

frac{5+sqrt{3}}{7-4sqrt{3}} -frac{5+sqrt{3}}{7+4sqrt{3}} = (47+27sqrt{3})-(23-13sqrt{3})

                                = 24+40 sqrt{3} = 8(3+5sqrt{3})

 

Alternative Method

 frac{5+sqrt{3}}{7-4sqrt{3}} - frac{5+sqrt{3}}{7+4sqrt{3}}    = frac{(5+sqrt{3})(7+4sqrt{3}) -(5+sqrt{3})(7-4sqrt{3})}{(7-4sqrt{3})(7+4sqrt{3})}

                                    = frac{35+20sqrt{3}+7sqrt{3}+12-35+20sqrt{3}-7sqrt{3}+12}{49-48}

                                    = frac{40sqrt{3}+24}{1} = 8(8+5sqrt{3}) .

show that    frac{[x^{a+b}]^{2}: : [x^{b+c}]^{2}: : [x^{c+a}]^{2}}{(x^{a}x^{b}x^{c})^{4}}= 1

frac{[x^{a+b}]^{2}[x^{b+c}]^{2}[x^{c+a}]^{2}}{(x^{a}x^{b}x^{c})^{4}}  = frac{x^{2a+2b}x^{2b+2c}x^{2c+2a}}{x^{4a}x^{4b}x^{4c}}

                                           = frac{x^{4a+4b+4c}}{x^{4a+4b+4c}}

                                           = 1

Simplify    2: sqrt[4]{81}-8sqrt[3]{216}+15sqrt[5]{32}+sqrt{225}-sqrt[4]{16}

2: sqrt[4]{81}-8sqrt[3]{216}+15sqrt[5]{32}+sqrt{225}-sqrt[4]{16}

                       = 2left ( 3^{4} right )^{frac{1}{4}}-8left ( 6^{3} right )^{frac{1}{3}}+15left ( 2^{5} right )^{frac{1}{5}}+15-left ( 2^{4} right )^{frac{1}{4}}

                       = 2times 3-8times 6+15times 2+15-2

                       = 6-48+30+15-2

                       = 51-50

                       = 1

Simplify  frac{sqrt{2}}{sqrt{5}+2} -frac{2}{sqrt{10}-2sqrt{2}} +frac{8}{sqrt{2}}

 frac{sqrt{2}}{sqrt{5}+2}-frac{2}{sqrt{10}-2sqrt{2}}+frac{8}{sqrt{2}}

                            = frac{sqrt{2}}{(sqrt{5}+2)}times frac{({sqrt{5}-2})}{(sqrt{5}-2)} -frac{2}{(sqrt{10}-2sqrt{2})}      times frac{sqrt{10}+2sqrt{2}}{sqrt{10}+2sqrt{2}}+frac{8}{sqrt{2}}times frac{sqrt{2} }{sqrt{2}}

 

                           = frac{sqrt{10}-2sqrt{2}}{5-4} -frac{2 (sqrt{10}+2sqrt{2})}{10-8} + frac{8sqrt{2}}{2}

                          = sqrt{10}-2sqrt{2} - frac{2(sqrt{10}+2sqrt{2})}{2} + 4sqrt{2}

                          = sqrt{10}-2sqrt{2} - {sqrt{10}-2sqrt{2}}{} + 4sqrt{2}

 

Simplify   frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{2}{sqrt{3}+sqrt{5}}

frac{1}{1+sqrt{2}}+frac{1}{sqrt{2}+sqrt{3}}+frac{2}{sqrt{3}+sqrt{5}}

                                            = frac{1}{left ( sqrt{2}+1right )}times frac{left ( sqrt{2}-1 right )}{left ( sqrt{2}-1 right )}+frac{1}{sqrt{3}+sqrt{2}}times frac{left ( sqrt{3}-sqrt{2} right )}{left ( sqrt{3} -sqrt{2}right )}+frac{2}{left ( sqrt{5} +sqrt{3}: right )}times frac{left ( sqrt{5}-sqrt{3} right )}{left ( sqrt{5}-sqrt{3} right )}

                                           = frac{left ( sqrt{2} -1right )}{2-1}+frac{sqrt{3}-sqrt{2}}{3-2}+frac{2left ( sqrt{5}-sqrt{3} right )}{5-3}

                                           = sqrt{2}-1+sqrt{3}-sqrt{2}+sqrt{5}-sqrt{3}= sqrt{5}-1

Simplify frac{1}{sqrt{3}+sqrt{2}} - frac{2}{sqrt{5}-sqrt{3}}- frac{3}{sqrt{2}-sqrt{5}}

 

frac{1}{sqrt{3}+sqrt{2}}-frac{2}{sqrt{5}-sqrt{3}}-frac{3}{sqrt{2}-sqrt{5}}

                             = frac{1}{sqrt{3}+sqrt{2}}times frac{sqrt{3}-sqrt{2}}{sqrt{3}-sqrt{2}}-frac{2}{sqrt{5}-sqrt{3}}  times frac{sqrt{5}+sqrt{3}}{sqrt{5}+sqrt{3}}-frac{3}{sqrt{2}-sqrt{5}}times frac{sqrt{2}+sqrt{5}}{sqrt{2}+sqrt{5}}

 

                            = frac{sqrt{3}-sqrt{2}}{1}-frac{2(sqrt{5}+sqrt{3})}{2}-frac{3(sqrt{2}+sqrt{5})}{-3}

                            = sqrt{3}-sqrt{2}-sqrt{5}-sqrt{3}+sqrt{2}+sqrt{5} = 0

Simplify   (sqrt{3}+1)(1 -sqrt{12})+frac{9}{(sqrt{3}+sqrt{12})}

(sqrt{3}+1) (1-sqrt{12})+frac{9}{(sqrt{3}+sqrt{12})}

                                           =(sqrt{3}-6+1-sqrt{12})+frac{9}{(sqrt{12}+sqrt{3})}times frac{(sqrt{12}-sqrt{3})}{(sqrt{12}-sqrt{3})}

                                          =(sqrt{3}-6+1-sqrt{12})+ frac{9(sqrt{12}+sqrt{3)}}{12-3}

                                          = sqrt{3}-5-sqrt{12}+sqrt{12}-sqrt{3}

                                          = -5.

Evaluate   frac{sqrt{5}+sqrt{2}}{sqrt{5}-sqrt{2}}   given that sqrt{10}=3.162

frac{sqrt{5}+sqrt{2}}{sqrt{5}-sqrt{2}} = frac{(sqrt{5}+sqrt{2}) }{sqrt{5}-sqrt{2}}times frac{(sqrt{5}+sqrt{2})}{sqrt{5}+sqrt{2}}     

                     = frac{(sqrt{5})^{2}+(sqrt{2})^{2}+2times sqrt{5}times sqrt{2}}{(sqrt{5})^{2}-(sqrt{2})^{2}}

                     = frac{5+2+2sqrt{10}}{5-2}

                    = frac{7+2times 3.162}{3}

                    = frac{7+6.324}{3} =frac{13.324}{3}

                    =4.441 (approx)

 if: : sqrt{2} =1.414 : : and : : sqrt{3} = 1.732: : then :, calculate: : frac{4}{3sqrt{3}-2sqrt{2}}+frac{3}{3sqrt{3}+2sqrt{2}} 

frac{4}{3sqrt{3}-2sqrt{2}}+frac{3}{3sqrt{3}+2sqrt{2}}=frac{21sqrt{3}+2sqrt{2}}{19}

                                                         =frac{21(1.732)+2(1.414)}{19}

                                                         = frac{39.2}{19}

                                                         =2.063

 

Alternative Method 

frac{4}{3sqrt{3}-2sqrt{2}}+frac{3}{3sqrt{3}+2sqrt{2}}     =frac{4(3sqrt{3}+2sqrt{2})+3(3sqrt{3}-2sqrt{2})}{(3sqrt{3}-2sqrt{2})(3sqrt{3}+2sqrt{2})}

                                                            =frac{12sqrt{3}+8 sqrt{2}+9sqrt{3}-6sqrt{2}}{27-8}

                                                            =frac{21 sqrt{3}+2sqrt{2}}{19}

                                                           =frac{21times 1.732+2times 1.414}{19}

                                                          =frac{36.372 +2.828}{19}     

                                                          = frac{39.2}{19} =2.063

 

Find the value of left ( x-frac{1}{x} right )^{3} , if x =1+sqrt{2}

 x =1+sqrt{2}

frac{1}{x} =frac{1}{1+sqrt{2}}times frac{1-sqrt{2}}{1-sqrt{2}} = frac{1-sqrt{2}}{1-2}

                                                 = sqrt{2}-1

therefore x-frac{1}{x} =(1+sqrt{2})-(sqrt{2}-1)

                   = 1+ sqrt{2}-sqrt{2}+1=2

left ( x-frac{1}{x} right )^{3}=2 ^{3}= 8

 

Alternative Method 

x = 1+sqrt{2}

frac{1}{x} =frac{1}{sqrt{2 +1}} times (frac{sqrt{2}-1}{sqrt{2}-1})

    = (frac{sqrt{2}-1}{2-1}) =sqrt{2}-1

therefore x-frac{1}{x }= (1+sqrt{2})-(sqrt{2}-1)

                  = 1+sqrt{2} -sqrt{2}+1 =2

therefore left ( x - frac{1}{x} right )^{3}=2^{3} =8

if   x = 9+4sqrt{5}   then find the value of  sqrt{x} -frac{1}{sqrt{x}}

  x = 9+4sqrt{5}

     = 5+4+4sqrt{5}

     =(sqrt{5}+2)^{2}

sqrt{x} = sqrt{5}+2

frac{1}{sqrt{x}} =sqrt{5}-2

sqrt{x}-frac{1}{sqrt{x}} =sqrt{5}+2-(sqrt{5}-2)

                     =sqrt{5}+2-sqrt{5}+2

                    =4

 

Alternative Method 

  x = 9+4 sqrt{5} =5+4+4sqrt{5}

       = (sqrt{5})^{2} + (2)^{2}+2times 2sqrt{5}

   x = (sqrt{5}+2)^{2}

sqrt{x }=sqrt{5}+2

frac{1}{sqrt{x}}=frac{1}{(sqrt{5}+2)} times frac{(sqrt{5}-2)}{(sqrt{5}-2)}=frac{sqrt{5}-2}{5-4} =sqrt{5}-2

                                therefore sqrt{x }-frac{1}{sqrt{x}}=(sqrt{5}-2)-(sqrt{5}-2)
                                                           =sqrt{5}+2-sqrt{5}+2 =4

Find a and b if   frac{1-sqrt{3}}{1+sqrt{3}} = a+b

 a+b =frac{1-sqrt{3}}{1+sqrt{3}}

            = frac{1-sqrt{3}}{1+sqrt{3}}times frac{1-sqrt{3}}{1-sqrt{3}}

           =frac{(1-sqrt{3})^{2}}{1-3}

           =frac{1+3-2sqrt{3}}{-2}

           =frac{4-2sqrt{3}}{-2}

a+b = -2+sqrt{3}

        a = -2

        b = sqrt{3}

Find the values of a and b when a+bsqrt{15} = frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}}

 frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}} =frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}} times frac{sqrt{5}+sqrt{3}}{sqrt{5}+sqrt{3}}

                =frac{8+2sqrt{15}}{2}= 4+sqrt{15}

a+b sqrt{15} =4+sqrt{15}

a= 4 : ,b = 1

Alternative method

frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}} = frac{sqrt{5}+sqrt{3}}{sqrt{5}-sqrt{3}} times frac{(sqrt{5}+sqrt{3})}{sqrt{5+sqrt{3}}}

                   = frac{5+3+2times sqrt{5}times sqrt{3}}{5-3}

                   =frac{8+2sqrt{15}}{2}

                  =frac{2(4+sqrt{15})}{2}

                  = 4+sqrt{15}

  a+b sqrt{15} = 4+sqrt{15}

Compairing both sides we get 

a= 4, b= 1

Height of students of  class x are given in the following distribution :

Find the model of height 

Height (in cm)                  150- 155            155 -160           160- 165            165 - 170            170-175

Number of students              15                      8                     20                     12                      5

 

Class interval                                Frequency 

 150 -155                                          15

155 - 160                                           8 

160 -165                                          20

165 -170                                         12

170- 175                                          5


Total                                              60

 

Here,

Modal class = 160 -165

l = 160, f_{1} = 20, f_{0} = 8, f_{2} = 12, h = 5

Mode  = l + left ( frac{f_{1}-f_{0}}{2f_{1}-f_{0}-f_{2}} right )times h

= 160 + left ( frac{20-8}{40-8-12} right ) times 5

= 160 left ( frac{12}{20} right )times 5

= 163

Model height  = 163 cm .

If    x= frac{sqrt{p + 2q}+sqrt{p-2q}}{sqrt{p+2q}-sqrt{p-2q}}   then show that  qx^{2}-px +q=0

           x =frac{left [ sqrt{p+2q}+sqrt{p-2q} right ]^{2}}{p+2q-p+2q}

             = frac{1}{4q} (2p+2sqrt{p^{2}-4q^{2}})

2qx -p =sqrt{p^{2}-4q^{2}}

4q(qx^{2}-px) =4q^{2}

qx^{2}-px+q=0

 

Alternative Method 

  x = frac{sqrt{p+2q}+sqrt{p-2q}}{sqrt{p+2q}-sqrt{p-2q}} times frac{sqrt{p+2q}+sqrt{p-2q}}{sqrt{p+2q}+sqrt{p-2q}}

    = frac{(p+2q)+(p-2q)+2times sqrt{p+2q}times sqrt{p-2q}}{(p+2q)-(p-2q)}

   = frac{2p +2sqrt{p^{2}-4q^{2}}}{p+2q-p+2q}

   =frac{2(p+2sqrt{p^{2}-4q^{2}})}{4q}

2qx = p+sqrt{p^{2}-4q^{2}}

2qx-p=sqrt{p^{2}-4q^{2}}

Squaring the both sides we get 

4q^{2}x^{2}+p^{2}-4pqx =p^{2}-4q^{2}

        4q (qx^{2}-px)=-4q^{2}

        qx^{2}-px+q =0

Rationalize the denominator of frac{1}{(sqrt{2}+sqrt{3})-sqrt{4}}

frac{1}{(sqrt{2}+sqrt{3})-sqrt{4}}times frac{(sqrt{2}+sqrt{3})+sqrt{4}}{(sqrt{2}+sqrt{3})+sqrt{4}}  =frac{sqrt{2}+sqrt{3}+sqrt{4}}{(sqrt{2}+sqrt{3})^{2}-{4}} = frac{sqrt{2}+sqrt{3}+sqrt{4}}{(2+3+2sqrt{6})-4}

                                                                                 =frac{sqrt{2}+sqrt{3}+sqrt{4}}{1+2sqrt{6}}times frac{1-2sqrt{6}}{1-2sqrt{6}}

                                                                                 =frac{sqrt{2}+ sqrt{3}+sqrt{4}-2sqrt{12}-2sqrt{18}-4sqrt{6}}{1^{2}-(2sqrt{6})^{2}}

                                                                                =frac{sqrt{2}+sqrt{3}+sqrt{4}-4sqrt{3}-6sqrt{2}-4sqrt{6}}{1-24}

                                                                               =frac{-5sqrt{2}-3sqrt{3}+2-4sqrt{6}}{-23}

                                                                               =frac{+5sqrt{2}+3sqrt{3}+4sqrt{6}-2}{23}

Prove that : frac{1}{3-sqrt{8}} - frac{1}{sqrt{8}-sqrt{7}}+frac{1}{sqrt{7}-sqrt{6}}-frac{1}{sqrt{6}-sqrt{5}}+frac{1}{sqrt{5}-2} =5

LHS: =: frac{1}{3-sqrt{8}}-frac{1}{sqrt{8}-sqrt{7}}+ frac{1}{sqrt{7}-sqrt{6}}-frac{1}{sqrt{6}-sqrt{5}}+frac{1}{sqrt{5}-2}

              =frac{3+sqrt{8}}{(3)^{2}-(sqrt{8})^{2}}-frac{sqrt{8}+sqrt{7}}{(sqrt{8})^{2}-(sqrt{7})^{2}}+frac{(sqrt{7}+sqrt{6})}{(sqrt{7})^{2}-(sqrt{6})^{2}}   -frac{(sqrt{6}+sqrt{5})}{(sqrt{6})^{2}-(sqrt{5})^{2}} +frac{(sqrt{6}+2)}{(sqrt{5})^{2}-(2)^{2}}

              =frac{3+sqrt{8}}{9-8}- frac{(sqrt{8}+sqrt{7})}{8-7}+frac{(sqrt{7}+sqrt{6})}{7-6}-frac{(sqrt{6}+sqrt{5})}{6-5}  +frac{(sqrt{5}+2)}{5-4}left [ because a^{2} -b^{2}=(a-b)(a+b) right ]

              = 3+sqrt{8}-sqrt{8}-sqrt{7}-sqrt{7}+sqrt{6}-sqrt{6}-sqrt{5}+sqrt{5}+2

              = 3+2 =5 =RHS

Prove that   frac{1}{3+sqrt{7}}+frac{1}{sqrt{7}+sqrt{5}}+frac{1}{sqrt{5}+sqrt{3}}+frac{1}{sqrt{3}+1}=1.

  frac{1}{3+sqrt{7}}= frac{1}{3+sqrt{7}}times frac{3-sqrt{7}}{3-sqrt{7}} =frac{3-sqrt{7}}{9-7} = frac{3-sqrt{7}}{2}

frac{1}{sqrt{7}+sqrt{5}} = frac{1}{sqrt{7}+sqrt{5}}times frac{sqrt{7}-sqrt{5}}{sqrt{7}-sqrt{5}}=frac{sqrt{7}-sqrt{5}}{2}

frac{1}{sqrt{5}+sqrt{3}} = frac{1}{sqrt{5}+sqrt{3}}times frac{sqrt{5}-sqrt{3}}{sqrt{5}-sqrt{3}}=frac{sqrt{5}-sqrt{3}}{5-3}

                     = frac{sqrt{5}-sqrt{3}}{2}

frac{1}{sqrt{3}+1} = frac{1}{sqrt{3}+1} times frac{sqrt{3}-1}{sqrt{3}-1}=frac{sqrt{3}-1}{3-1}=frac{sqrt{3}-1}{2}

   LHS = frac{3-sqrt{7}}{2} + frac{sqrt{7}-sqrt{5}}{2} + frac{sqrt{5}-sqrt{3}}{2}+frac{sqrt{3}-1}{2}

               =frac{3-1}{2}

              = frac{2}{2} = 1 = RHS

Evaluate   frac{15}{sqrt{10}+sqrt{20}+sqrt{40}-sqrt{5}-sqrt{80}}   given that  sqrt{5}=2.2 :: and: sqrt{10}=3.2

Denominator = sqrt{10}+2sqrt{5}+2sqrt{10}-sqrt{5}-4sqrt{5} =3(sqrt{10}-sqrt{5})

     frac{15}{3(sqrt{10}sqrt{5})} = frac{5times (sqrt{10}+sqrt{5})}{10-5}

                             =sqrt{10}+sqrt{5}

                             =3.2+2.2= 5.4

 

Alternative Method 

frac{15}{sqrt{10}+sqrt{20}+sqrt{40}-sqrt{5}-sqrt{80}}    =frac{15}{sqrt{10}+2sqrt{5}+2sqrt{10}-sqrt{5}-4sqrt{5}}

                                                                       =frac{15}{3sqrt{10}-3sqrt{5}}

                                                                       =frac{15}{3sqrt{10}-3sqrt{5}}times frac{(sqrt{10}+sqrt{5})}{(sqrt{10}+sqrt{5})}

                                                                      = frac{5times (sqrt{10}+sqrt{5})}{10-5}

                                                                      =frac{5(3.2+2.2)}{5} =5.4

Find the  values of a and b in   frac{3- sqrt{5} }{3+2sqrt{5}} = asqrt{5} - frac{b}{11}

 asqrt{5}-frac{b}{11} = frac{(3-sqrt{5})}{(3+2sqrt{5})}times frac{(3-2sqrt{5})}{(3-2sqrt{5})}

                        =frac{9-6sqrt{5}-3sqrt{5}+2times 5}{(3)^{2}-(2sqrt{5})^{2}}

                       =frac{9-9sqrt{5}+10}{9-20}

                       =frac{19-9sqrt{5}}{-11}

                      =frac{: : : 19}{-11}-frac{9sqrt{5}}{-11}=frac{9sqrt{5}}{11}-frac{19}{11}

asqrt{5}-frac{b}{11} = frac{9}{11}sqrt{5}-frac{19}{11}

On Comparing both sides we get 

a = frac{9}{11}, , ; , b =19

  if: x=4-sqrt{15}: : then: find: the: value: of left ( x +frac{1}{x} right )^{2}

                  x = 4- sqrt{15}

                 frac{1}{x} =frac{1}{4-sqrt{15}}times frac{4+sqrt{15}}{4+sqrt{15}}

                 frac{1}{x} = frac{4+sqrt{15}}{16-15}

                  frac{1}{x} = 4+sqrt{15}

left ( x+frac{1}{x} right )^{2} = (4-sqrt{15}+4+sqrt{15})^{2}

                      = (8)^{2}

                     =64

If   a = frac{2-sqrt{5}}{2+sqrt{5}}, b = frac{2+sqrt{5}}{2-sqrt{5}},   then find  (a+b)^{3}

 a = frac{2-sqrt{5}}{2+sqrt{5}}times frac{2-sqrt{5}}{2-sqrt{5}} = frac{(2-sqrt{5})^{2}}{-1}= -(4-4sqrt{5}+5)

                                                  =4sqrt{5}-9

b = frac{2+sqrt{5}}{2-sqrt{5}}times frac{2+sqrt{5}}{2+sqrt{5}} =frac{(2+sqrt{5})^{2}}{-1}

                                                =-(4+4sqrt{5}+5)

                                                =-9-4sqrt{5}

                                    a+b = -18

                              (a+b)^{3} =(-18)^{3}= -5832

 

 Alternative Method

a+b = frac{2-sqrt{5}}{2+sqrt{5}}+frac{2+sqrt{5}}{2-sqrt{5}}

          =frac{(2-sqrt{5})^{2}+(2+sqrt{5})^{2}}{(2+sqrt{5})(2-sqrt{5})}

         =frac{4+5-4sqrt{5}+4+5+4sqrt{5}}{4-5}

         =frac{18}{-1} = -18

(a+b)^{3} =(-18)^{3}= -5832

if: : x=frac{sqrt{3}+sqrt{2}}{sqrt{3-sqrt{2}}} : : and : : y = frac{sqrt{3}-sqrt{2}}{{sqrt{3}+sqrt{2}}} : : find: : the: : x^{2 + y^{2}}

             x =frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}}

  and: : y = frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}}

x^{2} +y^{2} = left ( frac{sqrt{3}+sqrt{2}}{sqrt{3}-sqrt{2}} right )^{2}+left ( frac{sqrt{3}-sqrt{2}}{sqrt{3}+sqrt{2}} right )^{2}

                = left ( frac{3+2+2sqrt{6}}{3+2-2sqrt{6}} right )+left ( frac{3+2-2sqrt{6}}{3+2+2sqrt{6}} right )

               = frac{5+2sqrt{6}}{5-2sqrt{6}}+frac{5-2sqrt{6}}{5+2sqrt{6}}

               =frac{(5+2sqrt{6})^{2}+(5-2sqrt{6})^{2}}{(5)^{2}-(2sqrt{6})^{2}}

              = frac{25+24+20sqrt{6}+25+24-20sqrt{6}}{25-24}

              = 98

Prove that  frac{1}{sqrt{4}+sqrt{5}}+frac{1}{sqrt{5}+sqrt{6}}+frac{1}{sqrt{6}+sqrt{7}}+frac{1}{sqrt{7}+sqrt{8}}+frac{1}{sqrt{8}+sqrt{9}}=1.

 frac{1}{sqrt{4}+sqrt{5}}=frac{1}{(sqrt{5}+sqrt{4})}times frac{(sqrt{5}-sqrt{4})}{(sqrt{5}-sqrt{4)}}

                      =frac{sqrt{5}-sqrt{4}}{5-4} = sqrt{5}-sqrt{4}

frac{1}{sqrt{5}+sqrt{6}}= frac{1}{(sqrt{6}+sqrt{5})}times frac{(sqrt{6}-sqrt{5})}{(sqrt{6}-sqrt{5})}

                  = frac{(sqrt{6}-sqrt{5})}{6-5}=sqrt{6}-sqrt{5}

frac{1}{sqrt{6}+sqrt{7}}=frac{1}{sqrt{7}+sqrt{6}} times frac{(sqrt{7}-sqrt{6})}{(sqrt{7}-sqrt{6})}=frac{sqrt{7}-sqrt{6}}{7-6}            

                      = sqrt{7}-sqrt{6}

frac{1}{sqrt{7}+sqrt{8}}=frac{1}{(sqrt{8}+sqrt{7})}times frac{(sqrt{8}-sqrt{7})}{(sqrt{8}-sqrt{7})} =frac{(sqrt{8}-sqrt{7})}{8-7}

                     = sqrt{8}-sqrt{7}

frac{1}{sqrt{8}+sqrt{9}}=frac{1}{sqrt{9}+sqrt{8}}times frac{(sqrt{9}-sqrt{8})}{(sqrt{9}-sqrt{8})}=frac{sqrt{9}-sqrt{8}}{9-8}

                  = sqrt{9}-sqrt{8}

Now LHS = sqrt{5}-sqrt{4}+sqrt{6}-sqrt{5}+sqrt{7}-sqrt{6}+sqrt{8}-sqrt{7}+sqrt{9}-sqrt{8}

                 =-sqrt{4}+sqrt{9}

                 =-2+3=1: : RHS

Thus LHS =  RHS

if  x =3-2sqrt{2}  then find the value of  x^{4}-frac{1}{x^{4}}

Given x: = 3-2sqrt{2}

 frac{1}{x}= frac{1}{(3-2sqrt{2})}times frac{(3+ 2sqrt{2})}{(3+2sqrt{2})}: : (: rationalizing: )

 

frac{1}{x}= frac{(3+2sqrt{2})}{9-8}= 3+2sqrt{2}

 

frac{1}{x^{2}}= (3+2sqrt{2})^{2}=9+8+12sqrt{2}=17+12sqrt{2}

 

x^{2}= (3-2sqrt{2})^{2}=9+8-12sqrt{2}=17-12sqrt{2}

 

Now , x^{4}-frac{1}{x^{4}}=left ( x^{2}-frac{1}{x^{2}} right )left ( x^{2}+frac{1}{x^{2}} right )

 

                          =left [ (17-12sqrt{2})-(17+12sqrt{2}) right ] left [ (17-12sqrt{2}) +(17+12sqrt{2})right ]

                         = 17-12sqrt{2}-17-12sqrt{2}left ( 17-12sqrt{2}+17+12sqrt{2} right )

                         =(-24sqrt{2})times 34=-816sqrt{2}

if      a =frac{sqrt{5}+sqrt{2}}{sqrt{5}-sqrt{2}}    and    b =frac{sqrt{5}-sqrt{2}}{sqrt{5}+sqrt{2}}    find the value  of    frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}

 a^{2}+ab+b^{2} =(a+b)^{2}-ab

a^{2}-ab+b^{2} =(a-b)^{2}+ab

  a = frac{sqrt{5}+sqrt{2}}{sqrt{5}-sqrt{2}}times frac{sqrt{5}+sqrt{2}}{sqrt{5}+sqrt{2}}

     = frac{(sqrt{5}+sqrt{2})^{2}}{5-2}

     = frac{5+2+2sqrt{10}}{3}= frac{7+2sqrt{10}}{3}   

 b = frac{sqrt{5}-sqrt{2}}{sqrt{5}+sqrt{2}}times frac{sqrt{5}-sqrt{2}}{sqrt{5}-sqrt{2}}

   =frac{7-2sqrt{10}}{3}

(a+b)^{2}-ab = left [ frac{7+2sqrt{10}}{3}+frac{7-2sqrt{10}}{3} right ]^{2}-left [ frac{7+2sqrt{10}}{3} right ]left [ frac{7-2sqrt{10}}{3} right ]

                        = left ( frac{14}{3} right )^{2}-left ( frac{49-40}{9} right )

                       = frac{196}{9}-frac{9}{9}

                      = frac{187}{9}

(a-b)^{2}-ab = left ( frac{7+2sqrt{10}}{3}+frac{7-2sqrt{10}}{3} right )^{2}+ left ( frac{7+2sqrt{10}}{3} right )left ( frac{7-2sqrt{10}}{3} right )

                         = left ( frac{4sqrt{10}}{3} right )^{2}+frac{49-40}{9}

                        = frac{160}{9}+frac{9}{9} =frac{169}{9}

therefore frac{a^{2}+ab+b^{2}}{a^{2}-ab+b^{2}}= frac{(a+b)^{2}-ab}{(a-b)^{2}+ab}

                             = frac{-187/9}{169/9}

                             =frac{187}{169}

Find the value of    frac{3^{30}+3^{29}+3^{28}}{3^{31}+3^{30}-3^{29}}

frac{3^{30}+3^{29}+3^{28}}{3^{31}+3^{30}-3^{29}} = frac{3^{38}(3^{2}+3^{1}+1)}{3^{29}(3^{2}+3^{1}-1)}

 

                                = frac{(9+3+1)}{3(9+3-1)}

 

                               = frac{13}{3times 11}=frac{13}{33}

Find the value of      frac{4}{(216)^{frac{-2}{3}}}+frac{1}{(216)^{frac{-3}{4}}}+frac{2}{(343)^{frac{-1}{5}}}

frac{4}{(216)^{frac{-2}{3}}}+frac{1}{(216)^{frac{-3}{4}}}+frac{2}{(343)^{frac{-1}{5}}}      = frac{4}{(6^{3})^{frac{-2}{3}}}+frac{1}{(4^{4})frac{-3}{4}}+frac{2}{(3^{5})^{frac{-1}{5}}}

 

                                                                     =frac{4}{6^{-2}}+frac{1}{4^{-3}}+frac{2}{3^{-1}}

 

                                                                    =frac{4}{36^{-1}}+frac{1}{64^{-1}}+frac{2}{3^{-1}}

 

                                                                    = 4times 36+1times 64+2times 3

 

                                                                    = 214

Show that   frac{1}{1+x^{a-b}}+frac{1}{1+x^{b-a}}= 1

frac{1}{1+x^{a-b}} +frac{1}{1+x^{b-a}}   =frac{x^{b}}{x^{b}+x^{a}} + frac{x^{a}}{x^{a}+x^{b}}   

                                             =frac{x^{b}+x^{a}}{x^{a}+x^{b}}

 

Alternative Method

 

frac{1}{1+x^{a b}}+frac{1}{1+x ^{ba}}= frac{1}{1+frac{x^{a}}{x^{b}}} +frac{1}{1+frac{x^{b}}{x^{a}}}

                                      =frac{1}{frac{x^{b}+x^{a}}{x^{b}}}+frac{1}{frac{x^{a}+x^{b}}{x^{a}}}

                                      =frac{x^{b}}{x^{b}+x^{a}} +frac{x^{a}}{x^{a}+x^{b}}

                                      = frac{(x^{b}+x^{a})}{(x^{a}+x^{b})}

Prove that    left ( frac{x^{a^{2}}}{x^{b^{2}}} right )^{frac{1}{a+b}}= left ( frac{x^{b^{}2}}{x ^{c^{}2}} right )^{frac{1}{b+c}}.left ( frac{x^{c^{}2}}{x^{a^{}2}} right )^{frac{1}{c+a}}=1

 

LHS = left ( frac{x^{a^{}2}}{x^{b^{}2}} right ) ^{frac{1}{a+b}}.left ( frac{x^{b^{}2}}{x^{c^{}2}} right )^{frac{1}{b+c}}.left ( frac{x^{c^{}2}}{x^{a^{}2}} right )^{frac{1}{c+a}}

            =left ( x^{a^{}2-b^{}2} right )^{frac{1}{a+b}}.left ( x^{b^{}2-c^{}2} right )^{frac{1}{b+c}}.left ( x^{c^{}2-a^{}2} right )^{frac{1}{c+a}}

           = x^{frac{a^{}2-b^{}2}{a+b}}.x^{frac{b^{}2-c^{}2}{a+b}}.x^{frac{c^{}2-a^{}2}{c+a}}

           = x^{a-b}. x^{b-c}.x^{c-a}

           =x^{0}

          = 1 = RHS

(i)Find the six rational  numbers bbetween 3 and 4 

(ii) Which mathematical concept is used in this problem

(iii) Which value is depticted in this question 

 

(i) We known that between two rational numbers  x and y such that x< y there is a rational number frac{x+y}{2}.

               ie,       3< frac{7}{2}< 4

Now a rational number between 3 and   frac{7}{2}< 4   is :

frac{1}{2}left ( 3+frac{7}{2} right )=frac{1}{2}times left ( frac{6+7}{2} right )=frac{13}{4}

A rational nummber berween  frac{7}{2}: : and : : 4 : : is

frac{1}{2}left ( frac{7}{2}+4 right )=frac{1}{2}times left ( frac{7+8}{2} right )=frac{15}{4}

 3< frac{13}{4}< frac{7}{2}< frac{15}{4}< 4

Further a rational number between 3 and  frac{13}{4}: : is :

frac{1}{2}left ( 3+frac{13}{4} right )=frac{1}{2}left ( frac{12+13}{4} right )=frac{25}{8}

A rational number between  frac{15}{4}: : and: : 4: : is :

frac{1}{2}left ( frac{15}{4}+4 right )=frac{1}{2}times frac{15+16}{4}=frac{31}{8}

A rational number between  frac{31}{8}: : and: : 4: : is

frac{1}{2}left ( frac{31}{8}+4 right ) = frac{1}{2}times left ( frac{31+32}{8} right )=frac{63}{16}

therefore 3< frac{25}{8}< frac{13}{4}< frac{7}{2}< frac{15}{4}< frac{31}{8}< frac{63}{16}< 4

Hence , six rational numbers between 3 and 4 are 

frac{25}{8}, frac{13}{4},frac{7}{2}, frac{15}{4}, frac{31}{8}: : and: : frac{63}{16}

(ii) Number syatem 

(iii) Rationality is always welcomed 

Two classmates Salma and Anil simplified two different expressions during te revision hour and explained to each other their simplifications. Salma explains simplification of   frac{sqrt{2}}{sqrt{5}+sqrt{3}}    and Anil explains simplification of   sqrt{28}+sqrt{98}+sqrt{147}

Write the both simplification. What value does it depict ?

 

Justify      frac{sqrt{2}}{sqrt{5}+sqrt{3}}= frac{sqrt{2}}{sqrt{5}+sqrt{3}}times frac{(sqrt{5}-sqrt{3})}{(sqrt{5}-sqrt{3})}

                               = frac{sqrt{10}-sqrt{6}}{(sqrt{5})^{2}-(sqrt{3})^{2}}= frac{sqrt{10}-sqrt{6}}{5-3}

                               =frac{sqrt{10}-sqrt{6}}{2}

Again    sqrt{28}+ sqrt{98}+sqrt{147}

                               =sqrt{2times 2times 7}+sqrt{2times 7times 7}+sqrt{3times 7times 7}

                               = 2sqrt{7}+7sqrt{2}+7sqrt{3}

Co-operative learning among classmates with out any gender and religious bias.

If f(x) = x^{3}-3x^{2}+3x-4: : find : : f (2)+f(-2)+f(0).

    f(x)=x^{3}-3x^{2}+3x-4

    f(2)=(2)^{3}-3(2)^{2}+3(2)-4

               =8-12+6-4

    f(2)=-2

f(-2)=(-2)^{3}-3(-2)^{2}+3(-2)-4

               =-8-12-6-4

f(-2)=-30

    f(0)=-4

therefore f(2)+f(-2)+f(0)=-2-30-4=-36

If   f(x)=5x^{2}-4x+5   find  f(1)+f(-1)+f(0). 

    f(x)= 5x^{2}-4x+5

    f(1)=5-4+5

               =6

f(-1)=5(-1)^{2}-4(-1)+5

              =5+4+5

              =14

   f(0)=5

therefore f(1)+f(-1)+f(0)=6+14+5=25

if: : : f(x)=x^{2}-5x+7,: : : evaluate: : : f(2)-f(-1)+fleft ( frac{1}{3} right )

            f(x)= x^{2}-5x+7

 Then  f (2)=2^{2}-5times 2+7=1

        f (-1)=(-1)^{2}-5(-1)+7=13

      f left ( frac{1}{3} right )=left ( frac{1}{3} right )^{2}-5left ( frac{1}{3} right )+7=frac{49}{9}

f(2)-f(-1)+f left ( frac{1}{3} right )=1-13+frac{49}{9}=frac{-59}{9}

Using remainder theorem, factorize  6x^{3}-25x^{2}+32x-12.

Factors of   12=(pm 1,pm 2,pm 3,pm 4,pm 6,pm 12,)

               p(x)=6x^{3}-25x^{2}+32x-12

               p(2)= 6(2)^{3}-25(2)^{2}+32times 2-12

                      = 48-100+64-12

                      =112-112=0

 therefore x = 2  is a zero of  p(x ): : or: : (x-2)  is a factor  p(x)

6x^{3}-25x^{2}+32x-12

                   = 6x^{2}(x-2)-13x(x-2)+6x(x-2)

                   = (x-2)(6x^{2}-13+6)

                   =(x-2)(6x^{2}-9x-4x+6)

                   =(x-2)left [ 3x(2x-3) -2(2x-3)right ]

                   =(x-2)(2x-3)(3x-2).

Factorize : x^{3}-2x^{2}-5x-6

factor of  6 =  6=(pm 1,pm 2,pm 3,pm 6)

p(x)= x^{3}+2x^{2}-5x-6

p(-1)= (-1)^{3}+2(-1)^{2}-5(-1)-6

=-1+2+5-6

=7-7=0

because x = -1 is zero of p (x) of (x+1) is a factor of p (x)

therefore x^{3}+2x^{2}-5x-6

=x^{2}(x+1)+x(x+1)-6(x+1)

= (x+1)[x^{2}+x-6]

= (x+1)[x(x+3)-2(x+3)]

= (x+1)(x+3)(x-2)

 

factorize x^{3}+13x^{2}+32x+20

(x+2)  is a factor of x^{3}+13x^{2}+32x +20

because [p(-2)=0]

x^{3}+13x^{2}+32x+20 = (x+2)(x^{2}+11x+10)

rightarrow x^{2}+ 11x+10= x^{2}+10x+x+10

= x (x+10)+1(x+10)

= (x+1) (x+10)

Factors are: (x+2)   (x+1)  (x+10)

Alternative method 

Factors  of  20 =(pm 1,pm 2,pm 3,pm 5,pm 10,pm 20)

p(x)=x^{3}+13x^{2}+32x+20

p(-1)= (-1)^{3}+13(-1)^{2}+32(-1)+20

=-1+13+32+20

33-33=0

therefore x = -1 is a zero of p(x), and (x+1) is a factor of p(x)

Then 

x^{3}+13x^{2}+32x+20

= x^{2}(x+1)+12x(x+1)+20 (x+1)

= (x+1)(x^{2}+12x+20)

= (x+1) [x (x+10)+2 (x+10)]

= (x+1)(x+2) (x+10)

Simplify  frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}

frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}

Both numerator and denominator are of the form a3+b3+c3

We  know that when a+b+c = 0

then a3+b3+c3 = 3abc

For numerator, a2-b2+b2-c2+c2-a2= 0

For denominator     a-b+b-c+c-a= 0 

 

therefore frac{(a^{2}-b^{2})^{3} +(b^{2}-c^{2})^{3}+(c^{2}-a^{2})^{3}}{(a-b)^{3}+(b-c)^{3}+ (c-a)^{3}}

 

=frac{3times (a^{2}-b^{2})(b^{2}-c^{2})(c^{2}-a^{2})}{3(a-b) (b-a)(b-c)}

 

=frac{ (a-b)(a+b)(b-c)(b+c)(c-a)(c+a)}{(a-b) (b-c)(c-a)}

 

= (a+b) (b+c) (c+a)

 

 

 

 

 

If x+frac{1}{x} = 5evaluate   x^{2}+frac{1}{x^{2}}

 x+frac{1}{x} = 5

On squaring both sides , we get 

left (x+frac{1}{x} right )^{2}= 5^{2}

rightarrow x^{2}+left ( frac{1}{x} right )^{2}+2times x+frac{1}{x} = 25

[because (a+b)^{2}=a^{2}+b^{2}+2ab)]                           

Rightarrow x^{2}+ frac{1}{x^{2}} +2= 25

Rightarrow x^{2}+ frac{1}{x^{2}} = 25-2

Rightarrow x^{2}+ frac{1}{x^{2}} =23

Ram has two rectangles having their areas as given below:

(1) 25a2-35a+12                   (II) 35y2+13y-12

(i) Given possible expressions for the length and breadth of each rectangle.

(ii) Which mathematical concept is used in this problem?

(iii) Which value is depicted in this problem

 

(i) Possible length and breadth of the rectangle are the factors of its given area.

Area    =  25a^{2}- 35a + 12

          = 25a^{2}-15a - 20 a + 12

          =5a (5a-3) -4 (5a-3)

          = (5a-4) -4 (5a-3)

So possible length and breadth are (5a -3) and (5a - 4) units, respectively.

(ii)  Area  = 35y^{2}+13y-12

              = 35y^{2}+28y-15y-12

              = 7y (5y+4)-3(5y+4)

              = ( 7y - 3) (5y+4)

So, possible length and breadth are (7y-3) and (5y+4) units respectively. 

(ii) Factorization of polynomials 

(iii) Expression of one's desires and news is very necessary.

The auto rikshaw fare in a city is charged RS . 10 for first kilometer and @Rs.4 per kilometer for subsequent distance covered. Write the linear equation to express the above statment. Draw the graph of the linear  equation. 

 

Total distance covered =  x km. 

Total fare = y km. 

Fare for the first kilometer 

Subsequent distance  = (x-1) km

Fare for the subsequent distance = Rs. 4(x-1)

According to the question, 

 y = 10+4 (x-1)

Rightarrowy = 10+4x -4 

Rightarrow y =4x +6

Table of solutions

X 0 1 -1
Y 6 10 2