T_r = r/r+1= 1 - 1/r+1

 

the given sum is 1+1+1+...n times -( 1/2 +1/3 +1/4...n terms)

 

the second part is an HP i dont think you can find the sum of a HP

srujana it's like your method but check it.

let the condition be met at 2 hrs.xminutes

 

in 60 min hr hand moves 30 degrees

 

in x min it moves x/2 defrees.

 

in 60 min min hand moves 360 degrees

in x min it moves 6x degrees. 

 

now 60 + x/2 = 6x - 180

 

11x/2 = 240

x= 480/11 = 43.636....

it matches with

mapqrs  ans

 

^{[ 7]}7⁷.7⁷.7⁷.7⁷.7⁷.7⁷.7⁷ = 7⁷ is it not the ans. i did not see the ( )

sec = x + 1/4x

          

             4x²+1                       4x² - 1                  1 - 4x²

sec = --------------  => tan  = -----------   (or)    -------------

               4x                             4x                     4x

 

 tan  = x - 1/4x (or) 1/4x -x

 

so sec + tan   = 2x (or) 1/2x

 

so tan x = tan( (sec + tan  )/2) or  tan( (1/2(sec + tan  ))

 

actually there are two sol. but I just posted 1 before 

 

tan((sec+tan)/2)

 

ob[]={ {1,7},{15,25},{35,45},{55,65}};

point *a;

a=ob;   // now a will contain the adress of ob[0].

a++;   // now a will contain the adress of ob[1]. 

a->x++; // now ob[1]->x is incrimented ie.15 becomes 16

a++;   // now a will contain the adress of ob[2]. ----------(1)

a->y++; // now ob[2]->y is incrimented ie.45 becomes 46

cout<<ob->x<<" "<<a-y<<endl;//ob contains adress of ob[0] so ob->x is 1

and a->y is 46 from (1)

cout<<ob->y<<" "<<a->y<<endl; //now ob->y is 7 and a->y is 46 from (1)

  so it is

 

 1 46

3) 1 46

    7 46

I hope to find some better questions. hear I go again  a[x++] is like a[x] but later x is incrimented .in a[x]++ the a[x] is incrimented.

9 8

5 4

 

then some crap will be printed because the pointer goes beyond the scope  of the array.

ofcoooourse they are different in the first case all the fees are added and if it is greater than 1500 then the avg of all the fees is printed . in the second case the fees of each grade is added and if the sum is > 1500 then the avg. of feesof that grade is printed 

SCIENTIST135  I think one big assumption your are making is that the foot of the perpendicular meets the line at the mid point then the intersecpts made should be equal but they are not

 

is | 25/cos100 |=| 25/sin100 |???