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Is answer is around 20. Something .....If yes then u may apply my methodU can do it by putting the value of x in the given eq.... I have tried it and it workU can also do it like... first find out...
There are total 7 relatives in both of mens as well as wives....according to questions we have to select 3 relatives from both of them so 7c3 +7c3 so you will get answer 70....-Shubham Pitroda
No...VIKRAM ENIMIREDDY.....tn and an is different form....tn is nth term and by an we can find the first term and second term.......Solution:first put 1 and 2 in an simultaneously where a1 is first...
You know both the tangents...Their point of intersection is (5,0)...You can use the formula of pair of tangents from a point drwn from (x`, y`)....that is SS` - T² = 0......S= x²/a² + y²/b² - 1 =...
Tangent at any point to the ellipse (acos ß, bsin ß) is xcosß / a + ysinß / b = 1....It mmets yhe axes in points (a secß,0) and (0,b coseß).....Lets the mid point be (h,k) then h= a secß / 2 and k = b...
Hello Ujjawal, you can refer to Arihant Publications books.....Skills in Mathematics Algebra by Dr SK Goyal... Its is a very good book complete in all topics of algebra including binomial theorem,...
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The hyperbola can be written as:x²/ (144/25) + y²/ (81/25) = 1Now taking out e by using b²=a²( e² - 1 ) where a²= 144/25 and b²= 81/25....e comes out to be 5/4......focii of hyperbola are ( -+ 3, 0...
let the roots of x^2-x+3a=0 be 2p and q respectively and the roots for x^2-x+a=0 be p and r respectively. then, 2p + q = 1 p + r = 1 2pq = a pr = 3a solving all the above eqns. we get, p = 5/11 q =...
Hello Aditi, your inequality is incomplete as after4^x - a2^x - a+3 nothing is displayed on the page... There is a problem in the page when you use signs like grater than or less than because anything...
Hello Suryanshu, i think the language of your question is a bit unclear..According to the question posted by you a point on the given parabola will pass through itself only...which will give the same...
Make graph for the expression|log|x|| = 2 - |x|....Plot log x then make its symmetry about y axis and then the portion of graph where y is negative, take its mod so that the whole graph of of |log|x||...
Using the quadratic formula:Roots =[ ( 1- a ) +- √ { ( 1- a )² + 4( a + 2 ) } ] / 2For the roots to be integral firstly discriminat should be a perfect square and then overall numerator should be...
12 is the remainder.When 12 gets divided by 7 remainder is 5 and when 12 gets divided by 9 remainder is 3 ;as it gave in question. So,when 12 gets divided by 63 remainder is 12 . This can also done by...
Answer options provided by you are not visible...hope the simplification of the expression may help you get your answer..
Hello vaibhav your problem is not visible fully and therefore i am unable to help you with the solution. I request you to resend it again...
The web page is not accepting inequalities` sign like greater than or less than.-> X = 4 - 4r ( where -4r belongs to (-4,0) ). For X to be an integer -4r should be an integer possible integral values...
The formula for area of parallelogram is base * corresponding altitude.Given one side 12cm.Corresponding altitude 6cm.Area of parallelogram is 12*6 = 72 cm²As area remains the same 72 cm² = 8 cm * x....
@ kavach Kindly let me know which questions u are talking about ?/ kindly mention the ques beacuse logarithm have many questions of diffrent pattern , so, explaining all of them will be a little...
Firstly, U should have provided the options either or LCM value to get the HCF. because absence of any one of themm cannot solve. However the concept would be: HCF is the factor of LCM therefore it...
Lim h->0 [ f(x+h) - f(x) ] / h = f`(x) Lim h->0 [ f(x) + f(h) - f(x) ] / h = f`(x) As f(x+y) = f(x) + f(y) Lim h->0 f(h) / h = f`(x) Transforming into differential eq. y/x = dy/dx (variable...
Sorry for not having an appealing solution to this. But by inspection one of the solution is a=b=c=0. And other solution can be c=0 which reduces the expressio to a² + 2b² - 3ab = 0 which gives a=1...
LHS=n!/[r!(n-r)!]+n!/[(r-1)!(n-r+1)!]=n!/[r(r-1)!(n-r)!]+...
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