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(x2+x/(2x+1)1/2) dx


3 years ago

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Answers : (1)

                                        

Dear,


 please find the attachment below


 


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jitender


980_36248_indefinite integral.JPG

3 years ago

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if alpha ,beeta are the roots of x square+px-q=0 and gama,delta are roots of x square+px+r=0 , then the value of (alpha-gama)(alpha-delta) is (a)- p+q (b)- q-r (c)- r-q (d)- q+r
 
 
Hint: alpha + beta = -p, alpha*beta = -q. gamma + delta = -p and gamma*delta = r. Simply the given equation and replace the values with the above given values. Thanks.
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Vijay Mukti 7 days ago
image is attached. Find eqn of normal to the curve at x=0
 
 
dy/dx= (1+x) y * y/(1+x) + sin2x/(1- sin 4 x) at x=0 dy/dx= 1 slope of normal =-1 y=-x
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Sher Mohammad one year ago
 
dy/dx= (1+x) y * y/(1+x) + sin2x/(1- sin 4 x) at x=0 dy/dx= 1 slope of normal =-1 y=-x
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Sher Mohammad one year ago
 
Hello student, Please find answer to your question
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Jitender Singh one year ago
If both roots of the quadratic equation x 2 + 2 (k-1)x + (k + 5) = 0 are positive then possible integral values of ‘k’ are ​a) -5 b) -4 c) -3 d) -1 (Multiple answers correct) Please explain.
 
 
sum of the roots equals 2 (1-k) and this should be grater than 0 secondly prodect of roots k + 5 should be greater than 0 hence there is only one value -5 which could not be substituted...
 
grenade 2 months ago
 
the answer is B C D i will give the prood later
 
grenade 2 months ago
 
the question is based over location of roots
 
grenade 2 months ago
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