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  • 2 answers

Piyush Kumar . 8 months, 3 weeks ago

Let x=cost then t= cos^-1x and by putting x=cost we get sin^-1(2sint•cost)=sin-1(sint2t)=2t and t=cos^-1x so 2cos^-1x hence proved

M.V.Priya Shree Priya Shree 9 months, 2 weeks ago

Y=cos2x sin3x.cos4x sin5x,find dy/dx.
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Aniket Agarwal 9 months, 3 weeks ago

tan x^5
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Marvellous Kyndiah 10 months ago

If the matrix [0 -1 3x 1 y -5 -6 5 0] is a skew symmetric matrix, find 6x+y?
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Piyush Kumar . 8 months, 3 weeks ago

|A|^n-1

Tania Jasrotia 9 months, 3 weeks ago

adj A ÷|A|

Umnesh Kumar 10 months ago

What is the A^-1
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Parveen Kumar 10 months, 3 weeks ago

One one onto Let X1,X2 €R s.t. F(X1)=f(X2) X1^3=X2^3 X1=X2 Hence f is one one Let x^3=y x=y^1/3 For each value of y there exist x € R Hence function is onto.
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Arun Singh 11 months ago

Yes, it is an equivalence relation . Because For Reflexive a=b. So (a,a)€R And for symmetric (a,b)€R, a=b Then (b,a) €R For transitive (a,b)€R a=b -1 (b,c)€R b=c -2 From eq 1 and 2 a=c Then (a,c)€R So it is an equivalence relation
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Ankit Singh 6 months, 3 weeks ago

Sum batana hai n

..... ..... 11 months, 1 week ago

Karna kya hai Esme ??

..... ..... 11 months, 1 week ago

Poora question sahi se bhejiye ???
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Twinklesharma Twinkle 7 months, 3 weeks ago

Yes ...

Shanika Sharma 11 months, 1 week ago

Determinant of skew symmetric matrix is always zero. For Example = Let's take a general skew symmetric matrix |0 b c| |-b 0 e| |-c - e. 0| 0(a+e²)-b(0-b)-(-ec)+c(be-0) =0-bec+cbe=0 Therefore determinant of skew symmetric matrix is always zero.

..... ..... 11 months, 1 week ago

No.
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Shanika Sharma 11 months, 1 week ago

Given A={1,2,3_____,13,14} R={(x, y) :3x-y=0} R={(1,3),(2,6),(3,9),(4,12)} 1)Since 1€A but (1,1)does not belong to R Therefore, R is not reflexive 2)since (1,3)€R but (3,1) does not belong to R Therefore R is not symmetric 3) since (1,3)&(3,9)€R but (1,9) does not belong to R Therefore R is not transitive

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