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Question Number 137635 Answers: 0 Comments: 1

Question Number 137634 Answers: 1 Comments: 0

x=2^(p ) and y=2^(q ) . Evaluate in terms of x and/ or y (i)2^(p+q) (ii) 2^(2q ) (iii) 2^(p−1)

$x = 2^{p} a n d y = 2^{q} . E v a l u a t e i n t e r m s o f x a n d / o r y$ $(i) 2^{p + q} (i i) 2^{2 q} (i i i) 2^{p - 1}$

Question Number 137627 Answers: 1 Comments: 0

Question Number 137626 Answers: 0 Comments: 0

Question Number 137625 Answers: 0 Comments: 0

Question Number 137624 Answers: 0 Comments: 0

(dy/dx) = ((6xy+y^2 )/(x+y))

$\frac{dy}{dx} = \frac{6 xy + y^{2}}{x + y}$

Question Number 137618 Answers: 1 Comments: 2

Question Number 137615 Answers: 0 Comments: 1

∫_0 ^( (π/2)) ln(((1+tanx)/(1−tanx)))^(xcos(8x)) dx

$\int_{0}^{\frac{π}{2}} l n {(\frac{1 + t a n x}{1 - t a n x})}^{x c o s (8 x)} d x$

Question Number 137613 Answers: 1 Comments: 0

Two particles of mass m and 4m are connected by a light inelastic string of length 3a, which passes through a small smooth fixed ring. The heavier particle hangs at rest at a distance 2a beneath the ring, while the lighter particle describes a horizontal circle with constant speed. Find (a) The distance of the plane of the circle below the ring. (b) The angular speed of the lighter particle.

$Two particles of mass m and 4 m are connected by a light inelastic$ $string of length 3 a, which passes through a small smooth fixed ring .$ $The heavier particle hangs at rest at a distance 2 a beneath the ring,$ $while the lighter particle describes a horizontal circle with constant$ $speed .$ $Find$ $(a) The distance of the plane of the circle below the ring .$ $(b) The angular speed of the lighter particle .$

Question Number 137611 Answers: 1 Comments: 0

Question Number 137610 Answers: 0 Comments: 0

........ mathematical analysis (II).... prove that :: 𝛀=∫_0 ^( 1) (1/(1+x))ln(((x^2 +2x+1)/(1+x+x^2 )))=Σ_(n=1) ^∞ (1/(n^2 (((2n)),(( n)) )))=(π^2 /(18))..

$. . . . . . . . m a t h e m a t i c a l a n a l y s i s (I I) . . . .$ $p r o v e t h a t ::$ $Ω = \int_{0}^{1} \frac{1}{1 + x} l n (\frac{x^{2} + 2 x + 1}{1 + x + x^{2}}) = \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{2} (\begin{matrix} 2 n \\ n \end{matrix})} = \frac{π^{2}}{18} . .$

Question Number 137605 Answers: 0 Comments: 0

.....nice calculus... prove that:: Σ_(n=0) ^∞ tan^(−1) ((1/F_n )).tan^(−1) ((1/F_(n+1) ))=(π^2 /4) F_n is fibonacci sequence....

$. . . . . n i c e c a l c u l u s . . .$ $p r o v e t h a t ::$ $\underset{n = 0}{\sum^{\infty}} {t a n}^{- 1} (\frac{1}{F_{n}}) . {t a n}^{- 1} (\frac{1}{F_{n + 1}}) = \frac{π^{2}}{4}$ $F_{n} i s f i b o n a c c i s e q u e n c e . . . .$

Question Number 137598 Answers: 0 Comments: 0

Question Number 137597 Answers: 2 Comments: 0

Find the minimum value of x^(2) +y^(2) +z^(2) , subject to the condition 2x+3y+5z=30?

Question Number 137594 Answers: 1 Comments: 0

(x+(√(x^2 +1)))(y+(√(y^4 +4)))=9 x(√(y^4 +4))+y(√(x^2 +1))=?

$(x + \sqrt{x^{2} + 1}) (y + \sqrt{y^{4} + 4}) = 9$ $x \sqrt{y^{4} + 4} + y \sqrt{x^{2} + 1} = ?$

Question Number 137592 Answers: 3 Comments: 0

......advanced.....calculus.... 𝛀=Σ_(n=1) ^∞ ((ψ′′(n))/n)=??? I havefound :: Ω=−(π^4 /(36)) ... !

$. . . . . . a d v a n c e d . . . . . c a l c u l u s . . . .$ $Ω = \underset{n = 1}{\sum^{\infty}} \frac{ψ^{″} (n)}{n} = ? ? ?$ $I h a v e f o u n d :: Ω = - \frac{π^{4}}{36} . . .!$

Question Number 137591 Answers: 0 Comments: 1

a=(4)^(1/3) +(2)^(1/3) +(1)^(1/3) (3/a)+(3/a^2 )+(1/a^3 )=?

$a = \sqrt[3]{4} + \sqrt[3]{2} + \sqrt[3]{1}$ $\frac{3}{a} + \frac{3}{a^{2}} + \frac{1}{a^{3}} = ?$

Question Number 137590 Answers: 0 Comments: 1

x=1+((π+(π+1)^2 +(π+2)^3 +(π+3)^4 )/(4+5^2 +6^3 +7^4 )) (√(x+2(√(x−1))))+(√(x−2(√(x−1))))=?

$x = 1 + \frac{π + {(π + 1)}^{2} + {(π + 2)}^{3} + {(π + 3)}^{4}}{4 + 5^{2} + 6^{3} + 7^{4}}$ $\sqrt{x + 2 \sqrt{x - 1}} + \sqrt{x - 2 \sqrt{x - 1}} = ?$

Question Number 137588 Answers: 1 Comments: 0

For a positive number n , let f(n) be the value of f(n)=((4n+(√(4n^2 −1)))/( (√(2n+1)) +(√(2n−1)))) calculate f(1)+f(2)+f(3)+...+f(40).

$F o r a p o s i t i v e n u m b e r n, l e t$ $f (n) b e t h e v a l u e o f$ $f (n) = \frac{4 n + \sqrt{4 n^{2} - 1}}{\sqrt{2 n + 1} + \sqrt{2 n - 1}}$ $c a l c u l a t e f (1) + f (2) + f (3) + . . . + f (40) .$

Question Number 137585 Answers: 2 Comments: 0

Find the cube of the number N= (√(7(√(3(√(7(√(3(√(7(√(3(√(7(√(3...))))))))))))))))

$F i n d t h e c u b e o f t h e n u m b e r$ $N = \sqrt{7 \sqrt{3 \sqrt{7 \sqrt{3 \sqrt{7 \sqrt{3 \sqrt{7 \sqrt{3 . . .}}}}}}}}$

Question Number 137584 Answers: 1 Comments: 0

Given { ((a_(2n) = a_n .a_2 +1)),((a_(2n+1) = a_n .a_2 −2 )) :} and { ((a_7 = 2)),((0<a_1 <1)) :}. Find a_(25) =?

$G i v e n {\begin{cases} a_{2 n} = a_{n} . a_{2} + 1 \\ a_{2 n + 1} = a_{n} . a_{2} - 2 \end{cases} a n d$ ${\begin{cases} a_{7} = 2 \\ 0 < a_{1} < 1 \end{cases} . F i n d a_{25} = ?$

Question Number 137582 Answers: 1 Comments: 0

A=(((1/3)((2)^(1/3) −1)((2)^(1/3) +1)^3 ))^(1/3)

$A = \sqrt[3]{\frac{1}{3} (\sqrt[3]{2} - 1) {(\sqrt[3]{2} + 1)}^{3}}$

Question Number 137579 Answers: 1 Comments: 0

(−1)×(1/(π.i)) =?

$(- 1) \times \frac{1}{π . i} = ?$

Question Number 137575 Answers: 0 Comments: 0

Question Number 137574 Answers: 0 Comments: 0

Question Number 137568 Answers: 0 Comments: 2

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