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Question Number 139992 Answers: 0 Comments: 1

Question Number 139455 Answers: 1 Comments: 0

# calculus# evaluate: 𝛗:=Σ_(k=1) ^∞ (((−1)^(k−1) Γ ((k/2)))/(k Γ(((k+1)/2)))) =?

$You can't use 'macro parameter character #' in math mode$ $e v a l u a t e :$ $ϕ := \underset{k = 1}{\sum^{\infty}} \frac{{(- 1)}^{k - 1} Γ (\frac{k}{2})}{k Γ (\frac{k + 1}{2})} = ?$

Question Number 138976 Answers: 1 Comments: 0

Question Number 138366 Answers: 2 Comments: 0

If x^2 +x^(−2) =(√(2+(√(2+(√2))))) x^(16) +x^(−16) =? Any help

$If x^{2} + x^{- 2} = \sqrt{2 + \sqrt{2 + \sqrt{2}}}$ $x^{16} + x^{- 16} = ?$ $Any help$

Question Number 137970 Answers: 1 Comments: 0

Prove that ▽^2 𝛗=−4𝛑Gρ φ=Potential of Gravitational field ρ=Density G=Universal Gravitational Constant

$P r o v e t h a t$ $▽^{2} ϕ = - 4 π G ρ$ $ϕ = P o t e n t i a l o f G r a v i t a t i o n a l f i e l d$ $ρ = D e n s i t y G = U n i v e r s a l G r a v i t a t i o n a l C o n s t a n t$

Question Number 136300 Answers: 3 Comments: 0

Let vector a^→ , b^→ and c^→ such that ∣a^→ ∣=∣b^→ ∣=((∣c^→ ∣)/2) and a^→ ×(a^→ ×c^→ )+b^→ =0 find the acute angle between a^→ and c^→ .

$L e t v e c t o r \vec{a}, \vec{b} a n d \vec{c} s u c h t h a t$ $∣ \vec{a} ∣=∣ \vec{b} ∣= \frac{∣ \vec{c} ∣}{2} a n d \vec{a} \times (\vec{a} \times \vec{c}) + \vec{b} = 0$ $f i n d t h e a c u t e a n g l e b e t w e e n \vec{a} a n d \vec{c} .$

Question Number 135965 Answers: 0 Comments: 2

Vector Three vectors satisfy a.b = b.c = c.a = -1 and a + b + c = 0. What is the magnitude of vector a, b , and c?

$V e c t o r$ Three vectors satisfy a.b = b.c = c.a = -1 and a + b + c = 0. What is the magnitude of vector a, b , and c?

Question Number 135821 Answers: 4 Comments: 0

....nice ..... calculus.... prove that :: 𝛗=∫_0 ^( 1) (((ln(1−x))/(1−(√(1−x)))))dx=4(1−ζ(2))

$. . . . n i c e . . . . . c a l c u l u s . . . .$ $p r o v e t h a t ::$ $ϕ = \int_{0}^{1} (\frac{l n (1 - x)}{1 - \sqrt{1 - x}}) d x = 4 (1 - ζ (2))$

Question Number 135790 Answers: 0 Comments: 0

Find the component form of the vector that reprecents the velocity of an airplane descending at speed of 150 miles per hour at angle 20° below the horizontal

$F i n d t h e c o m p o n e n t f o r m o f$ $t h e v e c t o r t h a t r e p r e c e n t s t h e$ $v e l o c i t y o f a n a i r p l a n e d e s c e n d i n g$ $a t s p e e d o f 150 m i l e s p e r h o u r$ $a t a n g l e 20 ° b e l o w t h e h o r i z o n t a l$

Question Number 135423 Answers: 1 Comments: 0

If a^→ =(4,2,−1), b^→ =(m,1,1) c^→ =(3^ −1,0) are three vectors then find the value of m such that a^→ ,b^→ and c^→ are coplanar and find a^→ ×(b^→ ×c^→ ).

$I f \vec{a} = (4, 2, - 1), \vec{b} = (m, 1, 1)$ $\vec{c} = (\bar{3} - 1, 0) a r e t h r e e v e c t o r s$ $t h e n f i n d t h e v a l u e o f m s u c h$ $t h a t \vec{a}, \vec{b} a n d \vec{c} a r e c o p l a n a r a n d$ $f i n d \vec{a} \times (\vec{b} \times \vec{c}) .$

Question Number 133925 Answers: 2 Comments: 0

Given vector a^→ = i^ −2j^ +k^ , b^→ = 2i^ +j^ −2k^ , c^→ =−i^ +3j^ −k^ and d^→ = 2j^ −2k^ . Find the value of (a^→ ×b^→ )×(c^→ ×d^→ ).

$Given vector \vec{a} = \hat{i} - 2 \hat{j} + \hat{k},$ $\vec{b} = 2 \hat{i} + \hat{j} - 2 \hat{k}, \vec{c} = - \hat{i} + 3 \hat{j} - \hat{k}$ $and \vec{d} = 2 \hat{j} - 2 \hat{k} . Find the value of$ $(\vec{a} \times \vec{b}) \times (\vec{c} \times \vec{d}) .$

Question Number 133857 Answers: 2 Comments: 0

.....#advanced ............... calculus#..... prove that ::: 𝛗=∫_0 ^( 1) ((ln^2 (1−x))/x)dx=^? 2ζ(3) =^(1−x=t) ∫_0 ^( 1) ((ln^2 (t))/(1−t))dt=∫_0 ^( 1) Σ_(n=0) ^∞ ln^2 (t).t^n dt =Σ_(n=0) ^∞ {[(t^(n+1) /(n+1))ln^2 (t)]_0 ^1 −(2/(n+1))∫_0 ^( 1) t^n ln(t) =−2Σ_(n=0) ^∞ (1/(n+1)){[(t^(n+1) /(n+1))ln(t)]_0 ^1 −(1/(n+1))∫_0 ^( 1) t^n dt} =2Σ_(n=0) ^∞ (1/((n+1)^3 ))=2Σ_(n=1) ^∞ (1/n^3 )=2ζ(3) .................... 𝛗=2ζ(3) .................... ..........m.n.july.1970.........

$You can't use 'macro parameter character #' in math mode$ $p r o v e t h a t ::: ϕ = \int_{0}^{1} \frac{{l n}^{2} (1 - x)}{x} d x \overset{?}{=} 2 ζ (3)$ $\overset{1 - x = t}{=} \int_{0}^{1} \frac{{l n}^{2} (t)}{1 - t} d t = \int_{0}^{1} \underset{n = 0}{\sum^{\infty}} {l n}^{2} (t) . t^{n} d t$ $= \underset{n = 0}{\sum^{\infty}} {{[\frac{t^{n + 1}}{n + 1} {l n}^{2} (t)]}_{0}^{1} - \frac{2}{n + 1} \int_{0}^{1} t^{n} l n (t)$ $= - 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{n + 1} {{[\frac{t^{n + 1}}{n + 1} l n (t)]}_{0}^{1} - \frac{1}{n + 1} \int_{0}^{1} t^{n} d t}$ $= 2 \underset{n = 0}{\sum^{\infty}} \frac{1}{{(n + 1)}^{3}} = 2 \underset{n = 1}{\sum^{\infty}} \frac{1}{n^{3}} = 2 ζ (3)$ $. . . . . . . . . . . . . . . . . . . . ϕ = 2 ζ (3) . . . . . . . . . . . . . . . . . . . .$ $. . . . . . . . . . m . n . j u l y . 1970 . . . . . . . . .$