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Question Number 156905 Answers: 2 Comments: 0

find the value of x and y, x:3:5=8:y:9

$f i n d t h e v a l u e o f x a n d y, x : 3 : 5 = 8 : y : 9$

Question Number 156904 Answers: 1 Comments: 0

find the value of x and y , x:3:5=2:y:10

$f i n d t h e v a l u e o f x a n d y, x : 3 : 5 = 2 : y : 10$

Question Number 156900 Answers: 0 Comments: 0

Ω_1 = 1 - (π/2) +Σ_(n=2) ^∞ (- (1/𝛑))^n ∙ (1/(n+1)) Ω_2 = 1 - (π/2) + Σ_(n=2) ^∞ (- (1/e))^n ∙ (1/(n+1)) A) Ω_1 < Ω_2 B) Ω_1 = Ω_2 C) Ω_1 > Ω_2

$Ω_{1} = 1 - \frac{π}{2} + \underset{n = 2}{\sum^{\infty}} {(- \frac{1}{π})}^{n} \cdot \frac{1}{n + 1}$ $Ω_{2} = 1 - \frac{π}{2} + \underset{n = 2}{\sum^{\infty}} {(- \frac{1}{e})}^{n} \cdot \frac{1}{n + 1}$ $A) Ω_{1} < Ω_{2} B) Ω_{1} = Ω_{2} C) Ω_{1} > Ω_{2}$

Question Number 156895 Answers: 0 Comments: 0

(dy/dx)−(x/y)+x^3 cos y = 0

$\frac{d y}{d x} - \frac{x}{y} + x^{3} \cos y = 0$

Question Number 156915 Answers: 1 Comments: 0

Question Number 156914 Answers: 2 Comments: 0

Question Number 156891 Answers: 1 Comments: 0

Question Number 156887 Answers: 1 Comments: 0

tan 2x tan 3x tan 5x =1

$\tan 2 x \tan 3 x \tan 5 x = 1$

Question Number 156869 Answers: 2 Comments: 0

Question Number 156867 Answers: 1 Comments: 0

Question Number 156864 Answers: 0 Comments: 4

φ := ∫_0 ^( 1) (( ln (1−x^( 2) ))/(1+ x^( 2) )) dx = proof : φ = ∫_0 ^( 1) (( ln(1−x ))/(1+x^( 2) ))dx + (π/8)ln(2) .... I= ∫_0 ^( 1) ((ln ( 1−x ))/(1+x^( 2) ))dx =^(x=tan(t)) ∫_0 ^( (π/4)) ln( cos(t)−sin(t))dt−∫_0 ^( (π/4)) ln(cos(t))dt = ∫_0 ^( (π/4)) ln((√2) )dt +∫_0 ^( (π/4)) ln(sin((π/4) −t))dt−(G/2) +(π/4)ln(2) =((3π)/8) ln(2)−(G/2) −(G/2) −(π/4) ln(2)=(π/8)ln(2)−G φ = (π/4)ln(2) − G ■ m.n

$ϕ := \int_{0}^{1} \frac{l n (1 - x^{2})}{1 + x^{2}} d x =$ $p r o o f :$ $ϕ = \int_{0}^{1} \frac{l n (1 - x)}{1 + x^{2}} d x + \frac{π}{8} l n (2)$ $. . . . I = \int_{0}^{1} \frac{l n (1 - x)}{1 + x^{2}} d x$ $\overset{x = t a n (t)}{=} \int_{0}^{\frac{π}{4}} l n (c o s (t) - s i n (t)) d t - \int_{0}^{\frac{π}{4}} l n (c o s (t)) d t$ $= \int_{0}^{\frac{π}{4}} l n (\sqrt{2}) d t + \int_{0}^{\frac{π}{4}} l n (s i n (\frac{π}{4} - t)) d t - \frac{G}{2} + \frac{π}{4} l n (2)$ $= \frac{3 π}{8} l n (2) - \frac{G}{2} - \frac{G}{2} - \frac{π}{4} l n (2) = \frac{π}{8} l n (2) - G$ $ϕ = \frac{π}{4} l n (2) - G ◼ m . n$