loving questions of the form “if ... then find the sum/product/etc. of... so please solve these: (1) if γ and λ are the solutions of x^2 +x−12=0 then find coshλ−cotγ (2) if a+b=2 and a−b=0 then find ∫x^((a+b)/(2ab)) ln(−e^(iπa) −x)ln(e^(cos^(−1) b) −x)dx are you intelligent enough? then please please please sir or madam help me!!! I need an answer urgentliest!!!! good luck! (c) by HeR MaJε∫tY 20200913


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Question Number 113393 by Her_Majesty last updated on 13/Sep/20

$l o v i n g q u e s t i o n s o f t h e f o r m$ $‘ ‘ i f . . . t h e n f i n d t h e s u m / p r o d u c t / e t c . o f . . .$ $s o p l e a s e s o l v e t h e s e :$ $(1)$ $i f γ a n d λ a r e t h e s o l u t i o n s o f$ $x^{2} + x - 12 = 0 t h e n f i n d c o s h λ - c o t γ$ $(2)$ $i f a + b = 2 a n d a - b = 0 t h e n f i n d$ $\int x^{\frac{a + b}{2 a b}} l n (- e^{i π a} - x) l n (e^{{c o s}^{- 1} b} - x) d x$ $a r e y o u i n t e l l i g e n t e n o u g h ?$ $t h e n p l e a s e p l e a s e p l e a s e s i r o r m a d a m$ $h e l p m e!!! I n e e d a n a n s w e r urgentliest!!!!$ $g o o d l u c k!$ $(c) b y H e R M a J ϵ \int t Y 20200913$
Commented by john santu last updated on 13/Sep/20
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Commented by malwaan last updated on 13/Sep/20
$x^2 +x−12=0 ⇒(x−3)(x+4)=0 ∴ solution set = {3;−4} (i)γ=3;λ=−4⇒cosh(−4)−cot(3)=34.32349 (ii)γ=−4;λ=3⇒cosh(3)−cot(−4)=10.93135$
$x^{2} + x - 12 = 0$ $\Rightarrow (x - 3) (x + 4) = 0$ $∴ s o l u t i o n s e t = {3; - 4}$ $(i) γ = 3; λ = - 4 \Rightarrow c o s h (- 4) - c o t (3) = 34.32349$ $(i i) γ = - 4; λ = 3 \Rightarrow c o s h (3) - c o t (- 4) = 10.93135$
Commented by malwaan last updated on 13/Sep/20

$(2) a + b = 2; a - b = 0$ $\Rightarrow a = 1; b = 1$ $∴ \int x l n (1 - x) l n (1 - x) d x =$ $\int x {[l n (1 - x)]}^{2} d x = . . .$
Commented by Rasheed.Sindhi last updated on 13/Sep/20

$P e r h a p s H e r M a j e s t y w a n t e d t o$ $g i v e a m e s s a g e!$
Commented by malwan last updated on 13/Sep/20

$w h a t i s t h e p o i n t p l e a s e ?$ $I a m s u r e y o u {d o n}^{,} t n e e d h e l p$ $b u t w e [a t l e a s t m e] d o$
Commented by Her_Majesty last updated on 13/Sep/20

$o n e p o i n t i s,$ $(y o u p o s t a q u e s t i o n) \Rightarrow (y o u n e e d h e l p)$ $n o n e e d t o p o s t t h e s a m e q u e s t i o n o v e r a n d$ $o v e r a g a i n . l$ $a n o t h e r p o i n t i s, p e o p l e w h o p o s t q u e s t i o n s$ $t o s h o w u s w e a r e u n a b l e t o s o l v e t h e m w h i l e$ $t h e y t h e m s e l v e s a r e a b l e t o, r u i n t h e f o r u m .$ $a 3^{r d} p o i n t i s, {i t}^{'} s u s e l e s s t o s o l v e i . e . a$ $p o l y n o m e a n d a s k f o r i . e . ‘ ‘ 7 x_{1} - 13 x_{2} + 1 / x_{3}^{″}$ $b e c a u s e {t h e r e}^{'} s n o t a k e a w a y f r o m t h i s .$ $i n d e e d t h e r e i s i f y o u a s k f o r ‘ ‘ 1 / x_{1} + 1 / x_{2} +$ $+ 1 / x_{3}^{″} . . . \Rightarrow \exists (s i l l y q u e s t i o n s)$ $4^{t h} . i f y o u p o s t s o m e t h i n g t o d a y e v e r y b o d y$ $c a n s e e i t w a s Y O U a n d T O D A Y . n o n e e d t o$ $s t a t e t h e s e f a c t s a g a i n .$
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