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AllQuestion and Answers: Page 1624

Question Number 40489 Answers: 2 Comments: 3

Question Number 40479 Answers: 1 Comments: 0

Question Number 40474 Answers: 2 Comments: 0

When x^7 −97x^6 −199x^5 +99x^4 − 2x+190 is divided by x−99 find the remainder.

$W h e n x^{7} - 97 x^{6} - 199 x^{5} + 99 x^{4} -$ $2 x + 190 i s d i v i d e d b y x - 99 f i n d$ $t h e r e m a i n d e r .$

Question Number 40469 Answers: 0 Comments: 4

Question Number 40457 Answers: 1 Comments: 0

Question Number 40442 Answers: 2 Comments: 0

Solve : y^4 dx + 2xy^3 dy = ((ydx− xdy)/(x^3 y^3 )).

$Solve :$ $y^{4} d x + 2 x y^{3} dy = \frac{yd x - x dy}{x^{3} y^{3}} .$

Question Number 40434 Answers: 0 Comments: 3

prove that ln(x) is irrational for x natural

$p r o v e t h a t \ln (x) i s i r r a t i o n a l f o r x n a t u r a l$

Question Number 40426 Answers: 0 Comments: 0

Question Number 40422 Answers: 1 Comments: 1

Solve: ydx − xdy +log xdx =0

$Solve :$ $yd x - x d y + \log x d x = 0$

Question Number 40418 Answers: 2 Comments: 1

Question Number 40417 Answers: 0 Comments: 0

Question Number 40407 Answers: 1 Comments: 0

let f(x)= x^3 −x−1 1) prove that ∃ α ∈ ]1,2[ /f(α)=0 2) use the newton method with x_0 =(3/2) to find a better value for α (take onlly 5 terms)

$l e t f (x) = x^{3} - x - 1$ $1) p r o v e t h a t \exists α \in] 1, 2 [/ f (α) = 0$ $2) u s e t h e n e w t o n m e t h o d w i t h x_{0} = \frac{3}{2}$ $t o f i n d a b e t t e r v a l u e f o r α (t a k e o n l l y 5 t e r m s)$

Question Number 40406 Answers: 1 Comments: 0

Question Number 40399 Answers: 3 Comments: 0

Solve : (2(√(xy)) −x)dy + ydx = 0.

$Solve :$ $(2 \sqrt{x y} - x) d y + y d x = 0 .$

Question Number 40397 Answers: 1 Comments: 0

Solve : (dy/dx) = ((sin y + x)/(sin 2y − xcos y)) .

$Solve :$ $\frac{dy}{d x} = \frac{\sin y + x}{\sin 2 y - x \cos y} .$

Question Number 40396 Answers: 0 Comments: 1

A block lying on a horizontal conv− eyor belt moving at a constant velocity receives a velocity 5m/s at t=0 sec. relative to the ground in the direction opposite to the dir− ction of motion of the conveyor. Aftert=4sec,the velocity of the block becomes equal to the velocity of the belt . the coefficient of friction between the block and the belt is 0.2 . then the velocity of the conveyor belt is: (g=10m/s^2 ) (A) 13 m/s (B) −13m/s (C)3m/s (D) 6m/s

$A b l o c k l y i n g o n a h o r i z o n t a l c o n v -$ $e y o r b e l t m o v i n g a t a c o n s t a n t$ $v e l o c i t y r e c e i v e s a v e l o c i t y 5 m / s$ $a t t = 0 s e c . r e l a t i v e t o t h e g r o u n d$ $i n t h e d i r e c t i o n o p p o s i t e t o t h e d i r -$ $c t i o n o f m o t i o n o f t h e c o n v e y o r .$ $A f t e r t = 4 s e c, t h e v e l o c i t y o f t h e$ $b l o c k b e c o m e s e q u a l t o t h e v e l o c i t y$ $o f t h e b e l t . t h e c o e f f i c i e n t o f$ $f r i c t i o n b e t w e e n t h e b l o c k a n d t h e$ $b e l t i s 0.2 . t h e n t h e v e l o c i t y o f t h e$ $c o n v e y o r b e l t i s : (g = 10 m / s^{2})$ $(A) 13 m / s (B) - 13 m / s$ $(C) 3 m / s (D) 6 m / s$