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Question Number 38857 Answers: 0 Comments: 0

Question Number 38856 Answers: 0 Comments: 0

Question Number 38855 Answers: 0 Comments: 0

Question Number 38854 Answers: 0 Comments: 0

Question Number 38853 Answers: 0 Comments: 1

Question Number 38847 Answers: 0 Comments: 0

Question Number 38836 Answers: 0 Comments: 0

Question Number 38734 Answers: 0 Comments: 1

Find the domain of the function f (x) = 1 − cos^2 x

$F i n d t h e d o m a i n o f t h e f u n c t i o n$ $f (x) = 1 - {c o s}^{2} x$

Question Number 38562 Answers: 1 Comments: 0

((√2) +i)(1−(√(2i)) )

$(\sqrt{2} + i) (1 - \sqrt{2 i})$

Question Number 38559 Answers: 1 Comments: 0

in a geometric series, the first term =a, common ratio=r. If S_n denotes the sum of the n terms and U_n =Σ_(n=1) ^n S_(n,) then rS_n +(1−r)U_(n ) equals to (a) 0 (b) n (c) na (d)nar

$i n a g e o m e t r i c s e r i e s, t h e f i r s t t e r m$ $= a, c o m m o n r a t i o = r . I f S_{n} d e n o t e s$ $t h e s u m o f t h e n t e r m s a n d U_{n} = \underset{n = 1}{\sum^{n}} S_{n,}$ $t h e n {r S}_{n} + (1 - r) U_{n} e q u a l s t o$ $(a) 0 (b) n (c) n a (d) n a r$

Question Number 38535 Answers: 0 Comments: 0

Given the function f(x) where f(x)= { ((∫x^2 + 1 ,for {x:x D(f) 2)),((∫x^3 − 1,for y = f′(x))) :} a) Evaluate f(2) if f(a)= 2 + a^(n−1) find the value of a hence the domain of f(x).

$G i v e n t h e f u n c t i o n$ $f (x) w h e r e$ $f (x) = {\begin{cases} \int x^{2} + 1, f o r {x : x D (f) 2 \\ \int x^{3} - 1, f o r y = f^{'} (x) \end{cases}$ $a) E v a l u a t e f (2)$ $i f f (a) = 2 + a^{n - 1}$ $f i n d t h e v a l u e o f a$ $h e n c e t h e d o m a i n o f f (x) .$

Question Number 38534 Answers: 0 Comments: 0

∫∫_R (2x + 3y)^2 dA=??

$\int \int_{R} {(2 x + 3 y)}^{2} d A = ? ?$

Question Number 38517 Answers: 2 Comments: 0

simlify A= (1/((2−(√5))^4 )) + (1/((2+(√5))^4 )) B = (1/((3−(√2))^6 )) +(1/((3+(√2))^6 ))

$s i m l i f y$ $A = \frac{1}{{(2 - \sqrt{5})}^{4}} + \frac{1}{{(2 + \sqrt{5})}^{4}}$ $B = \frac{1}{{(3 - \sqrt{2})}^{6}} + \frac{1}{{(3 + \sqrt{2})}^{6}}$

Question Number 38515 Answers: 1 Comments: 0

Question ; x^3 + x^3 = A) x^9 B) x^6 C) x^3 D) 1 Give a reason for your answer.

$Q u e s t i o n;$ $x^{3} + x^{3} =$ $A) x^{9}$ $B) x^{6}$ $C) x^{3}$ $D) 1$ $G i v e a r e a s o n f o r y o u r a n s w e r .$

Question Number 38495 Answers: 4 Comments: 0

prove that tan 3a tan 2a tan a = tan 3a − tan 2a − tan a

$p r o v e t h a t$ $\tan 3 a \tan 2 a \tan a = \tan 3 a - \tan 2 a - \tan a$

Question Number 38488 Answers: 2 Comments: 0

find the value of x if 3^x = 9x

$find the value of x if$ $3^{x} = 9 x$

Question Number 39416 Answers: 1 Comments: 0

Given that f(x) is a cubic function and f(x) = x^3 − (x^2 /4) + 5x − 7 a) find one factor of f(x) b) find (d^2 y/dx^2 ) for f(x) c) hence Evaluate y = ∫_0 ^∞ f(x).

$G i v e n t h a t f (x) i s a c u b i c f u n c t i o n$ $a n d f (x) = x^{3} - \frac{x^{2}}{4} + 5 x - 7$ $a) f i n d o n e f a c t o r o f f (x)$ $b) f i n d \frac{d^{2} y}{{d x}^{2}} f o r f (x)$ $c) h e n c e E v a l u a t e y = \int_{0}^{\infty} f (x) .$

Question Number 38391 Answers: 1 Comments: 0

a+2b+3c=12 2ab+3ac+6bc=48 a+b+c=...

$a + 2 b + 3 c = 12$ $2 a b + 3 a c + 6 b c = 48$ $a + b + c = . . .$

Question Number 38420 Answers: 0 Comments: 0

In the figure below,a particle A of mass 2kg is lying on a rough wooden block.The particle A is connected by a light inextensible horizontal string passing over a smooth light fixed pulley at the edge of the block,to a particle B of mass 3kg which hangs freely. The coefficent of friction between the particle A and the surface of the block is μ. Given that the string is taut and the system is released from rest such that the particle move with an acceleration of 4ms^(−1) . Find a) the tension b) the value of μ.

$I n t h e f i g u r e b e l o w, a p a r t i c l e A o f$ $m a s s 2 k g i s l y i n g o n a r o u g h w o o d e n$ $b l o c k . T h e p a r t i c l e A i s c o n n e c t e d b y$ $a l i g h t i n e x t e n s i b l e h o r i z o n t a l s t r i n g$ $p a s s i n g o v e r a s m o o t h l i g h t f i x e d$ $p u l l e y a t t h e e d g e o f t h e b l o c k, t o a$ $p a r t i c l e B o f m a s s 3 k g w h i c h h a n g s$ $f r e e l y . T h e c o e f f i c e n t o f f r i c t i o n$ $b e t w e e n t h e p a r t i c l e A a n d t h e s u r f a c e$ $o f t h e b l o c k i s μ .$ $G i v e n t h a t t h e s t r i n g i s t a u t a n d t h e s y s t e m i s r e l e a s e d f r o m r e s t s u c h t h a t t h e p a r t i c l e m o v e w i t h a n$ $a c c e l e r a t i o n o f 4 {m s}^{- 1} . F i n d$ $a) t h e t e n s i o n$ $b) t h e v a l u e o f μ .$

Question Number 38419 Answers: 1 Comments: 0

Question Number 38373 Answers: 1 Comments: 7

Question Number 38366 Answers: 1 Comments: 0

At time t,the force acting on a particle P of mass 2kg is (2ti + 4j)N.P is initially at rest at the point with position vector (i + 2j). Find: a) the velocity of P when t = 2. b) the position vector when t = 2.

$A t t i m e t, t h e f o r c e a c t i n g o n a p a r t i c l e$ $P o f m a s s 2 k g i s (2 t i + 4 j) N . P$ $i s i n i t i a l l y a t r e s t a t t h e p o i n t w i t h$ $p o s i t i o n v e c t o r (i + 2 j) .$ $F i n d :$ $a) t h e v e l o c i t y o f P w h e n t = 2 .$ $b) t h e p o s i t i o n v e c t o r w h e n t = 2 .$

Question Number 38362 Answers: 0 Comments: 0

Question Number 38365 Answers: 1 Comments: 0

A particle P moves on a straightline from a fixed point O and the distance x from O after t seconds is given as x = (1/(4 )) t^4 − (3/2) t^2 + 2t. Find: a) the velocity of P when t = 2, b) the acceleration of P when t = 2, c) the time at which the speed P is Minimum.

$A p a r t i c l e P m o v e s o n a s t r a i g h t l i n e$ $f r o m a f i x e d p o i n t O a n d t h e d i s t a n c e$ $x f r o m O a f t e r t s e c o n d s i s g i v e n a s$ $x = \frac{1}{4} t^{4} - \frac{3}{2} t^{2} + 2 t .$ $F i n d :$ $a) t h e v e l o c i t y o f P w h e n t = 2,$ $b) t h e a c c e l e r a t i o n o f P w h e n t = 2,$ $c) t h e t i m e a t w h i c h t h e s p e e d P$ $i s M i n i m u m .$

Question Number 38296 Answers: 0 Comments: 7

i have a suggestion pls comment...we are all virtual friends common bond is mathematics so may know each other by posting our self photo...if administator give permission..

$i h a v e a s u g g e s t i o n p l s c o m m e n t . . . w e a r e a l l$ $v i r t u a l f r i e n d s c o m m o n b o n d i s m a t h e m a t i c s$ $s o m a y k n o w e a c h o t h e r b y p o s t i n g o u r s e l f$ $p h o t o . . . i f a d m i n i s t a t o r g i v e p e r m i s s i o n . .$

Question Number 38252 Answers: 0 Comments: 2

if x^2 + 3xy − y^2 = 3 find (dy/dx) at point (1,1) hence differentiate ((sin x)/(1 + x)) with respect to x.

$i f x^{2} + 3 x y - y^{2} = 3 f i n d$ $\frac{d y}{d x} a t p o i n t (1, 1) h e n c e$ $d i f f e r e n t i a t e \frac{s i n x}{1 + x} w i t h r e s p e c t$ $t o x .$

Pg 105 Pg 106 Pg 107 Pg 108 Pg 109 Pg 110 Pg 111 Pg 112 Pg 113 Pg 114

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