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Question Number 63470 Answers: 0 Comments: 0

Question Number 63466 Answers: 0 Comments: 7

If A=sin^(28) θ+cos^(36) θ then Ans: 0<A≤1

$If A = \sin^{28} θ + \cos^{36} θ then$ $Ans : 0 < A ⩽ 1$

Question Number 63460 Answers: 1 Comments: 2

factorise cosθ−cos3θ−cos5θ +cos7θ

$f a c t o r i s e$ $c o s θ - c o s 3 θ - c o s 5 θ + c o s 7 θ$

Question Number 63473 Answers: 0 Comments: 5

question lim_(x→0) ((sin(x+A)−sin(A−x))/(2x))

$q u e s t i o n$ $\lim_{x \to 0} \frac{s i n (x + A) - s i n (A - x)}{2 x}$

Question Number 63457 Answers: 1 Comments: 1

((3/4))^x ((2/3))^y =((32)/(27)) find the value of x and yy y.

${(\frac{3}{4})}^{x} {(\frac{2}{3})}^{y} = \frac{32}{27} f i n d t h e v a l u e o f x a n d y y$ $y .$

Question Number 63449 Answers: 0 Comments: 2

Do you remember that π≈3.1415926535897932⋱ 38462643383279502884⋱ 197...

$D o y o u r e m e m b e r t h a t$ $π \approx 3.1415926535897932 ⋱$ $38462643383279502884 ⋱$ $197 . . .$

Question Number 63447 Answers: 1 Comments: 2

How to calculate ⌈(n) using gamma function ∀n∈R

$H o w t o c a l c u l a t e ⌈ (n)$ $u s i n g g a m m a f u n c t i o n$ $\forall n \in R$

Question Number 63438 Answers: 1 Comments: 2

Question Number 63433 Answers: 1 Comments: 1

∫_1 ^x x^2 −3x(√x)dx =((−716)/(15)) then calculate ∫_x ^(x+1) (1/(x+3))dx

$\underset{1}{\int^{x}} x^{2} - 3 x \sqrt{x} d x = \frac{- 716}{15}$ $t h e n c a l c u l a t e \underset{x}{\int^{x + 1}} \frac{1}{x + 3} d x$

Question Number 63430 Answers: 1 Comments: 0

Question Number 63428 Answers: 0 Comments: 2

It is given that S_n =Σ_(r=1) ^n (3r^(2 ) −3r−1). Use the the formulae of Σ_(r=1) ^n r^(2 ) and Σ_(r=1) ^n r to show that S_n =n^3 . sir Forkum Michael

$I t i s g i v e n t h a t S_{n} = \underset{r = 1}{\sum^{n}} (3 r^{2} - 3 r - 1) . U s e t h e t h e f o r m u l a e$ $o f \underset{r = 1}{\sum^{n}} r^{2} a n d \underset{r = 1}{\sum^{n}} r t o s h o w t h a t S_{n} = n^{3} .$ $s i r F o r k u m M i c h a e l$

Question Number 63427 Answers: 0 Comments: 4

(1) A plane contains the lines ((x+1)/2)=((4−y)/2)=((z−2)/3) and r= (2i+2j + 12k)+t(−i+2j +4k). find (a) the angle between these lines. (b) A cartesian equation of the plane. (2) Given the lines l_1 :((x−10)/3)=((y−1)/1)=((z−9)/4) l_2 :r=(−9j+13k)+μ(i+2j−3k) where μ is a parameter; l_3 :((x+10)/4)=((y+5)/3)=((z+4)/1). a) show that the point (4,−1,1) is common to l_1 and l_2 . Find b) the point of intersection of l_2 and l_3 . c) A vector parametric equation of the plane containing the lines l_2 and l_3 . sir Forkum Michael

$(1) A p l a n e c o n t a i n s t h e l i n e s \frac{x + 1}{2} = \frac{4 - y}{2} = \frac{z - 2}{3} a n d$ $r = (2 i + 2 j + 12 k) + t (- i + 2 j + 4 k) . f i n d$ $(a) t h e a n g l e b e t w e e n t h e s e l i n e s .$ $(b) A c a r t e s i a n e q u a t i o n o f t h e p l a n e .$ $(2) G i v e n t h e l i n e s l_{1} : \frac{x - 10}{3} = \frac{y - 1}{1} = \frac{z - 9}{4} l_{2} : r = (- 9 j + 13 k) + μ (i + 2 j - 3 k)$ $w h e r e μ i s a p a r a m e t e r; l_{3} : \frac{x + 10}{4} = \frac{y + 5}{3} = \frac{z + 4}{1} .$ $a) s h o w t h a t t h e p o i n t (4, - 1, 1) i s c o m m o n t o l_{1} a n d l_{2} . F i n d$ $b) t h e p o i n t o f i n t e r s e c t i o n o f l_{2} a n d l_{3} .$ $c) A v e c t o r p a r a m e t r i c e q u a t i o n o f t h e p l a n e c o n t a i n i n g t h e$ $l i n e s l_{2} a n d l_{3} .$ $s i r F o r k u m M i c h a e l$

Question Number 63426 Answers: 1 Comments: 5

show that (a) cos[2cos^(−1) (x) +sin^(−1) (x)]= −(√(1−x^2 )) (b) ((sinα + sinβ)/(cosα−cosβ))=cot(((β−α)/2)) (c) 2cos((π/3)+p)≊ 1−(√3) if p is small enough to neglect p^2 . (d) if θ =(1/2)sin^(−1) ((3/4)), show that sinθ−cosθ = ±(1/2) (e)write tan3A in terms of tanA (f) Factorise cosθ − cos3θ−cos5θ+cos7θ (g)i) verify that f(x)=((sin2θ+sin10θ)/(cos2θ+cos10θ))=((2tan3θ)/(1−tan^2 3θ)) ii) hence find in radians the general solution of f(x)=1 sir Forkum Michael

$s h o w t h a t$ $(a) c o s [2 {c o s}^{- 1} (x) + {s i n}^{- 1} (x)] = - \sqrt{1 - x^{2}}$ $(b) \frac{s i n α + s i n β}{c o s α - c o s β} = c o t (\frac{β - α}{2})$ $(c) 2 c o s (\frac{π}{3} + p) ≊ 1 - \sqrt{3} i f p i s s m a l l e n o u g h t o n e g l e c t p^{2} .$ $(d) i f θ = \frac{1}{2} {s i n}^{- 1} (\frac{3}{4}), s h o w t h a t s i n θ - c o s θ = \pm \frac{1}{2}$ $(e) w r i t e t a n 3 A i n t e r m s o f t a n A$ $(f) F a c t o r i s e c o s θ - c o s 3 θ - c o s 5 θ + c o s 7 θ$ $(g) i) v e r i f y t h a t f (x) = \frac{s i n 2 θ + s i n 10 θ}{c o s 2 θ + c o s 10 θ} = \frac{2 t a n 3 θ}{1 - {t a n}^{2} 3 θ}$ $i i) h e n c e f i n d i n r a d i a n s t h e g e n e r a l s o l u t i o n o f f (x) = 1$ $s i r F o r k u m M i c h a e l$

Question Number 63425 Answers: 0 Comments: 0

The probability that a vaccinated person(V) contracts a disease is (1/(20)). For a person vaccinated(V ′) , the probability of contracting a disease is (5/6). In a certain town 90%of thepopulation has been vaccinated against a disease. A person is selected at random from the town,find the probability that: (a) he has the disease, (b) he is vaccinated or he has the disease. sir Forkum Michael

$T h e p r o b a b i l i t y t h a t a v a c c i n a t e d p e r s o n (V) c o n t r a c t s a d i s e a s e$ $i s \frac{1}{20} . F o r a p e r s o n v a c c i n a t e d (V^{'}), t h e p r o b a b i l i t y o f c o n t r a c t i n g$ $a d i s e a s e i s \frac{5}{6} . I n a c e r t a i n t o w n 90 % o f t h e p o p u l a t i o n h a s$ $b e e n v a c c i n a t e d a g a i n s t a d i s e a s e . A p e r s o n i s s e l e c t e d a t$ $r a n d o m f r o m t h e t o w n, f i n d t h e p r o b a b i l i t y t h a t :$ $(a) h e h a s t h e d i s e a s e,$ $(b) h e i s v a c c i n a t e d o r h e h a s t h e d i s e a s e .$ $s i r F o r k u m M i c h a e l$

Question Number 63424 Answers: 0 Comments: 0

A colony of bacteria if left undisturbed will grow at a rate proportional to the number of bacteria, P present at time,t. However,a toxic substance is being added slowly such that at time t, the bacteria also die at the rate μPt where μ is a positive constant. (a) Show that at time t the rate of growth of the bacteria in the colony is governed by the differential equation (dP/dt)= (k−μt)p where k is apositive constant. when t=0, (dP/dt)=2P and when t=1, (dP/dt)=((19)/(10))P (b) show that (dP/dt)= (1/(10))(20−t)P. Sir Forkum Michael.

$A c o l o n y o f b a c t e r i a i f l e f t u n d i s t u r b e d w i l l g r o w a t a r a t e$ $p r o p o r t i o n a l t o t h e n u m b e r o f b a c t e r i a, P p r e s e n t a t t i m e, t .$ $H o w e v e r, a t o x i c s u b s t a n c e i s b e i n g a d d e d s l o w l y s u c h t h a t$ $a t t i m e t, t h e b a c t e r i a a l s o d i e a t t h e r a t e μ P t w h e r e μ i s$ $a p o s i t i v e c o n s t a n t .$ $(a) S h o w t h a t a t t i m e t t h e r a t e o f g r o w t h o f t h e b a c t e r i a i n$ $t h e c o l o n y i s g o v e r n e d b y t h e d i f f e r e n t i a l e q u a t i o n$ $\frac{d P}{d t} = (k - μ t) p w h e r e k i s a p o s i t i v e c o n s t a n t .$ $w h e n t = 0, \frac{d P}{d t} = 2 P a n d w h e n t = 1, \frac{d P}{d t} = \frac{19}{10} P$ $(b) s h o w t h a t$ $\frac{d P}{d t} = \frac{1}{10} (20 - t) P .$ $S i r F o r k u m M i c h a e l .$