Question and Answers Forum

All Questions Topic List

AllQuestion and Answers: Page 1625

Question Number 39635 Answers: 1 Comments: 1

calculate lim_(x→1) ((1+cos(πx))/(x^2 − sin(((πx)/2))))

$c a l c u l a t e {l i m}_{x \to 1} \frac{1 + c o s (π x)}{x^{2} - s i n (\frac{π x}{2})}$

Question Number 39633 Answers: 0 Comments: 3

find the value of f(x) = ∫_0 ^π ln(x^2 −2x cosθ +1)dθ with x fromR.

$f i n d t h e v a l u e o f$ $f (x) = \int_{0}^{π} l n (x^{2} - 2 x c o s θ + 1) d θ w i t h x f r o m R .$

Question Number 39632 Answers: 0 Comments: 0

let S_n = Σ_(k=1) ^n (1/(k(√k))) find a equivalent of S_n (n→+∞)

$l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{k \sqrt{k}}$ $f i n d a e q u i v a l e n t o f S_{n} (n \to + \infty)$

Question Number 39631 Answers: 0 Comments: 0

let S_n = Σ_(k=1) ^n (1/(√k)) find a equivalent of S_n when n →+∞

$l e t S_{n} = \sum_{k = 1}^{n} \frac{1}{\sqrt{k}}$ $f i n d a e q u i v a l e n t o f S_{n} w h e n n \to + \infty$

Question Number 39626 Answers: 2 Comments: 1

Question Number 39623 Answers: 0 Comments: 0

Two equal charges q_1 =q_(2 ) =−6μC are on the y−axis at y_1 =3cm and y_2 =−3cm. i)What is the magnitude and direction of the electric field on the x−axis at x= 4cm, ii)if a test charge q_0 =2μC is plqced at x=4cm,find the force the net charge experiences.

$T w o e q u a l c h a r g e s q_{1} = q_{2} = - 6 μ C$ $a r e o n t h e y - a x i s a t y_{1} = 3 c m a n d$ $y_{2} = - 3 c m .$ $i) W h a t i s t h e m a g n i t u d e a n d$ $d i r e c t i o n o f t h e e l e c t r i c f i e l d o n t h e$ $x - a x i s a t x = 4 c m,$ $i i) i f a t e s t c h a r g e q_{0} = 2 μ C i s p l q c e d$ $a t x = 4 c m, f i n d t h e f o r c e t h e n e t$ $c h a r g e e x p e r i e n c e s .$

Question Number 39622 Answers: 0 Comments: 0

Two charges Q_0 and 3Q_0 are at l distance apart.These two charges are free to move but do not because is a third charge nearby.What must be the charge for the first two to be in equilibrium?

$T w o c h a r g e s Q_{0} a n d 3 Q_{0} a r e a t l$ $d i s t a n c e a p a r t . T h e s e t w o c h a r g e s$ $a r e f r e e t o m o v e b u t d o n o t b e c a u s e$ $i s a t h i r d c h a r g e n e a r b y . W h a t m u s t$ $b e t h e c h a r g e f o r t h e f i r s t t w o t o b e$ $i n e q u i l i b r i u m ?$

Question Number 39612 Answers: 1 Comments: 4

Question Number 39611 Answers: 1 Comments: 0

write the expression of electrostatic force between two charges Q_(1 ) and Q_2 separated by distance r. for the following condition 1)in air 2)when die electric present between them 3)when die electric partially fill space betweenp them

$w r i t e t h e e x p r e s s i o n o f e l e c t r o s t a t i c f o r c e$ $b e t w e e n t w o c h a r g e s Q_{1} a n d Q_{2} s e p a r a t e d b y$ $d i s t a n c e r . f o r t h e f o l l o w i n g c o n d i t i o n$ $1) i n a i r$ $2) w h e n d i e e l e c t r i c p r e s e n t b e t w e e n t h e m$ $3) w h e n d i e e l e c t r i c p a r t i a l l y f i l l s p a c e b e t w e e n p$ $t h e m$

Question Number 39607 Answers: 3 Comments: 0

find the minimum and maximum value of the quadratic functions a) 4x^2 + 5x + 1 b) x + (2/x) = 3 c) x^2 − (x/4) + 6 hence draw each draw

$f i n d t h e$ $m i n i m u m a n d m a x i m u m v a l u e$ $o f t h e q u a d r a t i c f u n c t i o n s$ $a) 4 x^{2} + 5 x + 1$ $b) x + \frac{2}{x} = 3$ $c) x^{2} - \frac{x}{4} + 6$ $h e n c e d r a w e a c h d r a w$

Question Number 39591 Answers: 1 Comments: 0

Given the lines l_1 :−3mx + 3y = 9 and l_(2 ) : y = mx + c find the value of m and c if the point (1,2) lie on both lines. hence the tangent of the curve y = (mx + c)^2 when it moves across the x−axis

$G i v e n t h e l i n e s$ $l_{1} : - 3 m x + 3 y = 9$ $a n d l_{2} : y = m x + c$ $f i n d t h e v a l u e o f m a n d c i f$ $t h e p o i n t (1, 2) l i e o n b o t h l i n e s .$ $h e n c e t h e t a n g e n t o f t h e$ $c u r v e y = {(m x + c)}^{2}$ $w h e n i t m o v e s a c r o s s t h e x - a x i s$

Question Number 39588 Answers: 1 Comments: 0

if cos A= (3/5) and tan B = ((12)/5) where A and B are reflex angles find without using tables,the value of a) sin (A − B) b) tan(A−B) c) cos (A + B).

$i f c o s A = \frac{3}{5} a n d t a n B = \frac{12}{5}$ $w h e r e A a n d B a r e r e f l e x a n g l e s$ $f i n d w i t h o u t u s i n g t a b l e s, t h e$ $v a l u e o f$ $a) s i n (A - B) b) t a n (A - B)$ $c) c o s (A + B) .$

Question Number 39587 Answers: 4 Comments: 0

Solve for x in the range 0 ≤ x ≤2π the equations a) cos(x + (π/3)) = 0 b) sin x = cos x. c) sin 2x + 2sin x = 1 + cos x

$S o l v e f o r x i n t h e r a n g e 0 ⩽ x ⩽ 2 π$ $t h e e q u a t i o n s$ $a) c o s (x + \frac{π}{3}) = 0$ $b) s i n x = c o s x .$ $c) s i n 2 x + 2 s i n x = 1 + c o s x$

Question Number 39586 Answers: 2 Comments: 0

show that a) ((1 + 2sin2θ − cos2θ)/(1+sin2θ + cos 2θ)) = tan θ b) tan^2 A − tan^2 B = ((sin^2 A−sin^2 B)/(cos^2 A cos^2 B))

$s h o w t h a t$ $a) \frac{1 + 2 s i n 2 θ - c o s 2 θ}{1 + s i n 2 θ + c o s 2 θ} = t a n θ$ $b) {t a n}^{2} A - {t a n}^{2} B = \frac{{s i n}^{2} A - {s i n}^{2} B}{{c o s}^{2} A {c o s}^{2} B}$

Question Number 39582 Answers: 1 Comments: 0

Question Number 39573 Answers: 1 Comments: 1

Point charges 88μC,−55μC and 70μC are placed in a straight line. The central one is 0.75m from each of the others.Calculate the net force on each due to the other two.

$P o i n t c h a r g e s 88 μ C, - 55 μ C a n d$ $70 μ C a r e p l a c e d i n a s t r a i g h t l i n e .$ $T h e c e n t r a l o n e i s 0.75 m f r o m$ $e a c h o f t h e o t h e r s . C a l c u l a t e t h e$ $n e t f o r c e o n e a c h d u e t o t h e o t h e r$ $t w o .$

Question Number 39559 Answers: 2 Comments: 1

Question Number 39529 Answers: 0 Comments: 5

Question Number 39520 Answers: 1 Comments: 1

if (1+x)^n =Σ_(i=0) ^n a_i x^i and (1+x)^(n+1) =Σ_(i=0) ^(n+1) b_i x^i calculate ((∐_(i=0) ^n a_i )/(Π_(i=0) ^(n+1) b_i )) .

$i f {(1 + x)}^{n} = \sum_{i = 0}^{n} a_{i} x^{i} a n d$ ${(1 + x)}^{n + 1} = \sum_{i = 0}^{n + 1} b_{i} x^{i} c a l c u l a t e$ $\frac{∐_{i = 0}^{n} a_{i}}{\prod_{i = 0}^{n + 1} b_{i}} .$

Question Number 39519 Answers: 1 Comments: 1

simplify 1) A_n =(1/(√a)){ (((1+(√a))/2) )^n −(((1−(√a))/2))^n } with n natural integr and a>0 2) f(x)= (1/(√(2x+1))){ (((1+(√(2x+1)))/2))^n −(((1−(√(2x+1)))/2))^n }

$s i m p l i f y$ $1) A_{n} = \frac{1}{\sqrt{a}} {{(\frac{1 + \sqrt{a}}{2})}^{n} - {(\frac{1 - \sqrt{a}}{2})}^{n}} w i t h n n a t u r a l$ $i n t e g r a n d a > 0$ $2) f (x) = \frac{1}{\sqrt{2 x + 1}} {{(\frac{1 + \sqrt{2 x + 1}}{2})}^{n} - {(\frac{1 - \sqrt{2 x + 1}}{2})}^{n}}$

Question Number 39517 Answers: 0 Comments: 2

find radius of S(x)=Σ_(n=1) ^∞ (x^n /n^2 ) and calculate its sum 2) find Σ_(n=1) ^∞ (1/n^2 ) and Σ_(n=1) ^∞ (1/(n^2 2^n )) .

$f i n d r a d i u s o f S (x) = \sum_{n = 1}^{\infty} \frac{x^{n}}{n^{2}}$ $a n d c a l c u l a t e i t s s u m$ $2) f i n d \sum_{n = 1}^{\infty} \frac{1}{n^{2}} a n d \sum_{n = 1}^{\infty} \frac{1}{n^{2} 2^{n}} .$

Question Number 39524 Answers: 0 Comments: 0

Question Number 39511 Answers: 0 Comments: 2

Question Number 39508 Answers: 1 Comments: 1

Question Number 39486 Answers: 0 Comments: 7

sin θ=sin αsin (((θ+α)/2)) Express θ explicitly in terms of α.

$\sin θ = \sin α \sin (\frac{θ + α}{2})$ $E x p r e s s θ e x p l i c i t l y i n t e r m s o f α .$

Question Number 39475 Answers: 1 Comments: 0

Pg 1620 Pg 1621 Pg 1622 Pg 1623 Pg 1624 Pg 1625 Pg 1626 Pg 1627 Pg 1628 Pg 1629

Contact: info@tinkutara.com