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Question Number 40764 Answers: 1 Comments: 0

a piece of wire 40cm long is cut into two parts and each part is then bent into a square.if the sum of these squares is 68cm^2 find the lengths of the two pieces of wire.

$a piece of wire 40 cm long$ $is cut into two parts and each part$ $is then bent into a square . if the sum$ $of these squares is 68 {cm}^{2}$ $find the lengths of the two$ $pieces of wire .$

Question Number 40763 Answers: 2 Comments: 0

Question Number 40760 Answers: 0 Comments: 0

find ∫ ((√(1+x^2 ))/(√(1−x^3 ))) dx

$f i n d \int \frac{\sqrt{1 + x^{2}}}{\sqrt{1 - x^{3}}} d x$

Question Number 40759 Answers: 1 Comments: 2

calculate lim_(n→+∞) ((1−e^(−nx^2 ) )/(x^2 sin((π/n))))

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

Question Number 40745 Answers: 3 Comments: 0

Question Number 40743 Answers: 2 Comments: 0

Question Number 40740 Answers: 0 Comments: 5

Question Number 40738 Answers: 2 Comments: 2

Question Number 40732 Answers: 1 Comments: 2

Question Number 40717 Answers: 1 Comments: 1

∫(√(tanx/sinx.cosxdx))

$\int \sqrt{t a n x / s i n x . c o s x d x}$

Question Number 40716 Answers: 2 Comments: 0

∫(cosx−cos2x/1−cosx)dx

$\int (c o s x - c o s 2 x / 1 - c o s x) d x$

Question Number 40715 Answers: 1 Comments: 1

for x≥2 ∣x−2∣=

$for x ⩾ 2 ∣ x - 2 ∣=$

Question Number 40711 Answers: 1 Comments: 0

Two point charges,q_1 =0.4μC and q_2 =−0.3μC are placed at 10cm apart.Calculate (a)the potential at point A which is midway between them,and (b)point B which is 6cm from q_1 and 8cm from q_2

$T w o p o i n t c h a r g e s, q_{1} = 0.4 μ C a n d$ $q_{2} = - 0.3 μ C a r e p l a c e d a t 10 c m$ $a p a r t . C a l c u l a t e$ $(a) t h e p o t e n t i a l a t p o i n t A w h i c h i s$ $m i d w a y b e t w e e n t h e m, a n d$ $(b) p o i n t B w h i c h i s 6 c m f r o m q_{1}$ $a n d 8 c m f r o m q_{2}$

Question Number 40709 Answers: 1 Comments: 1

calculate lim_(x→0) ((cos(x−sinx)−1)/(x^2 ))

$c a l c u l a t e {l i m}_{x \to 0} \frac{c o s (x - s i n x) - 1}{x^{2}}$

Question Number 40700 Answers: 1 Comments: 0

A charge q_1 =2.0×10^(−9) C is placed at the point,(x=0,y=4cm) and another charge,q_2 =−3.0×10^(−9) C is located at the point,(x=3cm,y=4cm). If a third charge,q_3 =4.0×10^(−9) C is placed at the origin, a)obtain the x and y component of the total force on q_3 b)Calculate the magnitude and direction of the total force on q_(3.)

$A c h a r g e q_{1} = 2.0 \times 10^{- 9} C i s p l a c e d$ $a t t h e p o i n t, (x = 0, y = 4 c m) a n d$ $a n o t h e r c h a r g e, q_{2} = - 3.0 \times 10^{- 9} C$ $i s l o c a t e d a t t h e p o i n t, (x = 3 c m, y = 4 c m) .$ $I f a t h i r d c h a r g e, q_{3} = 4.0 \times 10^{- 9} C$ $i s p l a c e d a t t h e o r i g i n,$ $a) o b t a i n t h e x a n d y c o m p o n e n t o f$ $t h e t o t a l f o r c e o n q_{3}$ $b) C a l c u l a t e t h e m a g n i t u d e a n d$ $d i r e c t i o n o f t h e t o t a l f o r c e o n q_{3 .}$

Question Number 40699 Answers: 1 Comments: 0

Two point charges q_1 =1.5×10^(−9) C and q_2 =3.0×10^(−9) C are seperated by a distance of 200cm.Calculate the point at which the total electric field is zero.

$T w o p o i n t c h a r g e s q_{1} = 1.5 \times 10^{- 9} C$ $a n d q_{2} = 3.0 \times 10^{- 9} C a r e s e p e r a t e d$ $b y a d i s t a n c e o f 200 c m . C a l c u l a t e$ $t h e p o i n t a t w h i c h t h e t o t a l e l e c t r i c$ $f i e l d i s z e r o .$

Question Number 40686 Answers: 0 Comments: 6

Question Number 40684 Answers: 0 Comments: 1

∫((x^7 −1)/(logx))dx

$\int \frac{x^{7} - 1}{logx} dx$

Question Number 40675 Answers: 1 Comments: 2

Question Number 40667 Answers: 1 Comments: 2

find the value lim_(x→0) g(x) must have, if g complies the statement about limit. Suppose lim_(x→ −4) [x lim_(x→0) g(x)] = 2

$find the value \lim_{x \to 0} g (x) must have, if g complies the statement$ $about limit . Suppose \lim_{x \to - 4} [x \lim_{x \to 0} g (x)] = 2$

Question Number 40665 Answers: 1 Comments: 0

−5−( )=3 −5−x=3 x=3+5 x=8(give sign of greater number) −5−8=3

$- 5 - () = 3$ $- 5 - x = 3$ $x = 3 + 5$ $x = 8 (g i v e s i g n o f g r e a t e r n u m b e r)$ $- 5 - 8 = 3$

Question Number 40664 Answers: 1 Comments: 0

Factorise: x^(19) − x^(17) + x^(10) + x^8 + 1

$Factorise : x^{19} - x^{17} + x^{10} + x^{8} + 1$

Question Number 40662 Answers: 1 Comments: 5

Question Number 40661 Answers: 0 Comments: 4

1)find g(x)=∫_0 ^(π/2) ln(1−x^2 cos^2 θ)dθ with x from R 2) find the value of ∫_0 ^(π/2) ln(1−2 cos^2 θ)dθ and 3) find the value of A(α)=∫_0 ^(π/2) ln(1−cos^2 α cos^2 θ)dθ

$1) f i n d g (x) = \int_{0}^{\frac{π}{2}} l n (1 - x^{2} {c o s}^{2} θ) d θ w i t h x f r o m R$ $2) f i n d t h e v a l u e o f \int_{0}^{\frac{π}{2}} l n (1 - 2 {c o s}^{2} θ) d θ a n d$ $3) f i n d t h e v a l u e o f$ $A (α) = \int_{0}^{\frac{π}{2}} l n (1 - {c o s}^{2} α {c o s}^{2} θ) d θ$

Question Number 40660 Answers: 0 Comments: 0

let f(t) = ∫_0 ^∞ ((arctan(tcosx))/(1+x^2 ))dx 1) find another form of f(t) 2) calculate ∫_0 ^∞ ((arctan(2cosx))/(1+x^2 ))dx .

$l e t f (t) = \int_{0}^{\infty} \frac{a r c t a n (t c o s x)}{1 + x^{2}} d x$ $1) f i n d a n o t h e r f o r m o f f (t)$ $2) c a l c u l a t e \int_{0}^{\infty} \frac{a r c t a n (2 c o s x)}{1 + x^{2}} d x .$

Question Number 40658 Answers: 0 Comments: 1

find ∫_0 ^(π/4) ((x−1)/(2+cosx))dx .

$f i n d \int_{0}^{\frac{π}{4}} \frac{x - 1}{2 + c o s x} d x .$

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