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Question Number 77186 Answers: 1 Comments: 0

given { ((3^y −1= (6/2^x ))),(((3)^(y/x) = 2 )) :} find (1/x)+(1/y).

$given$ ${\begin{cases} 3^{y} - 1 = \frac{6}{2^{x}} \\ {(3)}^{\frac{y}{x}} = 2 \end{cases} find \frac{1}{x} + \frac{1}{y} .$

Question Number 77180 Answers: 1 Comments: 0

given a quadratic equation 3x^2 −x+(t^2 −4t+3)=0 has roots sin α and cos α. find the value (√(t^2 −4t+5)) .

$given a quadratic equation$ ${3 x}^{2} - x + (t^{2} - 4 t + 3) = 0 has$ $roots \sin α and \cos α . find the$ $value \sqrt{t^{2} - 4 t + 5} .$

Question Number 77160 Answers: 1 Comments: 1

Question Number 77149 Answers: 0 Comments: 2

Any reference to a book or video that coould help me solve Differential equations? please help

$Any reference to a book or video$ $that coould help me solve Differential equations ?$ $please help$

Question Number 77147 Answers: 0 Comments: 1

Σ_(r=1) ^∞ (1/r^k ) is divergent for: A. k ≤ 1 B. k > 2 C. k ≤ 2 D. 0 ≤ k < 2

$\underset{r = 1}{\sum^{\infty}} \frac{1}{r^{k}} i s d i v e r g e n t f o r :$ $A . k ⩽ 1$ $B . k > 2$ $C . k ⩽ 2$ $D . 0 ⩽ k < 2$

Question Number 76973 Answers: 2 Comments: 0

In a ABC triangle the side a=6 and c^2 −b^2 =66. Calculate the projections of sides b and c on a.

$I n a A B C t r i a n g l e t h e s i d e a = 6 a n d$ $c^{2} - b^{2} = 66 . C a l c u l a t e t h e p r o j e c t i o n s$ $o f s i d e s b a n d c o n a .$

Question Number 76922 Answers: 0 Comments: 1

Question Number 76819 Answers: 0 Comments: 4

one of the foci of the ellipse (x^2 /9) + (y^2 /4) = 1 is A. (4,0) B. (9,0) C. (5,0) D. ((√5) , 0)

$one of the foci of the ellipse$ $\frac{x^{2}}{9} + \frac{y^{2}}{4} = 1 is$ $A . (4, 0)$ $B . (9, 0)$ $C . (5, 0)$ $D . (\sqrt{5}, 0)$

Question Number 76817 Answers: 0 Comments: 2

A compound pendulum ocsillates through angles θ about its equilibrium position such that 8aθ^2 = 9g cosθ, a>0. its period is A. 2π(√((8a)/(9g))) B. ((3π)/8)(√(a/g)) C. 2π(√((9g)/(8a))) D. ((8π)/3)(√(a/g))

$A compound pendulum ocsillates$ $through angles θ about its equilibrium$ $position such that$ $8 a θ^{2} = 9 g c o s θ, a > 0 . its period is$ $A . 2 π \sqrt{\frac{8 a}{9 g}}$ $B . \frac{3 π}{8} \sqrt{\frac{a}{g}}$ $C . 2 π \sqrt{\frac{9 g}{8 a}}$ $D . \frac{8 π}{3} \sqrt{\frac{a}{g}}$

Question Number 76813 Answers: 1 Comments: 7

Σ_(k=1) ^(2n) (−1)^k = A. ∞ B. 1 C. −1 D. 0

$\underset{k = 1}{\sum^{2 n}} {(- 1)}^{k} =$ $A . \infty$ $B . 1$ $C . - 1$ $D . 0$

Question Number 76811 Answers: 0 Comments: 2

The eccentricity of the hyperbola (x^2 /(64)) − (y^2 /(36)) = 1 is A. (5/4) B. (3/4) C. (4/5) D. (4/3)

$The eccentricity of the hyperbola$ $\frac{x^{2}}{64} - \frac{y^{2}}{36} = 1 is$ $A . \frac{5}{4}$ $B . \frac{3}{4}$ $C . \frac{4}{5}$ $D . \frac{4}{3}$

Question Number 76809 Answers: 0 Comments: 3

Question Number 76802 Answers: 0 Comments: 0

3)αparticle of energy 5MeV pass through an ionisation chamber at the rate of 10 pwe second . Assum that all the energy is used in producing ion pairs,calculate the current produced. (35MeV is required for producing an ion pair and e=1.6×10^(-19) C) solution: Energy of α particles=5×10^6 eV Energy required for producing one ion pair=35eV No.of ion pairs produced by one α particle =((5×10^6 )/(35))=1.426×10^5 since 10 particle enter the chamber in one second =1.426×10^5 ×10=1.426×10^6 charge on dach ion=1.6×10^(-19) C Current=(1.426×10^6 )×(1.6×10^(-19) )C/s =2.287×10^(-13) A.

$3) α particle of energy 5 MeV pass$ $through an ionisation chamber at the$ $rate of 10 pwe second . Assum that all$ $the energy is used in producing ion$ $pairs, calculate the current produced .$ $(35 MeV is required for producing an$ $ion pair and e = 1.6 \times 10^{- 19} C)$ $solution :$ $Energy of α particles = 5 \times 10^{6} eV$ $Energy required for producing one$ $ion pair = 35 eV$ $No . of ion pairs produced by one α$ $particle = \frac{5 \times 10^{6}}{35} = 1.426 \times 10^{5}$ $since 10 particle enter the chamber in$ $one second$ $= 1.426 \times 10^{5} \times 10 = 1.426 \times 10^{6}$ $charge on dach ion = 1.6 \times 10^{- 19} C$ $Current = (1.426 \times 10^{6}) \times (1.6 \times 10^{- 19}) C / s$ $= 2.287 \times 10^{- 13} A .$

Question Number 76726 Answers: 3 Comments: 0

var(x) = 2 then var(2x −3)=? E(x) = 2 then E(2x −3) = ?

$var (x) = 2 then var (2 x - 3) = ?$ $E (x) = 2 then E (2 x - 3) = ?$

Question Number 76724 Answers: 1 Comments: 1

∫_0 ^3 ∣x^2 −1∣ dx ≡

$\int_{0}^{3} ∣ x^{2} - 1 ∣ dx \equiv$

Question Number 76723 Answers: 0 Comments: 0

the maclaurin expansion of ln (3 + 4x) is valid for A) −(3/4) ≤ x< (3/4) B) −(3/4)< x ≤ (3/4) C) −(1/4)< x ≤ (1/4) D) −(3/4)< x < (3/4)

$the maclaurin expansion of \ln (3 + 4 x) i s v a l i d f o r$ $A) - \frac{3}{4} ⩽ x < \frac{3}{4}$ $B) - \frac{3}{4} < x ⩽ \frac{3}{4}$ $C) - \frac{1}{4} < x ⩽ \frac{1}{4}$ $D) - \frac{3}{4} < x < \frac{3}{4}$

Question Number 76680 Answers: 0 Comments: 4

prove that: ∫_0 ^1 (1−x^7 )^(1/3) dx=∫_0 ^1 (1−x^3 )^(1/7) dx

$p r o v e t h a t :$ $\int_{0}^{1} {(1 - x^{7})}^{\frac{1}{3}} d x = \int_{0}^{1} {(1 - x^{3})}^{\frac{1}{7}} d x$

Question Number 76579 Answers: 1 Comments: 0

A uniform ladder of weight W and length 2a rest in limiting equilibrium with one end on a rough horizontal ground and the other end on a rough vertical wall. The coefficient of friction between the ladder and the ground and between the ladder and the wall are respectively μ and λ . If the ladder makes an angle θ with the ground where tan θ = (5/(12)), a) show that 5μ + 6λμ − 6 = 0. b) find the value of λ and μ given that λμ =(1/2).

$A uniform ladder of weight W and length$ $2 a rest in limiting equilibrium with one$ $end on a rough horizontal ground and the$ $other end on a rough vertical wall .$ $The coefficient of friction between the ladder$ $and the ground and between the ladder and$ $the wall are respectively μ and λ . If the ladder$ $makes an angle θ with the ground where \tan θ = \frac{5}{12},$ $a) show that 5 μ + 6 λ μ - 6 = 0 .$ $b) find the value of λ and μ given that λ μ = \frac{1}{2} .$

Question Number 76577 Answers: 1 Comments: 0

given a sequence defined by {((3n)/(2n+ 5))}_(n=1) ^∞ , does this sequence converge or diverge, explain

$g i v e n a s e q u e n c e d e f i n e d b y {\frac{3 n}{2 n + 5}}_{n = 1}^{\infty}, d o e s t h i s$ $s e q u e n c e c o n v e r g e o r d i v e r g e, e x p l a i n$

Question Number 76404 Answers: 0 Comments: 2

Hello have nice end of year good bless you all i respond note in y re message becsuse i have so many problemes that mack me feel no pleasur any more to do somthing i think its importante to say it i will back Soon i hop so Sorry for my English

$Hello have nice end of year good bless you$ $all i respond note in y re message becsuse i have so many problemes$ $that mack me feel no pleasur any more to do somthing$ $i think its importante to say it i will back Soon i hop so Sorry for$ $my English$

Question Number 76384 Answers: 0 Comments: 1

Question Number 76368 Answers: 3 Comments: 5

prove that 1. Σ_(r=1) ^n r = (1/2)n(n+1) 2. Σ_(r=1) ^n r^2 = (1/6)n(n+1)(2n + 1) 3. Σ_(r=1) ^n r^3 = (1/4)n^2 (n + 1)^2

$p r o v e t h a t$ $1 . \underset{r = 1}{\sum^{n}} r = \frac{1}{2} n (n + 1)$ $2 . \underset{r = 1}{\sum^{n}} r^{2} = \frac{1}{6} n (n + 1) (2 n + 1)$ $3 . \underset{r = 1}{\sum^{n}} r^{3} = \frac{1}{4} n^{2} {(n + 1)}^{2}$

Question Number 76367 Answers: 0 Comments: 4

prove that Σ_(r=1) ^∞ (1/r^2 ) = (π^2 /6)

$p r o v e t h a t \underset{r = 1}{\sum^{\infty}} \frac{1}{r^{2}} = \frac{π^{2}}{6}$

Question Number 76229 Answers: 1 Comments: 0

Let P(x) be polynomial in x with integral coefficients. If n is a solution of P(x)≡0(mod n) , and a≡b(mod n), prove that b is also a solution.

$L e t P (x) b e p o l y n o m i a l i n x w i t h i n t e g r a l$ $c o e f f i c i e n t s . I f n i s a s o l u t i o n o f$ $P (x) \equiv 0 (m o d n), a n d a \equiv b (m o d n),$ $p r o v e t h a t b i s a l s o a s o l u t i o n .$

Question Number 76232 Answers: 2 Comments: 2

how do we find ∫_0 ^(π/2) sinh^(−1) x dx and ∫_0 ^(π/2) cosh^(−1) xdx

$h o w d o w e f i n d$ $\underset{0}{\int^{\frac{π}{2}}} {s i n h}^{- 1} x d x a n d \underset{0}{\int^{\frac{π}{2}}} \cosh^{- 1} x d x$

Question Number 76110 Answers: 1 Comments: 0

hello solve in R tanx>(√3) please explain me if possible.

$h e l l o solve in R$ $\tan x > \sqrt{3}$ $p l e a s e e x p l a i n m e i f p o s s i b l e .$

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