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Question Number 177628 Answers: 1 Comments: 1

Question Number 177627 Answers: 1 Comments: 0

Question Number 177626 Answers: 3 Comments: 0

Question Number 177621 Answers: 1 Comments: 1

For all positive integer n. f(n) is defined as the remainder when n is divided by k where k is an integer greater than 1. f(k+25)=3 Quantity A: k Quantity B: 5

$$\mathrm{For}\:\mathrm{all}\:\mathrm{positive}\:\mathrm{integer}\:\mathrm{n}.\:\mathrm{f}\left(\mathrm{n}\right)\:\mathrm{is}\:\mathrm{defined} \\ $$$$\mathrm{as}\:\mathrm{the}\:\mathrm{remainder}\:\mathrm{when}\:\mathrm{n}\:\mathrm{is}\:\mathrm{divided}\:\mathrm{by}\:\mathrm{k} \\ $$$$\mathrm{where}\:\mathrm{k}\:\mathrm{is}\:\mathrm{an}\:\mathrm{integer}\:\mathrm{greater}\:\mathrm{than}\:\mathrm{1}.\: \\ $$$$\mathrm{f}\left(\mathrm{k}+\mathrm{25}\right)=\mathrm{3} \\ $$$$\mathrm{Quantity}\:\mathrm{A}:\:\mathrm{k} \\ $$$$\mathrm{Quantity}\:\mathrm{B}:\:\mathrm{5} \\ $$

Question Number 177619 Answers: 0 Comments: 2

Let r be a positive integer. List S consists of integers from −r to r where each integers occurs once. List T consists of the integers from −r to r where each of the two integers −r and r occurs twice and each of the other integers occur once. Quantity A: the standard deviation of the integers in S Quantity B: the standard deviation of the integers in T.

$${Let}\:{r}\:{be}\:{a}\:{positive}\:{integer}.\:{List}\:{S}\:{consists} \\ $$$${of}\:{integers}\:{from}\:−{r}\:{to}\:\:{r}\:{where}\:{each}\:{integers} \\ $$$${occurs}\:{once}.\:{List}\:{T}\:{consists}\:{of}\:{the}\:{integers}\:{from} \\ $$$$−{r}\:{to}\:{r}\:{where}\:{each}\:{of}\:{the}\:{two}\:{integers} \\ $$$$−{r}\:{and}\:{r}\:{occurs}\:{twice}\:{and}\:{each}\:{of}\:{the}\: \\ $$$${other}\:{integers}\:{occur}\:{once}. \\ $$$${Quantity}\:{A}:\:{the}\:{standard}\:{deviation}\:{of}\:{the} \\ $$$${integers}\:{in}\:{S} \\ $$$${Quantity}\:{B}:\:{the}\:{standard}\:{deviation}\:{of}\:{the} \\ $$$${integers}\:{in}\:{T}. \\ $$

Question Number 177606 Answers: 2 Comments: 0

Question Number 177605 Answers: 2 Comments: 0

Question Number 177604 Answers: 0 Comments: 0

Calculate 𝛗=∫_0 ^( 1) (( ln(x ))/(1 + x^( 2) )) dx =^? −G ∼ Solution ∼ 𝛗 = ∫_0 ^( 1) {Σ_(k=0) ^∞ (−1)^( k) x^( 2k) ln(x) }dx = Σ_(k=0) ^∞ (−1 )^( k) ∫_0 ^( 1) x^( 2k) ln(x) dx =^(i.b.p) Σ_(k=0) ^∞ (−1 )^( k) {[(x^( 1+2k) /(1+2k)) ln(x ) ]_0 ^1 −(1/(1+2k))∫_0 ^( 1) x^( 2k) dx } = Σ_(k=0) ^∞ (( (−1)^(1+k) )/(( 1 + 2k )^( 2) )) = − G ( Catalan constant ) ∴ 𝛗 = − G ■ m.n

$$ \\ $$$$\:\:\:\:\:\:\:\:\mathrm{Calculate} \\ $$$$\:\:\:\:\:\boldsymbol{\phi}=\int_{\mathrm{0}} ^{\:\mathrm{1}} \frac{\:\mathrm{ln}\left({x}\:\right)}{\mathrm{1}\:+\:{x}^{\:\mathrm{2}} }\:\mathrm{d}{x}\:\overset{?} {=}\:−\mathrm{G} \\ $$$$\:\:\:\:\sim\:\mathrm{Solution}\:\sim \\ $$$$\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\int_{\mathrm{0}} ^{\:\mathrm{1}} \left\{\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\right)^{\:{k}} {x}^{\:\mathrm{2}{k}} \mathrm{ln}\left({x}\right)\:\right\}\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:\:\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\:\left(−\mathrm{1}\:\right)^{\:{k}} \:\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{k}} \:\mathrm{ln}\left({x}\right)\:\mathrm{d}{x} \\ $$$$\:\:\:\:\:\:\:\:\overset{{i}.{b}.{p}} {=}\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\left(−\mathrm{1}\:\right)^{\:{k}} \left\{\left[\frac{{x}^{\:\mathrm{1}+\mathrm{2}{k}} }{\mathrm{1}+\mathrm{2}{k}}\:\mathrm{ln}\left({x}\:\right)\:\right]_{\mathrm{0}} ^{\mathrm{1}} −\frac{\mathrm{1}}{\mathrm{1}+\mathrm{2}{k}}\int_{\mathrm{0}} ^{\:\mathrm{1}} {x}^{\:\mathrm{2}{k}} \mathrm{d}{x}\:\right\} \\ $$$$\:\:\:\:\:\:\:\:=\:\underset{{k}=\mathrm{0}} {\overset{\infty} {\sum}}\frac{\:\left(−\mathrm{1}\right)^{\mathrm{1}+{k}} }{\left(\:\mathrm{1}\:+\:\mathrm{2}{k}\:\right)^{\:\mathrm{2}} }\:\:=\:−\:\mathrm{G}\:\left(\:\mathrm{C}{atalan}\:{constant}\:\right)\: \\ $$$$\:\:\:\:\:\:\:\:\therefore\:\:\:\:\:\:\:\:\:\:\boldsymbol{\phi}\:=\:\:−\:\mathrm{G}\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\blacksquare\:\mathrm{m}.\mathrm{n}\: \\ $$$$\:\:\:\:\: \\ $$

Question Number 177597 Answers: 2 Comments: 0

Question Number 177596 Answers: 1 Comments: 0

Question Number 177586 Answers: 1 Comments: 4

Question Number 177585 Answers: 0 Comments: 2

looks like my question was deleted. why? i cant find it

$${looks}\:{like}\:{my}\:{question}\:{was}\:{deleted}.\:{why}?\:{i}\:{cant}\:{find}\:{it} \\ $$

Question Number 177579 Answers: 2 Comments: 0

if x^3 +y^3 +((x+y)/4)=((15)/2), find maximum value of x+y.

$${if}\:{x}^{\mathrm{3}} +{y}^{\mathrm{3}} +\frac{{x}+{y}}{\mathrm{4}}=\frac{\mathrm{15}}{\mathrm{2}},\:{find}\:{maximum} \\ $$$${value}\:{of}\:{x}+{y}. \\ $$

Question Number 177578 Answers: 1 Comments: 6

Question Number 177570 Answers: 1 Comments: 5

lim_(Δx→cos(π/2)) ((sin^3 (Δx+x)−sin^3 x)/(2^(−1) Δx))=?

$$\underset{\Delta{x}\rightarrow{cos}\frac{\pi}{\mathrm{2}}} {\mathrm{lim}}\:\frac{{sin}^{\mathrm{3}} \left(\Delta{x}+{x}\right)−{sin}^{\mathrm{3}} {x}}{\mathrm{2}^{−\mathrm{1}} \Delta{x}}=? \\ $$

Question Number 177560 Answers: 2 Comments: 1

Question Number 177557 Answers: 0 Comments: 0

Question Number 177552 Answers: 0 Comments: 0

Question Number 177548 Answers: 2 Comments: 0

A projectile is fired with velocity(v_o ) such that it passes through two points both a distance(h) above the horizontal.show that if the gun is adjusted for the maximum range of the separation of two position is d=((v_o (√(v_o ^2 −4gh)))/g)

$$\mathrm{A}\:\mathrm{projectile}\:\mathrm{is}\:\mathrm{fired}\:\mathrm{with}\:\mathrm{velocity}\left(\mathrm{v}_{\mathrm{o}} \right) \\ $$$$\mathrm{such}\:\mathrm{that}\:\mathrm{it}\:\mathrm{passes}\:\mathrm{through}\:\mathrm{two} \\ $$$$\mathrm{points}\:\mathrm{both}\:\mathrm{a}\:\mathrm{distance}\left(\mathrm{h}\right)\:\mathrm{above}\:\mathrm{the} \\ $$$$\mathrm{horizontal}.\mathrm{show}\:\mathrm{that}\:\mathrm{if}\:\mathrm{the}\:\mathrm{gun}\:\mathrm{is}\:\mathrm{adjusted} \\ $$$$\mathrm{for}\:\mathrm{the}\:\mathrm{maximum}\:\mathrm{range}\:\mathrm{of}\:\mathrm{the} \\ $$$$\mathrm{separation}\:\mathrm{of}\:\mathrm{two}\:\mathrm{position}\:\mathrm{is} \\ $$$$\boldsymbol{\mathrm{d}}=\frac{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} \sqrt{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} ^{\mathrm{2}} −\mathrm{4}\boldsymbol{\mathrm{gh}}}}{\boldsymbol{\mathrm{g}}} \\ $$

Question Number 177546 Answers: 0 Comments: 0

A particle is projected vertically upward in a constant gravitation field with initial speed (v_0 ).show that there is retarding proportional to the square of the instantaneous speed,the speed of the partical when it returns on the initial position is ((v_o v_c )/( (√(v_o ^2 +v_c ^2 )))) where v_c is terminal speed

$$\mathrm{A}\:\mathrm{particle}\:\mathrm{is}\:\mathrm{projected}\:\mathrm{vertically} \\ $$$$\mathrm{upward}\:\mathrm{in}\:\mathrm{a}\:\mathrm{constant}\:\mathrm{gravitation} \\ $$$$\mathrm{field}\:\mathrm{with}\:\mathrm{initial}\:\mathrm{speed}\:\left(\mathrm{v}_{\mathrm{0}} \right).\mathrm{show} \\ $$$$\mathrm{that}\:\mathrm{there}\:\mathrm{is}\:\mathrm{retarding}\:\mathrm{proportional} \\ $$$$\mathrm{to}\:\mathrm{the}\:\mathrm{square}\:\mathrm{of}\:\mathrm{the}\:\mathrm{instantaneous} \\ $$$$\mathrm{speed},\mathrm{the}\:\mathrm{speed}\:\mathrm{of}\:\mathrm{the}\:\mathrm{partical} \\ $$$$\mathrm{when}\:\mathrm{it}\:\mathrm{returns}\:\mathrm{on}\:\mathrm{the}\:\mathrm{initial} \\ $$$$\mathrm{position}\:\mathrm{is}\:\:\:\frac{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} \boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{c}}} }{\:\sqrt{\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{o}}} ^{\mathrm{2}} +\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{c}}} ^{\mathrm{2}} }}\:\:\boldsymbol{\mathrm{where}}\:\boldsymbol{\mathrm{v}}_{\boldsymbol{\mathrm{c}}} \:\boldsymbol{\mathrm{i}}\mathrm{s} \\ $$$$\mathrm{terminal}\:\mathrm{speed} \\ $$

Question Number 177542 Answers: 3 Comments: 0

Question Number 177541 Answers: 0 Comments: 3

Question Number 177540 Answers: 2 Comments: 0

Question Number 177530 Answers: 3 Comments: 0

f (x ) + f ((( 1)/( ((1 −x^( 3) ))^(1/3) )) )= x^( 3) is given. find the value of: f (−1)=?

$$ \\ $$$${f}\:\left({x}\:\right)\:+\:{f}\:\left(\frac{\:\mathrm{1}}{\:\sqrt[{\mathrm{3}}]{\mathrm{1}\:−{x}^{\:\mathrm{3}} }}\:\right)=\:{x}^{\:\mathrm{3}} \\ $$$$\:\:\:\:\:\:{is}\:{given}.\:{find}\:{the}\:{value}\:{of}: \\ $$$$\:\:\:\:\:\:\:\: \\ $$$$\:\:\:\:\:\:\:{f}\:\left(−\mathrm{1}\right)=? \\ $$$$ \\ $$

Question Number 177519 Answers: 2 Comments: 0

Question Number 177525 Answers: 1 Comments: 3

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