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Question Number 95784    Answers: 0   Comments: 0

calculate Σ_(n=1) ^∞ (H_n /n^2 ) H_n =Σ_(k=1) ^n (1/k)

$$\mathrm{calculate}\:\sum_{\mathrm{n}=\mathrm{1}} ^{\infty} \:\frac{\mathrm{H}_{\mathrm{n}} }{\mathrm{n}^{\mathrm{2}} } \\ $$$$\mathrm{H}_{\mathrm{n}} =\sum_{\mathrm{k}=\mathrm{1}} ^{\mathrm{n}} \:\frac{\mathrm{1}}{\mathrm{k}} \\ $$

Question Number 95782    Answers: 0   Comments: 2

it looks version 2.077 has a problem. for very long loading

$$\mathrm{it}\:\mathrm{looks}\:\mathrm{version}\:\mathrm{2}.\mathrm{077}\:\mathrm{has}\:\mathrm{a}\: \\ $$$$\mathrm{problem}.\:\:\mathrm{for}\:\mathrm{very}\:\mathrm{long}\:\mathrm{loading} \\ $$

Question Number 95780    Answers: 0   Comments: 0

Question Number 95779    Answers: 2   Comments: 0

if plane 3x+4y+tz=2 and kx+6y+5z−2=0 are parallel. find the value of k and t

$$\mathrm{if}\:\mathrm{plane}\:\mathrm{3x}+\mathrm{4y}+\mathrm{tz}=\mathrm{2}\:\mathrm{and}\: \\ $$$$\mathrm{kx}+\mathrm{6y}+\mathrm{5z}−\mathrm{2}=\mathrm{0}\:\mathrm{are}\:\mathrm{parallel}. \\ $$$$\mathrm{find}\:\mathrm{the}\:\mathrm{value}\:\mathrm{of}\:\mathrm{k}\:\mathrm{and}\:\mathrm{t}\: \\ $$

Question Number 95773    Answers: 4   Comments: 0

Question Number 95772    Answers: 0   Comments: 3

Question Number 95770    Answers: 0   Comments: 8

the app doesn′t work properly for me any more. I open it and most of the time the home screen doesn′t show the forum. refreshing or trying to switch to the forum ends up in an endless turning loop...

$$\mathrm{the}\:\mathrm{app}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{work}\:\mathrm{properly}\:\mathrm{for}\:\mathrm{me}\:\mathrm{any} \\ $$$$\mathrm{more}.\:\mathrm{I}\:\mathrm{open}\:\mathrm{it}\:\mathrm{and}\:\mathrm{most}\:\mathrm{of}\:\mathrm{the}\:\mathrm{time}\:\mathrm{the} \\ $$$$\mathrm{home}\:\mathrm{screen}\:\mathrm{doesn}'\mathrm{t}\:\mathrm{show}\:\mathrm{the}\:\mathrm{forum}. \\ $$$$\mathrm{refreshing}\:\mathrm{or}\:\mathrm{trying}\:\mathrm{to}\:\mathrm{switch}\:\mathrm{to}\:\mathrm{the}\:\mathrm{forum} \\ $$$$\mathrm{ends}\:\mathrm{up}\:\mathrm{in}\:\mathrm{an}\:\mathrm{endless}\:\mathrm{turning}\:\mathrm{loop}... \\ $$

Question Number 95768    Answers: 0   Comments: 1

find x such that x≡3 (mod5) x≡5 (mod7) x≡7(mod11)

$$\mathrm{find}\:\mathrm{x}\:\mathrm{such}\:\mathrm{that}\: \\ $$$${x}\equiv\mathrm{3}\:\left(\mathrm{mod5}\right) \\ $$$${x}\equiv\mathrm{5}\:\left(\mathrm{mod7}\right) \\ $$$${x}\equiv\mathrm{7}\left(\mathrm{mod11}\right) \\ $$

Question Number 95767    Answers: 1   Comments: 0

Question Number 95766    Answers: 0   Comments: 1

Solve the equation 2^(2x) − 5x^2 + 4 = 0

$$\mathrm{Solve}\:\mathrm{the}\:\mathrm{equation} \\ $$$$\:\:\:\:\:\:\:\mathrm{2}^{\mathrm{2x}} \:\:−\:\:\mathrm{5x}^{\mathrm{2}} \:\:+\:\:\mathrm{4}\:\:\:=\:\:\:\mathrm{0} \\ $$

Question Number 95765    Answers: 0   Comments: 3

Sum the series: 2((1/(40)) + (1/(20)) + (1/(10)) + ... + n)

$$\mathrm{Sum}\:\mathrm{the}\:\mathrm{series}: \\ $$$$\:\:\:\:\:\:\:\:\:\mathrm{2}\left(\frac{\mathrm{1}}{\mathrm{40}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{20}}\:\:+\:\:\frac{\mathrm{1}}{\mathrm{10}}\:\:+\:\:...\:\:+\:\:\boldsymbol{\mathrm{n}}\right) \\ $$

Question Number 95763    Answers: 0   Comments: 3

∫ x (√(x^3 +1)) dx ?

$$\int\:{x}\:\sqrt{{x}^{\mathrm{3}} +\mathrm{1}}\:{dx}\:? \\ $$

Question Number 95760    Answers: 1   Comments: 0

(d/(d(x)))(W(x))=? W(x) is lambert W function

$$\frac{{d}}{{d}\left({x}\right)}\left({W}\left({x}\right)\right)=? \\ $$$$\:{W}\left({x}\right)\:{is}\:{lambert}\:{W}\:{function} \\ $$

Question Number 95748    Answers: 2   Comments: 0

y′′ = sin x−cos x

$$\mathrm{y}''\:=\:\mathrm{sin}\:\mathrm{x}−\mathrm{cos}\:\mathrm{x} \\ $$

Question Number 95742    Answers: 2   Comments: 2

(0/0)=2 ((100−100)/(100−100))=((10^2 −10^2 )/(10^2 −10^2 ))=(((10+10)(10−10))/(10(10−10))) ((20)/(10))=2 where is the mastike

$$\frac{\mathrm{0}}{\mathrm{0}}=\mathrm{2} \\ $$$$\frac{\mathrm{100}−\mathrm{100}}{\mathrm{100}−\mathrm{100}}=\frac{\mathrm{10}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }{\mathrm{10}^{\mathrm{2}} −\mathrm{10}^{\mathrm{2}} }=\frac{\left(\mathrm{10}+\mathrm{10}\right)\left(\mathrm{10}−\mathrm{10}\right)}{\mathrm{10}\left(\mathrm{10}−\mathrm{10}\right)} \\ $$$$\frac{\mathrm{20}}{\mathrm{10}}=\mathrm{2} \\ $$$$\mathrm{where}\:\mathrm{is}\:\mathrm{the}\:\mathrm{mastike} \\ $$

Question Number 95738    Answers: 1   Comments: 2

∫ ((p−tan x)/(p+tan x)) dx

$$\int\:\frac{\mathrm{p}−\mathrm{tan}\:\mathrm{x}}{\mathrm{p}+\mathrm{tan}\:\mathrm{x}}\:\mathrm{dx}\: \\ $$

Question Number 95846    Answers: 1   Comments: 0

A = { 0, 1, 2, 3, ... , 2020 } how many zero on A ?

$${A}\:=\:\left\{\:\mathrm{0},\:\mathrm{1},\:\mathrm{2},\:\mathrm{3},\:...\:,\:\mathrm{2020}\:\right\} \\ $$$${how}\:\:{many}\:\:{zero}\:\:{on}\:\:{A}\:? \\ $$

Question Number 95726    Answers: 0   Comments: 1

th forces F_1 ,F_2 ,F_3 act at points ith position vectors r_1 ,r_2 ,r_3 where F_1 = (4i + j + 2k)N r_1 = (6i + 4j + k) m F_2 = (i−2j + k)N r_2 = (i + 5j −2k) m F_3 = (−5i + j−3k)N r_3 = (i + j + k) m (a) show that this system reduces to a couple and find its magnitude. (b) if F_3 is removed and replaced with F_4 such that the system is now in equilibrium find the force vector F_4 . (c) the equation of the line of action of F_4 (d) the moment of F_4 about the origin.

$$\mathrm{th}\:\mathrm{forces}\:{F}_{\mathrm{1}} \:,{F}_{\mathrm{2}} ,{F}_{\mathrm{3}} \:\mathrm{act}\:\mathrm{at}\:\mathrm{points}\:\mathrm{ith}\:\mathrm{position}\:\mathrm{vectors}\:{r}_{\mathrm{1}} ,{r}_{\mathrm{2}} ,{r}_{\mathrm{3}} \:\:\mathrm{where} \\ $$$${F}_{\mathrm{1}} \:=\:\left(\mathrm{4}{i}\:+\:{j}\:+\:\mathrm{2}{k}\right){N}\:\:\:\:\:\:\:{r}_{\mathrm{1}} \:=\:\left(\mathrm{6}{i}\:+\:\mathrm{4}{j}\:+\:{k}\right)\:\mathrm{m} \\ $$$${F}_{\mathrm{2}} \:=\:\left({i}−\mathrm{2}{j}\:+\:{k}\right){N}\:\:\:\:\:\:\:\:\:\:\:{r}_{\mathrm{2}} \:=\:\left({i}\:+\:\mathrm{5}{j}\:−\mathrm{2}{k}\right)\:\mathrm{m} \\ $$$$\:{F}_{\mathrm{3}} \:=\:\left(−\mathrm{5}{i}\:+\:{j}−\mathrm{3}{k}\right){N}\:\:\:\:{r}_{\mathrm{3}} \:=\:\left({i}\:+\:{j}\:+\:{k}\right)\:\mathrm{m} \\ $$$$\left(\mathrm{a}\right)\:\mathrm{show}\:\mathrm{that}\:\mathrm{this}\:\mathrm{system}\:\mathrm{reduces}\:\mathrm{to}\:\mathrm{a}\:\mathrm{couple}\:\mathrm{and}\:\mathrm{find}\:\mathrm{its}\:\mathrm{magnitude}. \\ $$$$\left(\mathrm{b}\right)\:\mathrm{if}\:\mathrm{F}_{\mathrm{3}} \:\mathrm{is}\:\mathrm{removed}\:\mathrm{and}\:\mathrm{replaced}\:\mathrm{with}\:{F}_{\mathrm{4}} \:\:\mathrm{such}\:\mathrm{that}\:\mathrm{the}\:\mathrm{system}\:\mathrm{is}\: \\ $$$$\mathrm{now}\:\mathrm{in}\:\mathrm{equilibrium}\:\mathrm{find}\:\mathrm{the}\:\mathrm{force}\:\mathrm{vector}\:{F}_{\mathrm{4}} . \\ $$$$\left(\mathrm{c}\right)\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\mathrm{line}\:\mathrm{of}\:\mathrm{action}\:\mathrm{of}\:{F}_{\mathrm{4}} \\ $$$$\left(\mathrm{d}\right)\:\mathrm{the}\:\mathrm{moment}\:\mathrm{of}\:\mathrm{F}_{\mathrm{4}} \:\mathrm{about}\:\mathrm{the}\:\mathrm{origin}. \\ $$

Question Number 95723    Answers: 3   Comments: 1

Question Number 95722    Answers: 2   Comments: 0

use cylinder ring method y = 2x−1 y = −2x + 3 x = 2 y−axis

$${use}\:{cylinder}\:{ring}\:{method} \\ $$$$ \\ $$$${y}\:=\:\mathrm{2}{x}−\mathrm{1} \\ $$$${y}\:=\:−\mathrm{2}{x}\:+\:\mathrm{3} \\ $$$${x}\:=\:\mathrm{2}\: \\ $$$$ \\ $$$${y}−{axis}\: \\ $$$$ \\ $$$$ \\ $$

Question Number 95707    Answers: 1   Comments: 3

Question Number 95703    Answers: 1   Comments: 1

(1/(cos^2 10^o )) +(1/(sin^2 20^o )) + (1/(sin^2 40^o )) =?

$$\frac{\mathrm{1}}{\mathrm{cos}\:^{\mathrm{2}} \mathrm{10}^{\mathrm{o}} }\:+\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{20}^{\mathrm{o}} }\:+\:\frac{\mathrm{1}}{\mathrm{sin}\:^{\mathrm{2}} \mathrm{40}^{\mathrm{o}} }\:=? \\ $$

Question Number 95699    Answers: 1   Comments: 4

many digits of number 5^8 ×4^5 ×7

$$\mathrm{many}\:\mathrm{digits}\:\mathrm{of}\:\mathrm{number}\: \\ $$$$\mathrm{5}^{\mathrm{8}} ×\mathrm{4}^{\mathrm{5}} ×\mathrm{7}\: \\ $$

Question Number 95697    Answers: 2   Comments: 0

Question Number 95695    Answers: 1   Comments: 0

solve (x+1)y^′ −x^3 y = arctan(2x)

$$\mathrm{solve}\:\left(\mathrm{x}+\mathrm{1}\right)\mathrm{y}^{'} −\mathrm{x}^{\mathrm{3}} \mathrm{y}\:=\:\mathrm{arctan}\left(\mathrm{2x}\right) \\ $$

Question Number 95694    Answers: 1   Comments: 0

solve by laplace transform y^(′′) +3y^′ +2y =e^(−x) withy(0)=1 and y^′ (0) =2

$$\mathrm{solve}\:\mathrm{by}\:\mathrm{laplace}\:\mathrm{transform}\:\:\mathrm{y}^{''} \:+\mathrm{3y}^{'} +\mathrm{2y}\:=\mathrm{e}^{−\mathrm{x}} \:\:\mathrm{withy}\left(\mathrm{0}\right)=\mathrm{1}\:\mathrm{and}\:\mathrm{y}^{'} \left(\mathrm{0}\right)\:=\mathrm{2} \\ $$

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