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Question Number 225054 Answers: 0 Comments: 1

Question Number 225047 Answers: 2 Comments: 0

∫ x^x dx

$$\int\:{x}^{{x}} \:{dx} \\ $$

Question Number 225020 Answers: 1 Comments: 1

tgx+tgy+tgz=A tg^3 x+tg^3 y+tg^3 z=?

$$\:\:\:{tgx}+{tgy}+{tgz}={A} \\ $$$$\:\:\:{tg}^{\mathrm{3}} {x}+{tg}^{\mathrm{3}} {y}+{tg}^{\mathrm{3}} {z}=? \\ $$

Question Number 225003 Answers: 0 Comments: 22

The new symbols are in progress. New update will be release by end of this month with - Close integration symbols - Italic greek capital letters - Ability to upload GIF images If anyone has any other improvement suggestion please email us or comment here

$$\mathrm{The}\:\mathrm{new}\:\mathrm{symbols}\:\mathrm{are}\:\mathrm{in} \\ $$$$\mathrm{progress}.\:\mathrm{New}\:\mathrm{update}\:\mathrm{will}\:\mathrm{be}\:\mathrm{release} \\ $$$$\mathrm{by}\:\mathrm{end}\:\mathrm{of}\:\mathrm{this}\:\mathrm{month}\:\mathrm{with} \\ $$$$-\:\mathrm{Close}\:\mathrm{integration}\:\mathrm{symbols} \\ $$$$-\:\mathrm{Italic}\:\mathrm{greek}\:\mathrm{capital}\:\mathrm{letters} \\ $$$$-\:\mathrm{Ability}\:\mathrm{to}\:\mathrm{upload}\:\mathrm{GIF}\:\mathrm{images} \\ $$$$ \\ $$$$\mathrm{If}\:\mathrm{anyone}\:\mathrm{has}\:\mathrm{any}\:\mathrm{other}\:\mathrm{improvement} \\ $$$$\mathrm{suggestion}\:\mathrm{please}\:\mathrm{email}\:\mathrm{us}\:\mathrm{or}\:\mathrm{comment} \\ $$$$\mathrm{here} \\ $$

Question Number 224989 Answers: 0 Comments: 0

Calculate D_n = determinant (((x_1 +a_1 ^2 ),(a_1 a_2 ),…,(a_1 a_n )),((a_2 a_1 ),(x_2 +a_2 ^2 ),…,(a_2 a_n )),(⋮,⋮,⋱,⋮),((a_n a_1 ),(a_n a_2 ),…,(x_n +a_n ^2 )))

$$\mathrm{Calculate} \\ $$$${D}_{{n}} =\begin{vmatrix}{{x}_{\mathrm{1}} +{a}_{\mathrm{1}} ^{\mathrm{2}} }&{{a}_{\mathrm{1}} {a}_{\mathrm{2}} }&{\ldots}&{{a}_{\mathrm{1}} {a}_{{n}} }\\{{a}_{\mathrm{2}} {a}_{\mathrm{1}} }&{{x}_{\mathrm{2}} +{a}_{\mathrm{2}} ^{\mathrm{2}} }&{\ldots}&{{a}_{\mathrm{2}} {a}_{{n}} }\\{\vdots}&{\vdots}&{\ddots}&{\vdots}\\{{a}_{{n}} {a}_{\mathrm{1}} }&{{a}_{{n}} {a}_{\mathrm{2}} }&{\ldots}&{{x}_{{n}} +{a}_{{n}} ^{\mathrm{2}} }\end{vmatrix} \\ $$

Question Number 224985 Answers: 0 Comments: 0

A dairy man gives ada and amaka eight litre of milk to share and ada carry five litre container and amaka Carry three litre container how can ada and amaka share the eight litre of milk equally.

$$\mathrm{A}\:\mathrm{dairy}\:\mathrm{man}\:\mathrm{gives}\:\mathrm{ada}\:\mathrm{and}\: \\ $$$$\mathrm{amaka}\:\mathrm{eight}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{milk}\:\mathrm{to}\:\mathrm{share}\: \\ $$$$\mathrm{and}\:\mathrm{ada}\:\mathrm{carry}\:\mathrm{five}\:\mathrm{litre}\:\mathrm{container} \\ $$$$\mathrm{and}\:\mathrm{amaka}\:\mathrm{Carry}\:\mathrm{three}\:\mathrm{litre} \\ $$$$\mathrm{container}\:\mathrm{how}\:\mathrm{can}\:\mathrm{ada}\:\mathrm{and}\:\mathrm{amaka} \\ $$$$\mathrm{share}\:\mathrm{the}\:\mathrm{eight}\:\mathrm{litre}\:\mathrm{of}\:\mathrm{milk}\:\mathrm{equally}. \\ $$

Question Number 224982 Answers: 0 Comments: 4

We have made some improvement is search. Search matching is done using AI so you should now get better search results.

$$\mathrm{We}\:\mathrm{have}\:\mathrm{made}\:\mathrm{some}\:\mathrm{improvement} \\ $$$$\mathrm{is}\:\mathrm{search}. \\ $$$$\mathrm{Search}\:\mathrm{matching}\:\mathrm{is}\:\mathrm{done}\:\mathrm{using}\:\mathrm{AI} \\ $$$$\mathrm{so}\:\mathrm{you}\:\mathrm{should}\:\mathrm{now}\:\mathrm{get}\:\mathrm{better}\: \\ $$$$\mathrm{search}\:\mathrm{results}. \\ $$

Question Number 224966 Answers: 2 Comments: 2

cos(π/7) − cos((2π)/7) + cos((3π)/7) = ? Help me please

$$ \\ $$$$\:\:\:{cos}\frac{\pi}{\mathrm{7}}\:−\:{cos}\frac{\mathrm{2}\pi}{\mathrm{7}}\:+\:{cos}\frac{\mathrm{3}\pi}{\mathrm{7}}\:=\:? \\ $$$$\:\:\:{Help}\:{me}\:{please} \\ $$$$ \\ $$

Question Number 224951 Answers: 1 Comments: 0

Question Number 224948 Answers: 2 Comments: 1

∫ (√x)sin x dx

$$\int\:\sqrt{{x}}\mathrm{sin}\:{x}\:{dx} \\ $$

Question Number 224947 Answers: 0 Comments: 0

REDACTED

$${REDACTED}\: \\ $$

Question Number 224938 Answers: 0 Comments: 0

S;(x,y,z)∈R^3 ∣x^2 +y^2 +z^2 =r^2 Find Geodesic equation ((d )/dt)∙(dx^i /dt)+Γ_(jk) ^i (dx^j /dt)∙(dx^k /dt)=0 , Γ_(jk) ^i =g^(il) ((∂g_(kl) /∂x_j )−(∂g_(jl) /∂x_k )+(∂g_(jk) /∂x_l ))

$$\mathcal{S};\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} \mid{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\mathrm{Find}\:\mathrm{Geodesic}\:\mathrm{equation} \\ $$$$\frac{\mathrm{d}\:\:}{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{{i}} }{\mathrm{d}{t}}+\Gamma_{{jk}} ^{{i}} \frac{\mathrm{d}{x}^{{j}} }{\mathrm{d}{t}}\centerdot\frac{\mathrm{d}{x}^{{k}} }{\mathrm{d}{t}}=\mathrm{0}\:\:,\:\:\Gamma_{{jk}} ^{{i}} =\mathrm{g}^{{il}} \left(\frac{\partial\mathrm{g}_{{kl}} }{\partial{x}_{{j}} }−\frac{\partial\mathrm{g}_{{jl}} }{\partial{x}_{{k}} }+\frac{\partial\mathrm{g}_{{jk}} }{\partial{x}_{{l}} }\right) \\ $$

Question Number 224927 Answers: 1 Comments: 2

Question Number 224926 Answers: 1 Comments: 0

∫ ((sin x)/x) dx

$$\int\:\frac{\mathrm{sin}\:{x}}{{x}}\:{dx} \\ $$

Question Number 224924 Answers: 1 Comments: 1

∫_0 ^(2π) ((xsin^(2n) x)/(sin^(2n) x+cos^(2n) x)) dx=π^2 (prove)

$$\underset{\mathrm{0}} {\overset{\mathrm{2}\pi} {\int}}\frac{{x}\mathrm{sin}^{\mathrm{2}{n}} {x}}{\mathrm{sin}^{\mathrm{2}{n}} {x}+\mathrm{cos}^{\mathrm{2}{n}} {x}}\:{dx}=\pi^{\mathrm{2}} \:\left({prove}\right) \\ $$

Question Number 224922 Answers: 1 Comments: 0

Domain f(x)=(√(log _e (x^2 −6x+6)))

$${Domain} \\ $$$${f}\left({x}\right)=\sqrt{\mathrm{log}\:_{{e}} \left({x}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}\right)} \\ $$

Question Number 224920 Answers: 0 Comments: 0

∫_( 0) ^( 1) ((x ln(1 + x) Li_2 (x))/(1 + x^2 )) dx

$$\int_{\:\mathrm{0}} ^{\:\mathrm{1}} \:\frac{\mathrm{x}\:\mathrm{ln}\left(\mathrm{1}\:\:\:+\:\:\:\mathrm{x}\right)\:\mathrm{Li}_{\mathrm{2}} \left(\mathrm{x}\right)}{\mathrm{1}\:\:\:\:+\:\:\:\:\mathrm{x}^{\mathrm{2}} }\:\mathrm{dx} \\ $$

Question Number 224919 Answers: 0 Comments: 0

T(x,y);R^2 →R , T(x,y)=((x/a))^2 +((y/b))^2 K(x,y)=(4/(a^2 b^2 (1+((4x^2 )/a^4 )+((4y^2 )/b^4 ))^2 )) ∫_( S) dr^2 K=? S; sheprical coordinate and associate with Euler characteristic χ(T)

$${T}\left({x},{y}\right);\mathbb{R}^{\mathrm{2}} \rightarrow\mathbb{R}\:,\:\:{T}\left({x},{y}\right)=\left(\frac{{x}}{{a}}\right)^{\mathrm{2}} +\left(\frac{{y}}{{b}}\right)^{\mathrm{2}} \\ $$$${K}\left({x},{y}\right)=\frac{\mathrm{4}}{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+\frac{\mathrm{4}{x}^{\mathrm{2}} }{{a}^{\mathrm{4}} }+\frac{\mathrm{4}{y}^{\mathrm{2}} }{{b}^{\mathrm{4}} }\right)^{\mathrm{2}} } \\ $$$$\int_{\:\mathcal{S}} \mathrm{d}\boldsymbol{\mathrm{r}}^{\mathrm{2}} {K}=? \\ $$$$\mathcal{S};\:\mathrm{sheprical}\:\mathrm{coordinate} \\ $$$$\:\: \\ $$$$\:\mathrm{and}\:\mathrm{associate}\:\mathrm{with}\:\mathrm{Euler}\:\mathrm{characteristic}\:\chi\left({T}\right) \\ $$

Question Number 224908 Answers: 1 Comments: 2

Hello Everyone We recently did an update that cause view older button to be partially hidden behind the navigation button. We will release an update soon. workaround Tap + sign at top of screen to open editor and then press back. Then the button will be shown correctly Part of the button is still visible and you can press at the edge. Fix We will upload another update today amd it will be available to download in two to three days.

$$\mathrm{Hello}\:\mathrm{Everyone} \\ $$$$\mathrm{We}\:\mathrm{recently}\:\mathrm{did}\:\mathrm{an}\:\mathrm{update}\:\mathrm{that} \\ $$$$\mathrm{cause}\:\mathrm{view}\:\mathrm{older}\:\mathrm{button}\:\mathrm{to}\:\mathrm{be}\:\mathrm{partially} \\ $$$$\mathrm{hidden}\:\mathrm{behind}\:\mathrm{the}\:\mathrm{navigation}\:\mathrm{button}. \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{release}\:\mathrm{an}\:\mathrm{update}\:\mathrm{soon}. \\ $$$$ \\ $$$$\boldsymbol{\mathrm{workaround}} \\ $$$$\mathrm{Tap}\:+\:\mathrm{sign}\:\mathrm{at}\:\mathrm{top}\:\mathrm{of}\:\mathrm{screen}\:\mathrm{to}\:\mathrm{open} \\ $$$$\mathrm{editor}\:\mathrm{and}\:\mathrm{then}\:\mathrm{press}\:\mathrm{back}.\:\mathrm{Then}\:\mathrm{the} \\ $$$$\mathrm{button}\:\mathrm{will}\:\mathrm{be}\:\mathrm{shown}\:\mathrm{correctly} \\ $$$$\mathrm{Part}\:\mathrm{of}\:\mathrm{the}\:\mathrm{button}\:\mathrm{is}\:\mathrm{still}\:\mathrm{visible}\:\mathrm{and} \\ $$$$\mathrm{you}\:\mathrm{can}\:\mathrm{press}\:\mathrm{at}\:\mathrm{the}\:\mathrm{edge}. \\ $$$$\boldsymbol{\mathrm{Fix}} \\ $$$$\mathrm{We}\:\mathrm{will}\:\mathrm{upload}\:\mathrm{another}\:\mathrm{update}\:\mathrm{today} \\ $$$$\mathrm{amd}\:\mathrm{it}\:\mathrm{will}\:\mathrm{be}\:\mathrm{available}\:\mathrm{to}\:\mathrm{download} \\ $$$$\mathrm{in}\:\mathrm{two}\:\mathrm{to}\:\mathrm{three}\:\mathrm{days}. \\ $$

Question Number 224907 Answers: 1 Comments: 0

Let B={(x,y,z)∈R^3 :x^2 +y^2 +z^2 ≤1} and define u(x,y,z)=sin ((1−x^2 −y^2 −z^2 )^2 ) for (x,y,z)∈B. Then the value of ∫∫∫_(B) ((∂^2 u/∂x^2 )+(∂^2 u/∂x^2 )+(∂^2 u/∂z^2 ))dxdydz is ..........

$${Let}\:{B}=\left\{\left({x},{y},{z}\right)\in\mathbb{R}^{\mathrm{3}} :{x}^{\mathrm{2}} +{y}^{\mathrm{2}} +{z}^{\mathrm{2}} \leqslant\mathrm{1}\right\} \\ $$$${and}\:{define}\:{u}\left({x},{y},{z}\right)=\mathrm{sin}\:\left(\left(\mathrm{1}−{x}^{\mathrm{2}} −{y}^{\mathrm{2}} −{z}^{\mathrm{2}} \right)^{\mathrm{2}} \right) \\ $$$${for}\:\left({x},{y},{z}\right)\in{B}. \\ $$$${Then}\:{the}\:{value}\:{of} \\ $$$$\underset{{B}} {\int\int\int}\left(\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{x}^{\mathrm{2}} }+\frac{\partial^{\mathrm{2}} {u}}{\partial{z}^{\mathrm{2}} }\right){dxdydz}\:{is}\:.......... \\ $$

Question Number 224915 Answers: 1 Comments: 0

evaluate ∫_C (3y^2 +2z^2 )dx+(6x−10z)y dy +(4xz−5y^2 )dz along the portion from (1,0,1) to (3,4,5) of the curve C, which is the intersection of the two surfaces z^2 =x^2 +y^2 and z=y+1

$${evaluate}\: \\ $$$$\int_{{C}} \left(\mathrm{3}{y}^{\mathrm{2}} +\mathrm{2}{z}^{\mathrm{2}} \right){dx}+\left(\mathrm{6}{x}−\mathrm{10}{z}\right){y}\:{dy}\:+\left(\mathrm{4}{xz}−\mathrm{5}{y}^{\mathrm{2}} \right){dz} \\ $$$${along}\:{the}\:{portion}\:{from}\:\left(\mathrm{1},\mathrm{0},\mathrm{1}\right)\:{to}\:\left(\mathrm{3},\mathrm{4},\mathrm{5}\right)\:{of} \\ $$$${the}\:{curve}\:{C}, \\ $$$${which}\:{is}\:{the}\:{intersection}\:{of}\:{the} \\ $$$${two}\:{surfaces}\:{z}^{\mathrm{2}} ={x}^{\mathrm{2}} +{y}^{\mathrm{2}} \:{and}\:{z}={y}+\mathrm{1} \\ $$

Question Number 224892 Answers: 1 Comments: 1

lim_(n→0) ∫_0 ^1 e^x^2 sin (nx)dx=?

$$\underset{{n}\rightarrow\mathrm{0}} {\mathrm{lim}}\:\underset{\mathrm{0}} {\overset{\mathrm{1}} {\int}}\:{e}^{{x}^{\mathrm{2}} } \mathrm{sin}\:\left({nx}\right){dx}=? \\ $$

Question Number 224888 Answers: 0 Comments: 2

Question Number 224886 Answers: 1 Comments: 2

Find for acute θ, sin θ and cos θ in terms of 0<k<1, if ((sin θ(1−cos θ))/(cos θ(1−sin θ)))=k.

$${Find}\:{for}\:{acute}\:\theta,\:\mathrm{sin}\:\theta\:{and}\:\mathrm{cos}\:\theta \\ $$$${in}\:{terms}\:{of}\:\mathrm{0}<{k}<\mathrm{1}, \\ $$$${if}\:\:\:\:\frac{\mathrm{sin}\:\theta\left(\mathrm{1}−\mathrm{cos}\:\theta\right)}{\mathrm{cos}\:\theta\left(\mathrm{1}−\mathrm{sin}\:\theta\right)}={k}. \\ $$

Question Number 224884 Answers: 0 Comments: 0

solve partial differantial equation ψ=−k^2 g^(μν) ▽_α ∂_β ψ , k≠0

$$\mathrm{solve}\:\mathrm{partial}\:\mathrm{differantial}\:\mathrm{equation} \\ $$$$\psi=−{k}^{\mathrm{2}} {g}^{\mu\nu} \bigtriangledown_{\alpha} \partial_{\beta} \psi\:,\:{k}\neq\mathrm{0} \\ $$$$\: \\ $$

Question Number 224883 Answers: 0 Comments: 0

∫ vol(g^ )=∫_( V) (√(det g_(μν) )) dx^1 ∧dx^2 ∧dx^3 parametric Surface S^→ (u,v,w);R^3 →R^3 S^→ (r,θ,ρ) { ((rsin(θ)cos(ρ))),((rsin(θ)sin(ρ))),((rcos(θ))) :} find metric tensor g_(μν) = ((g_(11) ,g_(12) ,g_(13) ),(g_(21) ,g_(22) ,g_(23) ),(g_(31) ,g_(32) ,g_(33) ) ) Describe it in the same as ds^2 =g_(μν) dx^μ dx^ν ds^2 =(dr dθ dρ) ((g_(11) ,g_(12) ,g_(13) ),(g_(21) ,g_(22) ,g_(23) ),(g_(31) ,g_(32) ,g_(33) ) ) ((dr),(dθ),(dρ) ) and find volume V=∫ vol(g)

Pg 1 Pg 2 Pg 3 Pg 4 Pg 5 Pg 6 Pg 7 Pg 8 Pg 9 Pg 10

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