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## numerical integration – How to increase the precision (quadruple-precision or double precision) using Implicit Runge Kutta method

I’m trying to solve a system of 24 non-linear equations, using Implicit Runge Kutta method. Code is working very smoothly with MachinePrecision. But, I want higher precision as much as I can. First, I change all initial conditions upto 34 decimal places, for example, using ` EE = N(976/1000, 40)` for `energy = EE = 0.976`, and same for other initial conditions.

I’m calculating initial data using `FindRoot`, where I’m using `WorkingPrecision -> 39`, then I have precision `34` of some variables, and some have `36`, etc.
I don’t know how to convert all initial conditions in the same precision, therefore I used like this.

Is there any command that we can use that convert all initial data in the same precision?

Moreover, I want to solve this system with precision higher than the `MachinePrecision`. I have tried using precision ` AccuracyGoal -> 18, PrecisionGoal -> 18, etc` or using `WorkingPrecision -> 40`, but code gives errors.

Can anyone please help me to solve the system with precision higher than the `MachinePrecision` or `quadruple-precision `.

I’m posting my code here

``````n = 4;
AA(r_) := (1 - (2 M)/r); M = 1;

gtt(r_, θ_) := -AA(r); grr(r_, θ_) := 1/AA(r);
gθθ(r_, θ_) := r^2;
gϕϕ(r_, θ_) := r^2 Sin(θ)^2;
gUtt(r_, θ_) := 1/gtt(r, θ);
gUrr(r_, θ_) := 1/grr(r, θ);
gUθθ(r_, θ_) := 1/gθθ(r, θ);
gUϕϕ(r_, θ_) := 1/gϕϕ(r, θ);

glo = FullSimplify({ {gtt(r, θ), 0, 0, 0}, {0,
grr(r, θ), 0, 0}, {0, 0, gθθ(r, θ),
0}, {0, 0, 0, gϕϕ(r, θ)}});
gup = Simplify(Inverse(glo));
dglo = Simplify(Det(glo));
crd = {t, r, θ, ϕ};

Xup = {t(τ), r(τ), θ(τ), ϕ(τ)};
Vup = {Vt, Vr, Vθ, Vϕ};(* v^μ vector *)
Pup = {Pt(τ), Pr(τ), Pθ(τ), Pϕ(τ)};

Sup = {{Stt(τ), Str(τ), Stθ(τ),
Stϕ(τ)},
{Srt(τ), Srr(τ), Srθ(τ), Srϕ(τ)},
{Sθt(τ), Sθr(τ), Sθθ(τ),
Sθϕ(τ)},
{Sϕt(τ), Sϕr(τ), Sϕθ(τ),
Sϕϕ(τ)}};

christoffel =
Simplify(Table((1/2)*
Sum((gup((i, s)))*(D(glo((s, k)), crd((j)) ) +
D(glo((s, j)), crd((k)) ) - D(glo((j, k)), crd((s)) )), {s,
1, n}), {i, 1, n}, {j, 1, n}, {k, 1, n})
);

riemann = Simplify(
Table(
D(christoffel((i, j, l)), crd((k)) ) -
D(christoffel((i, j, k)), crd((l)) ) +
Sum(christoffel((s, j, l)) christoffel((i, k, s)) -
christoffel((s, j, k)) christoffel((i, l, s)),
{s, 1, n}), {i, 1, n}, {j, 1, n}, {k, 1, n}, {l, 1, n}) );

loriemann =
Simplify(Table(
Sum(glo((i, m))*riemann((m, j, k, l)), {m, 1, n}), {i, 1, n}, {j,
1, n}, {k, 1, n}, {l, 1, n}) );

EOM1 = Table( D(Xup((a)), τ) == Vup((a)) , {a, 1, n});

EOM2 = Table(
D(Pup((a)), τ) + !(
*UnderoverscriptBox((∑), (b = 1), (n))(
*UnderoverscriptBox((∑), (c =
1), (n))christoffel((()(a, b, c)()))*
Pup((()(b)()))*Vup((()(c)())))) == -(1/2) !(
*UnderoverscriptBox((∑), (b = 1), (n))(
*UnderoverscriptBox((∑), (c = 1), (n))(
*UnderoverscriptBox((∑), (d = 1), (n))riemann((()(a,
b, c, d)()))*Vup((()(b)()))*
Sup((()(c, d)()))))),
{a, 1, n});

EOM3 = Table(
D(Sup((a, b)), τ) + !(
*UnderoverscriptBox((∑), (c = 1), (n))(
*UnderoverscriptBox((∑), (d =
1), (n))christoffel((()(a, c, d)()))*
Sup((()(c, b)()))*Vup((()(d)())))) + !(
*UnderoverscriptBox((∑), (c = 1), (n))(
*UnderoverscriptBox((∑), (d =
1), (n))christoffel((()(b, c, d)()))*
Sup((()(a, c)()))*Vup((()(d)())))) ==
Pup((a))*Vup((b)) - Pup((b))*Vup((a)),
{a, 1, n}, {b, 1, n});

Wfactor = Simplify(4*μ^2 + !(
*UnderoverscriptBox((∑), (i = 1), (4))(
*UnderoverscriptBox((∑), (j = 1), (4))(
*UnderoverscriptBox((∑), (k = 1), (4))(
*UnderoverscriptBox((∑), (l =
1), (4))((loriemann((()(i, j, k,
l)()))*((Sup((()(i, j)()))))* ((Sup((()(k,
l)())))))))))));

Wvec = Simplify(Table(2/(μ*Wfactor)*(!(
*UnderoverscriptBox((∑), (i = 1), (4))(
*UnderoverscriptBox((∑), (k = 1), (4))(
*UnderoverscriptBox((∑), (m = 1), (4))(
*UnderoverscriptBox((∑), (l = 1), (4))Sup((()(j,
i)()))*
Pup((()(k)()))*((loriemann((()(i, k, l,
m)()))))*((Sup((()(l, m)()))))))))), {j,
1, n}));

NN = 1/Sqrt(1 - !(
*UnderoverscriptBox((∑), (i = 1), (4))(
*UnderoverscriptBox((∑), (k =
1), (4))((glo((()(i, k)()))))*Wvec((()(i)()))*
Wvec((()(k)())))));

{Vt, Vr, Vθ, Vϕ} = NN (Wvec + Pup);

EOM = Flatten(
Join({EOM1, EOM2, EOM3} /.
r -> r(τ) /. θ -> θ(τ) /.
Derivative(1)(r(τ))(τ) -> Derivative(1)(r)(τ) /.
Derivative(1)(θ(τ))(τ) ->
Derivative(1)(θ)(τ)));

(**********************************************************************************************************************************)

Sθϕ0 = (
LL Cot(θ0))/r0^2; Srθ0 = -(pθ0/
r0); Srϕ0 = -(1/
r0) (-LL + pϕ0/Sin(θ0)^2); Str0 = -(1/(
r0*pt0)) (pθ0^2 + pϕ0^2/Sin(θ0)^2 - LL*pϕ0);
Stθ0 =
1/(r0*pt0)*(pr0*pθ0 + (LL * pϕ0)/r0*
Cot(θ0)); Stϕ0 = -(1/(
r0*pt0))*(LL*pr0 - (pr0*pϕ0)/
Sin(θ0)^2 + (LL*pθ0)/r0*Cot(θ0));

glo0 = Simplify(glo /. {r -> r0, θ -> θ0});
gup0 = Simplify(Inverse(glo0));

plo0 = {pt0, pr0, pθ0, pϕ0};
Sup0 = {{0, Str0, Stθ0, Stϕ0},
{-Str0, 0, Srθ0, Srϕ0},
{-Stθ0, -Srθ0, 0, Sθϕ0},
{-Stϕ0, -Srϕ0, -Sθϕ0, 0}};

F0 = Simplify(μ^2 + !(
*UnderoverscriptBox((∑), (a = 1), (4))(
*UnderoverscriptBox((∑), (b =
1), (4))((((gup0((()(a,
b)()))))*((plo0((()(a)()))))*((plo0((()(b)
())))))))));
F1 = EE + pt0 -
M/r0^2*(1/(
r0*pt0)*(pθ0^2 + pϕ0^2/Sin(θ0)^2 -
LL*pϕ0));
F2 = Simplify(SS^2 - 1/2*(!(
*UnderoverscriptBox((∑), (a = 1), (4))(
*UnderoverscriptBox((∑), (b = 1), (4))(
*UnderoverscriptBox((∑), (c = 1), (4))(
*UnderoverscriptBox((∑), (d =
1), (4))((glo0((()(a, b)()))*
glo0((()(c, d)()))*Sup0((()(a, c)()))*
Sup0((()(b, d)()))))))))));

pth = N(-(μ^2 + (gup0((1, 1)) )*(- EE)^2 +
gup0((2, 2))  * (0)^2 + (gup0((4, 4))  )* (LL)^2)/(gup0((3,
3))  ), 40);
pθGR = N(If(pth < 0, 1, Sqrt(pth)), 40);

GiveMePSpoints(SS_, r0_, θ0_, EE_, LL_, konec_, color_) := (
Clear(Str0, Stθ0, Stϕ0, Srθ0, Srϕ0,
Sθϕ0, pt0, pr0, pθ0, pϕ0);
μ = N(1, 40); pr0 = N(0, 40);

{pt0, pr0, pθ0,
pϕ0} = {pt0, pr0, pθ0, pϕ0} /.
FindRoot({F0 == 0, F1 == 0,
F2 == 0}, {{pt0, -EE}, {pθ0, pθGR}, {pϕ0,
LL}}, WorkingPrecision -> 39);

Sθϕ0 = (LL Cot(θ0))/r0^2;
Srθ0 = -(pθ0/r0);
Srϕ0 = -(1/r0) (-LL + pϕ0/Sin(θ0)^2);
Str0 = -(1/(
r0*pt0)) (pθ0^2 + pϕ0^2/Sin(θ0)^2 -
LL*pϕ0);
Stθ0 =
1/(r0*pt0)*(pr0*pθ0 + (LL * pϕ0)/r0*Cot(θ0));
Stϕ0 = -(1/(
r0*pt0))*(LL*pr0 - (pr0*pϕ0)/
Sin(θ0)^2 + (LL*pθ0)/r0*Cot(θ0));

INT1 = {t(0) == 0,
r(0) == r0, θ(0) == θ0, ϕ(0) == 0};
INT2 = {Pt(0) == gUtt(r0, θ0) pt0,
Pr(0) == gUrr(r0, θ0) pr0,
Pθ(0) == gUθθ(r0, θ0) pθ0,
Pϕ(0) == gUϕϕ(r0, θ0) pϕ0};
INT3 = {{Stt(0) == 0, Str(0) == Str0, Stθ(0) == Stθ0,
Stϕ(0) == Stϕ0},
{Srt(0) == -Str0, Srr(0) == 0, Srθ(0) == Srθ0,
Srϕ(0) == Srϕ0},
{Sθt(0) == -Stθ0, Sθr(0) == -Srθ0,
Sθθ(0) == 0,
Sθϕ(0) == Sθϕ0},
{Sϕt(0) == -Stϕ0, Sϕr(0) == -Srϕ0,
Sϕθ(0) == -Sθϕ0, Sϕϕ(0) == 0}};

INT = Flatten(Join({INT1, INT2, INT3}));

SetSystemOptions(
"NDSolveOptions" -> "DefaultSolveTimeConstraint" -> 100.`);

data =
Reap(NDSolve(
Flatten(Join({EOM, INT})), {t, r, θ, ϕ, Pt, Pr,
Pθ, Pϕ, Stt, Str, Stθ, Stϕ, Srt, Srr,
Srθ, Srϕ,
Sθt, Sθr, Sθθ, Sθϕ,
Sϕt, Sϕr, Sϕθ, Sϕϕ}, {τ,
0, konec},
StartingStepSize ->
0.1, {Method -> {"EventLocator",
"Event" -> θ(τ) - Pi/2,
"EventAction" :> Sow({r(τ), Pr(τ)}),
"Method" -> {"FixedStep",
"Method" -> {"ImplicitRungeKutta", "DifferenceOrder" -> 10,
"ImplicitSolver" -> {"Newton",
AccuracyGoal -> MachinePrecision,
PrecisionGoal -> MachinePrecision,
"IterationSafetyFactor" -> 1}}}}}));

psdata = Take(data((-1, 1)));
ListPlot(psdata, PlotStyle -> {{PointSize(0.003), color}},
Frame -> True, Axes -> False, BaseStyle -> 15, ImageSize -> 350,
AspectRatio -> 1, PlotRange -> range)

);

θ0 =
N(Pi/2, 40); r0 =
4.2521489655172413793103448275862068965517241379310344827585`40.;
EE = N(976/1000, 40); LL = N(38/10, 40); SS = N(10^-4, 40);
konec = 7 10^5; range = {{4.25213, 4.25218}, {-15 10^-6, 15 10^-6}};
GiveMePSpoints(SS, r0, θ0, EE, LL, konec, Black)
``````

## differential forms – Double exterior derivative

In the proof of $$d(domega)=0$$ for all k-form $$omega=f_Idx^I=f_I wedge dx^I$$. We compute
$$d(domega)=d(d(f_I wedge dx^I))=df_Iwedge d(dx^I) + d(df_I)wedge dx^I=df_Iwedge d(dx^I)=0$$

Where the third equality is because the double exterior derivative of any 0-form is zero since mixed partial commutes. Now in the last equality, the proof goes:

$$d(dx^I)=d(1)wedge dx^I=0$$

But I don’t understand why $$d(dx^I)=d(1)wedge dx^I$$. I know that $$dx^I$$ will be the sum of the wedge of k number of 1-forms but I don’t think that is relevant here.

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## How should I approach this double angle problem?

sin a=-3/5, pi<a<3pi/2;cos b=12/13, 3pi/2<b<2pi

a)sin(a+b)
b)cos(a+b)
c)sin(a-b)
d)tan(a+b)
e)sin(2a)
f)cos(2b)
g)sin b/2
h)cos a/2
I know that you have to divide the pi<a<3pi/2 by two and that will give you the a/2. I just need to understand what to do afterward. Got this on a test tomorrow so any help would be appreciated. Also if you recommend any good websites where I can practice similar problems or videos on these types of questions that would help as well.

## functional analysis – Showing double orthogonal complement of \$X\$ is \$X\$

I know this has been asked many times but I know there’s a way using the orthogonal projection.

So if we have $$H$$ a Hilbert and then $$X subset H$$ a closed linear subspace, then for any $$x in H$$ there are unique $$u in X$$ , $$v in X^{perp }$$ such that $$x=u+v$$.

How can I use this to show that $$(X^{perp })^{perp } subset X$$?

## c# – Double em comparação

Olá, eu queria saber o porquê de o resultado estar dando como aprovado quando eu escrevo até 6.1, já que o código fala para só dar o aprovado quando o valor da nota for maior ou igual a 7, de 6.1 para cima está dando aprovado mas se eu escrever de 6 pra baixo dá como reprovado.

``````        double nota = Convert.ToDouble(Console.ReadLine());

if(nota >= 7)
{
}

``````

## dnd 5e – Double Range for Storm Sphere with Spell Sniper

If we break down the instruction in the Feat, I’d conclude that it won’t work:

• When you cast a spell that requires you to make an attack roll, the spell’s range is doubled.

Making a spell attack as a bonus action is not using the “Cast a spell” action, so the extra spell attacks wouldn’t count. And the stated spell does not require you to make an attack roll, it merely has some additional effects that do.

During the actual casting (which is what brings forth the sphere) no attack rolls are made, so you can’t benefit from spell sniper there.

After the spell is available, you are making attack rolls but they’re not coming from the “cast a spell” action, so you can’t benefit from spell sniper there either.

## dnd 5e – Can the arcane/druidic focus staff double as quarterstaff?

Probably.

The entries for Staff and Quarterstaff are different in the equipment list. Under the Arcane Focus section you get:

a specially constructed staff (p. 48).

this is distinctly different (and quite a bit more expensive) than a typical quarter staff (the arcane focus version runs 5gp vs 2sp for a quarterstaff). So no, a quarterstaff cannot double for an arcane foci staff, or the other way round (at least without further clarification).

In the Druidic Focus section you get

a staff drawn whole out of a living tree (p 151 PHB)

The druid is much closer to nature and uses a druidic focus rather than an arcane one. We see that the staff for the druidic focus has a far more…utilitarian description. There is nothing here that makes it not a quarterstaff (though it may be rougher than a typical staff). My guess is most Druids would have no problem wielding their staff as a quarterstaff. I mean who is going to argue with you, you have a quarterstaff.

Last thing, there is an argument to be made for allowing implement staves to be treated as quarterstaves under the improved weapon rules on p 46 of BD&D

In many cases, an improvised weapon is similar to an actual weapon and can be treated as such. For example, a table leg is akin to a club. At the DM’s option, a character proficient with a weapon can use a similar object as if it were that weapon and use his or her proficiency bonus.

This means that if your staff focus resembles a quarterstaff (for some value of resemble I guess), then it could be treated as one and your proficiency bonus would apply to attacks with it (and it’d get a d6 for the damage die).

## Double lines division representation

How to make a fraction to display a numerator on top of the denominator in Google Sheet?