Løs operatornamede^J=sqrt[3]{10^29/10^5}

\operatorname { de } ^ { J } = \sqrt[ 3 ] { \frac { 10 ^ { 29 } } { 10 ^ { 5 } } }

Løs for J

J=\frac{8\ln(10)}{\ln(d)+1}

d\neq \frac{1}{e}\text{ and }d>0

Løs for d

\left\{\begin{matrix}d=-\frac{100000000^{\frac{1}{J}}}{e}\text{, }&J\neq 0\text{ and }Numerator(J)\text{bmod}2=0\text{ and }Denominator(J)\text{bmod}2=1\\d=\frac{100000000^{\frac{1}{J}}}{e}\text{, }&J\neq 0\end{matrix}\right,

Spørrelek

\operatorname { de } ^ { J } = \sqrt[ 3 ] { \frac { 10 ^ { 29 } } { 10 ^ { 5 } } }

Lignende problemer fra nettsøk

finding complex n-th root

https://math.stackexchange.com/questions/2203005/finding-complex-n-th-root

You are wrong here (\sqrt[8]{5}cis\frac{\pi}{4})^8=5\left( \cos { 2\pi +i\sin { 2\pi } } \right) =5

Definition of Brinell hardness

https://physics.stackexchange.com/questions/440286/definition-of-brinell-hardness

I haven't been able to find the details of how John Brinell developed his hardness test, but here is my guess , based on how I would proceed if given the task: He wanted an easily replicatable test. ...

Why the map w=-1/2(z+1/z) maps upper half disk to upper half plane

https://math.stackexchange.com/q/2679727

There is an alternative approach, which avoids the hassle with (holomorphic branches of) square roots: We have f(z) = -\frac 12 (z + \frac 1z) = \frac{\left( \frac{z-1}{z+1}\right)^2 + 1}{\left( \frac{z-1}{z+1}\right)^2 - 1} = T_2(S(T_1(z)) ...

Is \mathbb Q[\sqrt{11} + 10^\frac{1}{3}] equal to \mathbb Q[\sqrt{11}, 10^\frac{1}{3}]?

https://math.stackexchange.com/questions/2445142/is-mathbb-q-sqrt11-10-frac13-equal-to-mathbb-q-sqrt11-10-fr

Let \newcommand{\al}{\alpha}\al=\sqrt{11}+10^{1/3}. Then (\al-\sqrt{11})^3=10. So \al^3+33\al-10=(3\al^2+11)\sqrt{11} and then \sqrt{11}\in\Bbb Q(\al).

The Effect of Outliers

https://stats.stackexchange.com/questions/84235/the-effect-of-outliers

The wiki page for the exponential distribution states that if Y\sim\mbox{Exp}(\lambda), then F^{-1}_q(\lambda)=-\frac{\ln(1-q)}{\lambda} so that \mbox{med}(Y)=\frac{\ln(2)}{\lambda} which ...

Show that the splitting field of x^8-3 has degree 32 over \mathbb{Q}

https://math.stackexchange.com/questions/1231994/show-that-the-splitting-field-of-x8-3-has-degree-32-over-mathbbq

To show that \Bbb Q(\sqrt{2}) \not\subset \Bbb Q(3^{1/8}), you can prove that Q(3^{1/8}) is not a degree four extension of \Bbb Q(\sqrt{2}). If that were the case, the polynomial x^8-3 would ...

Aksje

Kopiert til utklippstavle

Eksempler

Kvadratisk ligning

{ x } ^ { 2 } - 4 x - 5 = 0

Trigonometri

4 \sin \theta \cos \theta = 2 \sin \theta

\left[ \begin{array} { l l } { 2 } & { 3 } \\ { 5 } & { 4 } \end{array} \right] \left[ \begin{array} { l l l } { 2 } & { 0 } & { 3 } \\ { -1 } & { 1 } & { 5 } \end{array} \right]

Samtidig formel

\left. \begin{cases} { 8x+2y = 46 } \\ { 7x+3y = 47 } \end{cases} \right.

Differensiering

\frac { d } { d x } \frac { ( 3 x ^ { 2 } - 2 ) } { ( x - 5 ) }

Integrasjon

\int _ { 0 } ^ { 1 } x e ^ { - x ^ { 2 } } d x

Grenser

\lim _{x \rightarrow-3} \frac{x^{2}-9}{x^{2}+2 x-3}

Emner

Emner

Lignende problemer fra nettsøk

Aksje

Eksempler