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The standard relation between these moduli is $\frac{9}{Y} = \frac{3}{\eta} + \frac{1}{K}$.
- Rearranging for $K$: $\frac{1}{K} = \frac{9}{Y} - \frac{3}{\eta} = \frac{9\eta - 3Y}{Y\eta}$
- $K = \frac{Y\eta}{9\eta - 3Y}$.
This is a linear ODE with $P = 3\sec^2 x$ and $Q = \sec^2 x$.
Integrating factor: $e^{\int 3\sec^2 x\,dx} = e^{3\tan x}$
$\frac{d}{dx}(ye^{3\tan x}) = \sec^2 x \cdot e^{3\tan x}$
Integrate: $ye^{3\tan x} = \frac{1}{3}e^{3\tan x} + C$
At $x=\pi/4$, $\tan(\pi/4)=1$, $y=4/3$: $\frac{4}{3}e^3 = \frac{1}{3}e^3 + C \Rightarrow C = e^3$
So $y = \frac{1}{3} + e^3 \cdot e^{-3\tan x}$
At $x=-\pi/4$, $\tan(-\pi/4)=-1$: $y = \frac{1}{3} + e^3 \cdot e^{3} = \frac{1}{3} + e^3$... wait: $e^{-3(-1)}=e^3$, so $y = \frac{1}{3}+e^3 \cdot e^3 = \frac{1}{3}+e^6$.
Correct: $y(-\pi/4) = \mathbf{\frac{1}{3}+e^6}$. So correct option index is 0.
\(r_n = a_0 \cdot \dfrac{n^2}{Z}\) → \(r \propto n^2\)
\(E_n = -13.6 \cdot \dfrac{Z^2}{n^2}\) eV → \(E \propto -\dfrac{1}{n^2}\)
As orbit number increases, radius increases (electrons move farther out) while energy becomes less negative (electrons become less tightly bound). For hydrogen (Z=1): \(r_1 = 0.529\) Å (Bohr radius).
Who are the internal users of financial information?
- Bank
- Directors
- Employees
- Potential investors
Option B (2 and 3) is correct.
- Directors — manage the company, internal users.
- Employees — employed by the business, internal users.
Banks and potential investors are external users.
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