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3 years ago

15

In the coordinate plane, draw quadrilateral ABCD with A(–5, 0), B(2, –6), C(8, 1), and D(1, 7), then demonstrate that ABCD is a

rectangle.

1 answer:

emmainna [20.7K]3 years ago

6 0

Answer:

Step-by-step explanation:

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Please help quickly !!

Alisiya [41]

Answer:

answer is 26000000

Step-by-step explanation:

10+14/7×-8=26

26 × 1000000=26000000

plz follow and like and make brainliest answer

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3 years ago

Verify that the divergence theorem is true for the vector field f on the region

schepotkina [342]

$\mathbf f(x,y,z)=5x\,\mathbf i+xy\,\mathbf j+2xz\,\mathbf k$

$\mathrm{div}(\mathbf f)=\dfrac{\partial(5x)}{\partial x}+\dfrac{\partial(xy)}{\partial y}+\dfrac{\partial(2xz)}{\partial z}=5+x+2x=5+3x$

By the divergence theorem, the flux of $\mathbf f$ across the boundary of $E$ is given by

$\displaystyle\iint_{\mathcal S}\mathbf f\cdot\mathrm d\mathbf S=\iiint_E\mathrm{div}(\mathbf f)\,\mathrm dV$

$=\displaystyle\int_{z=0}^{z=2}\int_{y=0}^{y=2}\int_{x=0}^{x=2}(5+3x)\,\mathrm dx\,\mathrm dy\,\mathrm dz=64$

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3 years ago

(- 8) + 5 + 4 + (- 2) = resuelve

choli [55]

Answer: -1

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Step-by-step explanation:

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3 years ago

How to integrate ∫㏑x/χ<br> from 1 to e

Natali [406]

$\int_1^e\dfrac{\ln x}{x}dx,~\text{if}~u=\ln x\to x=e^u\to dx=e^udu\\\\\text{So, finding the new limits:}\begin{cases}x=1\to u=\ln1\to u=0\\x=e\to u=\ln e\to u=1\end{cases}\\\\
\int_1^e\dfrac{\ln x}{x}dx=\int_0^1\dfrac{u}{e^u}e^udu=\int_0^1u\,du=\left[\dfrac{u^2}{2}\right]_0^1=\dfrac{1^2}{2}-\dfrac{0^2}{2}=\dfrac{1}{2}\\\\
\boxed{\int_1^e\dfrac{\ln x}{x}dx=\dfrac{1}{2}}$

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3 years ago

A certain semiconductor device requires a tunneling probability of T = 10-5 for an electron tunneling through a rectangular barr

Goryan [66]

Answer:

Generally the barrier width is $a = 1.9322 *10^{-9} \ m$

Step-by-step explanation:

From the question we are told that

The tunneling probability required is $T = 1 * 10^{-5}$

The barrier height is $V_o = 0.4 eV$

The electron energy is $E = 0.08eV$

Generally the wave number is mathematically represented as

$k = \sqrt{ \frac{2 * m [V_o - E]}{\= h^2} }$

Here m is the mass of the electron with the value $m = 9.11 *10^{-31} \ kg$

h is is know as h-bar and the value is $\= h = 1.054*10^{-34} \ J \cdot s$

So

$k = \sqrt{ \frac{2 * 9.11 *10^{-31 } [0.4 - 0.04] * 1.6*10^{-19}}{[1.054*10^{-34}^2]} }$

=> $k = 3.073582 *10^{9} \ m^{-1}$

Generally the tunneling probability is mathematically represented as

$T = 16 * \frac{E}{V_o } * [1 - \frac{E}{V_o} ] * e^{-2 * k * a}$

So

$1.0 *10^{-5} = 16 * \frac{0.04}{0.4 } * [1 - \frac{0.04}{0.4} ] * e^{-2 * 3.0736 *10^{9} * a}$

=> $6.944*10^{-6}= e^{-2 * 3.0736 *10^{9} * a}$

Taking natural log of both sides

$ln[6.944*10^{-6}] = -2 * 3.0736 *10^{9} * a}$

=> $-11.8776 = -2 * 3.0736 *10^{9} * a}$

=> $a = 1.9322 *10^{-9} \ m$

4 0

2 years ago