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2 years ago

5

Reuben needs 280 programs for the school play on Thursday. How many boxes of programs will he need, given that each box contains

37 programs?

1 answer:

Tresset [83]2 years ago

4 0

The answer would be 8 boxes

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50 is 40% of what number

notsponge [240]

125
50=40%
25=20%
125=100%

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3 years ago

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The black graph is the graph of y=f(x). Choose the equation for the red graph.

Makovka662 [10]

Answer:

y-1 = f(x)

Step-by-step explanation:

Here, we want to choose the equation for the ref graph

The red graph as we can see is above the black

This means it is more positive

The difference between the two is just 1 unit

By the addition of 1 to the y-value of the black graph, we get the red

Thus, we have that;

y- 1 = f(x)

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2 years ago

Line parallel to y=2x has a slope of ?

loris [4]

In order to find the slope of a line parallel to this one, we must be familiar with the slope/intercept form, which is y= mx + b. M would be the slope. Therefore this line has a slope of 2. Parallel lines have the same slope, therefore<em> a line parallel to y=2x has a slope of </em><em>2</em><em>. </em>

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3 years ago

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X+2y+1=0,2x-3y-12=0 solve by substitution method

Kitty [74]

Answer:

Step-by-step explanation:

x=-1-2y

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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