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RSB [31]
2 years ago
15

Va rog e urgent am nevoie de raspuns

Mathematics
1 answer:
Lady_Fox [76]2 years ago
3 0

Answer: 4

<u>Explanation:</u>

f(x) = 2x - 1

f(√2) = 2√2 - 1

f(1) = 2(1) - 1

     =  2 - 1

     =     1

f(√3)  = 2√3 - 1

*******************************************************

\frac{f(\sqrt{2})-f(1)}{\sqrt{2}-1} +\frac{f(\sqrt{3})-f(\sqrt{2})}{\sqrt{3}-\sqrt{2}}

= \frac{(2\sqrt{2}-1)-1}{\sqrt{2}-1} +\frac{(2\sqrt{3}-1)-(2\sqrt{2}-1)}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1} +\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}

= \frac{2\sqrt{2}-2}{\sqrt{2}-1}(\frac{\sqrt{2}+1}{\sqrt{2}+1})+\frac{2\sqrt{3}-2\sqrt{2}}{\sqrt{3}-\sqrt{2}}(\frac{\sqrt{3}+\sqrt{2}}{\sqrt{3}+\sqrt{2}})

= \frac{4+2\sqrt{2}-2\sqrt{2}-2}{2 - 1} + \frac{6 +2\sqrt{6}-2\sqrt{6}-4}{3-2}

= \frac{2}{1} +\frac{2}{1}

= 4

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gayaneshka [121]

Answer:

C. 7.8

Step-by-step explanation:

This problem involves the use of a trigonometric ratio. This one specifically involves the use of cosine as it provides the leg adjacent to the angle and the hypotenuse.

cos(40) = 6/x

cos(40)*x=6/x*x

cos(40)*x=6

cos(40)/cos(40)*x=6/cos(40)

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The value of x is C. 7.8

4 0
2 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

       

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Just plug in your numbers:

M = 0.0375/12 * 30000/(1-(1+.0375/12)^60)

since 5 years is 60 months

6 0
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