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Ksivusya [100]
3 years ago
9

Can y'all help me please guys thank you because i am confessed

Mathematics
1 answer:
VashaNatasha [74]3 years ago
5 0
Well area triangler area is multiplied by 3 after you multiply the other numbers
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Evaluate 3r(t – v) if r = 2, t = 6, and v = 4
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Substitute\ data\\\\r=2,\ \ t=6 and\ \ v=4\\\\into \ equation\\\\ 3r(t-v)=3*2(6-4)=3*2*2=3*4=12
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Distribute and simplify the radicals below
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Answer:

i think A is answer you should to do

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3 years ago
The ground-state wave function for a particle confined to a one-dimensional box of length L is Ψ=(2/L)^1/2 Sin(πx/L). Suppose th
Hitman42 [59]

Answer:

(a) 4.98x10⁻⁵

(b) 7.89x10⁻⁶

(c) 1.89x10⁻⁴

(d) 0.5

(e) 2.9x10⁻²  

Step-by-step explanation:  

The probability (P) to find the particle is given by:

P=\int_{x_{1}}^{x_{2}}(\Psi\cdot \Psi) dx = \int_{x_{1}}^{x_{2}} ((2/L)^{1/2} Sin(\pi x/L))^{2}dx  

P = \int_{x_{1}}^{x_{2}} (2/L) Sin^{2}(\pi x/L)dx     (1)

The solution of the intregral of equation (1) is:

P=\frac{2}{L} [\frac{X}{2} - \frac{Sin(2\pi x/L)}{4\pi /L}]|_{x_{1}}^{x_{2}}  

(a) The probability to find the particle between x = 4.95 nm and 5.05 nm is:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{4.95}^{5.05} = 4.98 \cdot 10^{-5}    

(b) The probability to find the particle between x = 1.95 nm and 2.05 nm is:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{1.95}^{2.05} = 7.89 \cdot 10^{-6}  

(c) The probability to find the particle between x = 9.90 nm and 10.00 nm is:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{9.90}^{10.00} = 1.89 \cdot 10^{-4}    

(d) The probability to find the particle in the right half of the box, that is to say, between x = 0 nm and 50 nm is:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{50.00} = 0.5

(e) The probability to find the particle in the central third of the box, that is to say, between x = 0 nm and 100/6 nm is:

P=\frac{2}{100} [\frac{X}{2} - \frac{Sin(2\pi x/100)}{4\pi /100}]|_{0}^{16.7} = 2.9 \cdot 10^{-2}

I hope it helps you!

3 0
4 years ago
Y=2x[<img src="https://tex.z-dn.net/?f=x%5E%7B2%7D%20" id="TexFormula1" title="x^{2} " alt="x^{2} " align="absmiddle" class="lat
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X=4 because if y=2 then u muiptly it by 2

7 0
4 years ago
Find at least three different sequences beginning with the terms 3, 5, 7 whose terms are generated by a simple for- mula or rule
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1. Perhaps the simplest rule is one that makes this the start of an arithmetic sequence:

a_{n+1}=a_{n}+2

Then the sequence will continue ...

... 3, 5, 7, 9, 11, 13, 15, ...

2. We can multiply the two terms by some values, add the results, then add some constant. There are an infinite number of ways to choose such values. Here's one set of numbers that give the third term from the first two:

a_{n+2}=4a_n-a_{n+1}

Then the sequence will continue ...

... 3, 5, 7, 13, 15, 37, 23, ...

3. We can multiply adjacent values and add a constant.

a_{n+2}=a_{n+1}\cdot a_n-8

Then the sequence will continue ...

... 3, 5, 7, 28, 188, 5256, 988120, ...

5 0
3 years ago
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