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Mademuasel [1]
3 years ago
6

Could yall answer these in need to finish it in 7mins

Mathematics
2 answers:
san4es73 [151]3 years ago
5 0

Answer:

- 6 \\  - 9 \\ 6 \\ 42 \\  - 30 \\ 4 \\  - 6 \\  - 14

Murljashka [212]3 years ago
5 0

Answer:

look below

Step-by-step explanation:

- have you already answered this

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Evaluate the expression<br> please help me
Contact [7]

Answer:

19/9 i think

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
What ordered pair would be located 5 units to the right on the x-axis?
Akimi4 [234]

Answer:

(5, 0)

Step-by-step explanation:

The ordered pair located 5 units to the right on the x axis is (5, 0) which indicates the x-location first (x = 5) and the y-location in the second coordinate (y = 0)

8 0
3 years ago
The original price of a skateboard was reduced by $15. The new price is $49.
olasank [31]

Answer:

The original price of a skateboard is $64

Step-by-step explanation:

Let

x ----> the original price of a skateboard

y ----> the new price of a skateboard

we know that

The linear equation that represent this problem is equal to

y=x-15 ----> equation A

y=49 ---> equation B

substitute equation B in equation A and solve for x

49=x-15

Adds 15 both sides

49+15=x

64=x

Rewrite

x=$64

6 0
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
What de answer be? will give brainliest!
Nonamiya [84]

Answer:

I think the answer is A sorry if it is wrong

Step-by-step explanation:

8 0
3 years ago
Read 2 more answers
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