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3 years ago

What is the measure of ∠ABD?

Mathematics

1 answer:

yawa3891 [41]3 years ago

7 0

Answer: it will be 60

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What is the percent of change from 5 cups to 8 cups

Maksim231197 [3]

8cups- 5cups= 3cups - change

5- 100%

3-p

cross multiply

5p=3x100

5p=300 divide both sides by 5

p=60

Answer: an increase of 60%

4 0

3 years ago

Which equation has a solution of 54.73 for n?

mr Goodwill [35]

Answer:

n + 8 = 62.73

the first one

Step-by-step explanation:

5 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

Which of the following is equivalent to the distance between the points (3,2) and (2,0)?

mr Goodwill [35]

The distance between any 2 points P(a,b) and Q(c,d) in the coordinate plane, is given by the formula:

$|PQ|= \sqrt{ (a-c)^{2} + (b-d)^{2} }$

Thus, the distance between points (3,2) and (2,0) is:

$d= \sqrt{ (3-2)^{2} + (2-0)^{2}}=\sqrt{1 + 4}= \sqrt{5}$

Answer: $\sqrt{5}$ units

4 0

3 years ago

Order from least to greatest:

777dan777 [17]

Answer:

0, 1 , 1.3 , 2 , 2.7

Step-by-step explanation:

8 0

3 years ago