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3 years ago

1. Given the information below, can we prove AABC = ADEF? If yes, which postulate or

Mathematics

1 answer:

Degger [83]3 years ago

7 0

Answer:

C. Yes, we can prove the triangles congruent by SAS.

Step-by-step explanation:

In ∆ABC and ∆DEF,

BC = EF = 6 cm

angle A = angle D = 22°

AC = DF = 14 cm

Hence, the triangles are congruent.

( by SAS congruency criterion )

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The larger triangle is the image of the smaller triangle after a dilation. The center of the dilation is (−2,−2).

dolphi86 [110]

Dilation of 3. Just count the squares... the smaller one has 3 squares and the larger has 9.

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3 years ago

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Abdul is saving money to buy a game. So far he has saved $24, which is three-fourths of the total cost of the game. How much doe

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Answer:

$32

Step-by-step explanation:

24 divided by 3= 8

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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 is the scale factor of the dilation?

dusya [7]

Answer:

$\dfrac{1}{2}$

Step-by-step explanation:

It is given that triangle ACB is the preimage and triangle DFE is the image.

We need to find the scale factor of the dilation.

$\text{Scale factor}=\dfrac{\text{Side of image}}{\text{Corresponding side of preimage}}$

In the given triangles, side AC and DF are corresponding sides.

$\text{Scale factor}=\dfrac{DF}{AC}$

$\text{Scale factor}=\dfrac{6}{12}$

$\text{Scale factor}=\dfrac{1}{2}$

Therefore, the scale factor of the dilation is $\dfrac{1}{2}$ .

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4 years ago

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Daniel [21]

Answer:

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3 years ago

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