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Kisachek [45]
3 years ago
8

What is the area of triangle ABC with vertices A(x¹,y¹), B(x²,y²)and C (x³,y³)??????????​

Chemistry
1 answer:
AlexFokin [52]3 years ago
7 0

Answer:

Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|

Area = 2\ units^2

Explanation:

Given

A = (x_1,y_1)

B = (x_2,y_2)

C = (x_3,y_3)

Required

Determine the area

The area of a triangle is :

Area = \frac{1}{2}|A_x(B_y - C_y) + B_x(C_y - A_y) + C_x(A_y - B_y)|

By substituting values for the x and y coordinates of A, B and C;

We have:

Area = \frac{1}{2}|x_1(y_2 - y_3)+x_2(y_3 - y_1)+x_3(y_1 - y_2)|

So:

For instance

A = (0,3)

B= (2,1)

C = (2,3)

The area is:

Area = \frac{1}{2}|0(1-3) + 2(3-3) + 2(3-1)|

Area = \frac{1}{2}| 2*0 + 2*2|

Area = \frac{1}{2}| 0 + 4|

Area = \frac{1}{2}|4|

Area = \frac{1}{2} * 4

Area = 2\ units^2

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What is the number of valence electrons that most atoms prefer?
Nataly_w [17]

Answer: 8 electrons

Explanation: Valence electrons are the electrons in the outermost shell of an atom and they are important in chemical reactions.

Atoms are <em>happy</em> when they follow the octet rule which states that 2 electrons can go in the first shell and 8 electrons can go in the other shells.

Electrons are happy when they have a full outer shell which usually needs to be 8 electrons. However, some of the smaller elements like helium are happy because they can hold a max of 2 electrons and helium has 2 electrons since it has 2 protons in its core as well as 2 electrons in its energy levels.

4 0
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What is the average yearly rate of change of carbon-14 during the first 5000 years?
erica [24]

Answer:

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year

Explanation:

Given that the mass of the carbon 14 at the start = 5 gram

At the end of 5,000 years we will have;

A = A_0 \times e^{-\lambda \times t}

Where

A = The amount of carbon 14 left

A₀ = The starting amount of carbon 14

e = Constant = 2.71828

T_{1/2} = The half life

\lambda = 0.693/T_{1/2}

t = The time elapsed = 5000 years

λ = 0.693/T_{1/2} = 0.693/5730 = 0.0001209424

Therefore;

A = 5 × e^(-0.0001209424×5000) = 2.7312 grams

Therefore, the amount of carbon 14 decayed in the 5000 years is the difference in mass between the starting amount and the amount left

The amount of carbon 14 decayed = 5 - 2.7312 = 2.2688 grams

The average yearly rate of change of carbon-14 during the first 5000 years  is therefore;

2.2688 grams/(5000 years) = 0.0004538 grams per year

The average yearly rate of change of carbon-14 during the first 5000 years = 0.0004538 grams per year.

5 0
3 years ago
Match each transition metal ion with its condensed ground-state electron configuration. A [Ar]3d2 B [Ar]4s23d3 C [Kr]4d10 D [Xe]
madreJ [45]

Answer:

Mn^{2+} : - F . [Ar]3d^{5}

Hg^{2+} : - G. [Xe]4f^{14}5d^{10}

La^{3+} : - D. [Xe]

Fe^{3+} : - F. [Ar]3d^{5}

Ag^{+} : - C. [Kr]4d^{10}

Co^{3+} : - E. [Ar]3d^{6}

Explanation:

The electronic configuration of the element Mn is:-

[Ar]3d^{5}4s^2

For, Mn^{2+}, 2 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Hg is:-

[Xe]4f^{14}5d^{10}6s^2

For, Hg^{2+}, 2 electrons are lost, thus the configuration is:-

[Xe]4f^{14}5d^{10}

The electronic configuration of the element La is:-

[Xe]5d^{1}6s^2

For, La^{3+}, 3 electrons are lost, thus the configuration is:-

[Xe]

The electronic configuration of the element Fe is:-

[Ar]3d^{6}4s^2

For, Fe^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{5}

The electronic configuration of the element Ag is:-

[Kr]4d^{10}5s^1

For, Ag^{+}, 1 electron is lost, thus the configuration is:-

[Kr]4d^{10}

The electronic configuration of the element Co is:-

[Ar]3d^{7}4s^2

For, Co^{3+}, 3 electrons are lost, thus the configuration is:-

[Ar]3d^{6}

7 0
3 years ago
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