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siniylev [52]
2 years ago
8

The March 11, 2011, earthquake in Japan caused a tsunami that crashed into the (cant say on here because it looks like a bad wor

d) nuclear power plant on Japan's Pacific coast. Some people compare this nuclear disaster to the 1986 disaster at Chernobyl. Describe the results and the possible future effects of these nuclear accidents
Mathematics
1 answer:
Lesechka [4]2 years ago
4 0

A radioactive atmosphere around the area and an expensive, hard time cleaning everything up.

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X-2=3(x-4) what is the solution
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Find the approximate perimeter of ABC plotted below.
maksim [4K]

Answer:

B. 21.2

Step-by-step explanation:

Perimeter of ∆ABC = AB + BC + AC

A(-4, 1)

B(-2, 3)

C(3, -4)

✔️Distance between A(-4, 1) and B(-2, 3):

AB = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AB = \sqrt{(-2 - (-4))^2 + (3 - 1)^2} = \sqrt{(2)^2 + (2)^2)}

AB = \sqrt{4 + 4}

AB = \sqrt{16}

AB = 4 units

✔️Distance between B(-2, 3) and C(3, -4):

BC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

BC = \sqrt{(3 - (-2))^2 + (-4 - 3)^2} = \sqrt{(5)^2 + (-7)^2)}

BC = \sqrt{25 + 49}

BC = \sqrt{74}

BC = 8.6 units (nearest tenth)

✔️Distance between A(-4, 1) and C(3, -4):

AC = \sqrt{(x_2 - x_1)^2 + (y_2 - y_1)^2}

AC = \sqrt{(3 - (-4))^2 + (-4 - 1)^2} = \sqrt{(7)^2 + (-5)^2)}

AC = \sqrt{47 + 25}

AC = \sqrt{74}

AC = 8.6 units (nearest tenth)

Perimeter of ∆ABC = 4 + 8.6 + 8.6 = 21.2 units

8 0
3 years ago
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