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Bess [88]
3 years ago
8

The​ half-life of radium is 1690 years. If 20 grams are present​ now, how much will be present in 250 ​years?

Mathematics
1 answer:
Aneli [31]3 years ago
3 0

Answer:

  18.05 g

Step-by-step explanation:

The multiplier of the initial quantity is ...

  (1/2)^(t/1690)

so, for t=250, this becomes ...

  (1/2)^(250/1690) ≈ 0.902545

Then the quantity remaining of the initial 20 g is ...

  (20 g)(0.902545) ≈ 18.05 g

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PilotLPTM [1.2K]

Answer:

x = 1 , y = 1 , z = 0

Step-by-step explanation by elimination:

Solve the following system:

{-2 x + 2 y + 3 z = 0 | (equation 1)

-2 x - y + z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Subtract equation 1 from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x - 3 y - 2 z = -3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Multiply equation 2 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

2 x + 3 y + 3 z = 5 | (equation 3)

Add equation 1 to equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+3 y + 2 z = 3 | (equation 2)

0 x+5 y + 6 z = 5 | (equation 3)

Swap equation 2 with equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+3 y + 2 z = 3 | (equation 3)

Subtract 3/5 × (equation 2) from equation 3:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - (8 z)/5 = 0 | (equation 3)

Multiply equation 3 by 5/8:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y - z = 0 | (equation 3)

Multiply equation 3 by -1:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y + 6 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 6 × (equation 3) from equation 2:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+5 y+0 z = 5 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 2 by 5:

{-(2 x) + 2 y + 3 z = 0 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 2 × (equation 2) from equation 1:

{-(2 x) + 0 y+3 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Subtract 3 × (equation 3) from equation 1:

{-(2 x)+0 y+0 z = -2 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Divide equation 1 by -2:

{x+0 y+0 z = 1 | (equation 1)

0 x+y+0 z = 1 | (equation 2)

0 x+0 y+z = 0 | (equation 3)

Collect results:

Answer: {x = 1 , y = 1 , z = 0

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