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Andrews [41]
3 years ago
9

The stable nuclide ²⁰⁶₈₂Pb is formed from ²³⁸₉₂U by a long series of α and β⁻ decays. Which of the following nuclides could NOT

be involved in this decay series?
A) Th-234
B) U-239
C) Po-218
D) Bi-214
E) Rn-222
Mathematics
2 answers:
Makovka662 [10]3 years ago
8 0

Answer:

Rn-222

Step-by-step explanation:

this decay takes specified step therefore Rn-222is not a decaying process

Lana71 [14]3 years ago
6 0

Answer:

B) U-239 is the answer

hope it helps you

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DENIUS [597]

I thinks it C, this a difficult question but idk, that my closest guess...

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3 years ago
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sertanlavr [38]

Answer:

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Step-by-step explanation:

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3 years ago
Given z1 and z2 below. Calculate z1/z2 and write your answer in rcistheta form.
andre [41]

Answer:

  (2/9)cis(30°)

Step-by-step explanation:

We can express the two numbers in magnitude∠angle form, then find their ratio.

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So, the ratio is ...

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  z1/z2 = (2/9)∠30°

8 0
3 years ago
The sum of the first 20 positive odd integers
svet-max [94.6K]

Answer:

400

Step-by-step explanation:

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Hope this helps! :)
3 0
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