Answer:
61
Step-by-step explanation:
a = 17
d = 21 - 17 = 4
12th day: 17 + 4(12 - 1)
17 + 44
61
Answer:
1st- what scenario are talking about?
2nd- false, becuase if you hide under any objects there is the risk of it falling over you and hurting you
If the half-life is t, then every t days, the amount of the radioactive isotope will be cut in half.
(1/2)^(number of half-lives) = 3%
number of half-lives = ln(0.03) / ln(0.5)
This gives the number of half-lives as 5.06.
Then 300 days = (5.06)(length of 1 half-life)
length of 1 half-life = 300 / 5.06 = 59.29 days
You're looking for the largest number <em>x</em> such that
<em>x</em> ≡ 1 (mod 451)
<em>x</em> ≡ 4 (mod 328)
<em>x</em> ≡ 1 (mod 673)
Recall that
<em>x</em> ≡ <em>a</em> (mod <em>m</em>)
<em>x</em> ≡ <em>b</em> (mod <em>n</em>)
is solvable only when <em>a</em> ≡ <em>b</em> (mod gcd(<em>m</em>, <em>n</em>)). But this is not the case here; with <em>m</em> = 451 and <em>n</em> = 328, we have gcd(<em>m</em>, <em>n</em>) = 41, and clearly
1 ≡ 4 (mod 41)
is not true.
So there is no such number.
