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shtirl [24]
3 years ago
12

I am a number greater than 40,000 and less than 60,000. My ones digit and tens digit are the same. My ten-thousands digit is 1 l

ess than 3 times the sum of my ones digit and tens digit. My thousands digit is half my hundreds digit, and the sum of those two digits is 9. What number am I?
Mathematics
1 answer:
jarptica [38.1K]3 years ago
3 0
I am a number greater than 40,000 and less than 60,000:

40,000 < n < 60,000

This means that:

n = 10,000n₁ + 1,000n₂ + 100n₃ + 11n₄

And also:

4 ≤ n₁ < 6

0 ≤ n₂ ≤ 9

0 ≤ n₃ ≤ 9

0 ≤ n₄ ≤ 9

My ten thousands digit is 1 less than 3 times the sum of my ones digit and tens digit:

n₁ = 3*2n₄ - 1

n₁ = 6n₄ - 1

This means that:

n = 10,000*(6n₄-1) + 1,000n₂ + 100n₃ + 11n₄

n = 60,000n₄ - 10,000 + 1,000n₂ + 100n₃ + 11n₄

n = 60,011n₄ - 10,000 + 1,000n₂ + 100n₃

<span>My thousands digit is half my hundreds digit, and the sum of those two digits is 9:

n</span>₂ = 1/2 * n₃
<span>
n</span>₂ + n₃ = 9
<span>
Therefore:

n</span>₂ = 9 - n₃
<span>
Therefore:

9 - n</span>₃ = 1/2 * n₃
<span>
9 = 1/2 * n</span>₃ + n₃
<span>
9 = 1.5 * n</span>₃
<span>
Therefore:

n</span>₃ = 6
<span>
If n</span>₃=6, n₂=3.
<span>
This means that:

</span>n = 60,011n₄ - 10,000 + 1,000*3 + 100*6

n = 60,011n₄ - 10,000 + 3,000 + 600

n = 60,011n₄ - 6,400

Therefore:

0<n₄<2, so n₄=1.

If n₄=1:

n = 60,011 - 6,400

n = 53,611

Answer:

53,611
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The ​half-life of a radioactive element is 130​ days, but your sample will not be useful to you after​ 80% of the radioactive nu
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Answer:

We can use the sample about 42 days.

Step-by-step explanation:

Decay Equation:

\frac{dN}{dt}\propto -N

\Rightarrow \frac{dN}{dt} =-\lambda N

\Rightarrow \frac{dN}{N} =-\lambda dt

Integrating both sides

\int \frac{dN}{N} =\int\lambda dt

\Rightarrow ln|N|=-\lambda t+c

When t=0, N=N_0 = initial amount

\Rightarrow ln|N_0|=-\lambda .0+c

\Rightarrow c= ln|N_0|

\therefore ln|N|=-\lambda t+ln|N_0|

\Rightarrow ln|N|-ln|N_0|=-\lambda t

\Rightarrow ln|\frac{N}{N_0}|=-\lambda t.......(1)

                            \frac{N}{N_0}=e^{-\lambda t}.........(2)

Logarithm:

  • ln|\frac mn|= ln|m|-ln|n|
  • ln|ab|=ln|a|+ln|b|
  • ln|e^a|=a
  • ln|a|=b \Rightarrow a=e^b
  • ln|1|=0

130 days is the half-life of the given radioactive element.

For half life,

N=\frac12 N_0,  t=t_\frac12=130 days.

we plug all values in equation (1)

ln|\frac{\frac12N_0}{N_0}|=-\lambda \times 130

\rightarrow ln|\frac{\frac12}{1}|=-\lambda \times 130

\rightarrow ln|1|-ln|2|-ln|1|=-\lambda \times 130

\rightarrow -ln|2|=-\lambda \times 130

\rightarrow \lambda= \frac{-ln|2|}{-130}

\rightarrow \lambda= \frac{ln|2|}{130}

We need to find the time when the sample remains 80% of its original.

N=\frac{80}{100}N_0

\therefore ln|{\frac{\frac {80}{100}N_0}{N_0}|=-\frac{ln2}{130}t

\Rightarrow ln|{{\frac {80}{100}|=-\frac{ln2}{130}t

\Rightarrow ln|{{ {80}|-ln|{100}|=-\frac{ln2}{130}t

\Rightarrow t=\frac{ln|80|-ln|100|}{-\frac{ln|2|}{130}}

\Rightarrow t=\frac{(ln|80|-ln|100|)\times 130}{-{ln|2|}}

\Rightarrow t\approx 42

We can use the sample about 42 days.

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