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Keith_Richards [23]
3 years ago
8

What’s the answer/ how should I work it?

Mathematics
1 answer:
SIZIF [17.4K]3 years ago
3 0

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

Hi my lil bunny!

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

-\frac{6}{11} + (-\frac{5}{11}) =

=\frac{-6}{11} + \frac{5}{11}

= \frac{-6}{11} + \frac{-5}{11}

= -1

❧⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯⎯☙

●✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎❀✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎✴︎●

Have a great day/night!

❀*May*❀

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It’s A

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Read 2 more answers
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
gtnhenbr [62]

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.

7 0
3 years ago
Which of these expressions is equivalent to 3/6
zmey [24]
1/2, 6/12, 9,/18, and 15/30
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3 years ago
PLS HELP ME ASAP FOR 13 and 12! (SHOW WORK!!) + LOTS OF POINTS!!
svet-max [94.6K]
13=idk

14=first minutes into hour: 40/60=2/3
                                    speed=distance/time
                                                 =280/(2/3)
                                                    ans=420miles/hour

4 0
3 years ago
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