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Vlad1618 [11]
3 years ago
13

Initially 100 milligrams of a radioactive substance was present. After 8 hours the mass had decreased by 2%. If the rate of deca

y is proportional to the amount of the substance present at time t, determine the half-life of the radioactive substance. (Round your answer to one decimal place.)
Mathematics
1 answer:
4vir4ik [10]3 years ago
5 0

Answer:

The half life of the radioactive substance is 277 hours.

Step-by-step explanation:

initial mass, No = 100 mg

mass decayed = 2% = 2 mg

Mass remained , N = 98 mg

time, t = 8 hours

Let the half life is T.

Use the equation of radioactivity

N = No\times e^{\frac{-0.693 t}{T}}\\\\98 = 100 \times e^{\frac{-0.693\times 8}{T}}\\\\0.98 = e^{\frac{-5.54}{T}}\\\\ln 0.98 = -\frac{5.54}{T}\\\\-0.02= -\frac{5.54}{T}\\\\T =  277 hours  

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PLEASE HELP ASAP!!! I NEED CORRECT ANSWERS ONLY PLEASE!!!
Eduardwww [97]
Work and answer in picture: (4.2)

7 0
3 years ago
10th term of 4, 7, 10, 13........ (Please enter number only)<br><br> Hint <br> Use a+(n-1)d
monitta

Answer:

31

Step-by-step explanation:

a + (n-1)d  where a = first term;  n = which number we want;  d = common difference.

=> 4 + (10 - 1)3

=> 4 + (9)*3

=> 4 + 27

=> 31

4 0
3 years ago
You buy tea light candles and mints as party favours for a baby shower. the tea light candles come in packs of 12 for $3.50. The
Reptile [31]

Answer:

The least amount of money it can be spent is $125.

Step-by-step explanation:

First, we write the prime factorization of each number:

12= 2·2·3

15= 2·5·5

Then, we search each different factor which appears the greater number of times. The factor 2 appears in both factorizations so least common multiple is:

LCM= 2·2·3·5·5=300

Hence, the total quantity of packs of each thing is:

Candles: 300÷12=25

Mints: 300÷50=6

The least amount of money it can be spent is:

T=25×$3.50 + 6×$6.25= $87.5 + $37.5= $125

6 0
3 years ago
3/10=0.3<br> So...<br> 71/100=?<br> 32 6/100=?
Veronika [31]
71/100= .71 or 71%
32 6/100= 32.06
3 0
3 years ago
Solve y=f(x) for x. Then find the input when the output is -3.
DochEvi [55]

Answer:

Please check the explanation

Step-by-step explanation:

Given the function

f\left(x\right)\:=\:\left(x-5\right)^3-1

Given that the output = -3

i.e. y = -3

now substituting the value y=-3 and solve for x to determine the input 'x'

\:\:y=\:\left(x-5\right)^3-1

-3\:=\:\left(x-5\right)^3-1\:\:\:

switch sides

\left(x-5\right)^3-1=-3

Add 1 to both sides

\left(x-5\right)^3-1+1=-3+1

\left(x-5\right)^3=-2

\mathrm{For\:}g^3\left(x\right)=f\left(a\right)\mathrm{\:the\:solutions\:are\:}g\left(x\right)=\sqrt[3]{f\left(a\right)},\:\sqrt[3]{f\left(a\right)}\frac{-1-\sqrt{3}i}{2},\:\sqrt[3]{f\left(a\right)}\frac{-1+\sqrt{3}i}{2}

Thus, the input values are:

x=-\sqrt[3]{2}+5,\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}-i\frac{\sqrt[3]{2}\sqrt{3}}{2},\:x=\frac{\sqrt[3]{2}\left(1+5\cdot \:2^{\frac{2}{3}}\right)}{2}+i\frac{\sqrt[3]{2}\sqrt{3}}{2}

And the real input is:

x=-\sqrt[3]{2}+5

  • x=3.74
4 0
3 years ago
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