Recall that m(t) = m.(1/2)^t/h for radioactive decay, where h is the half-life. Suppose that a 500 g sample of phosphorus-32 decays to 356 g over 7 days. Calculate the half life of the sample.

<u>Answer:</u> The half life of the sample of phosphorus-32 is $14.28days^{-1}$

<u>Step-by-step explanation:</u>

The equation used to calculate the half life of the sample is given as:

$m(t)=m_o(1/2)^{t/h}$

where,

m(t) = amount of sample after time 't' = 356 g

$m_o$ = initial amount of the sample = 500 g

t = time period = 7 days

h = half life of the sample = ?

Putting values in above equation, we get:

$356=500\times (\frac{1}{2})^{7/h}\\\\h=14.28days^{-1}$

Hence, the half life of the sample of phosphorus-32 is $14.28days^{-1}$

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What is the domain of g(x) = sqrt x +1

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There can be no negative in a sqrt
so x+1 must be equal or greater then 0
x+1 $\geq$ 0
x $\geq$ -1
so all answers must be greater then or equal to -1
-1,0,1,2,3,4,5,6,7,8
(-1,infinity) is the domain

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Which of the inequalities represents the statement, "Two times a number x, less 15, is greater than or equal to 4 times the numb

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Therefore, the best and most correct answer among the choices provided by the question is the fourth choice "15 - 2x ≥ 4y". I hope my answer has come to your help.

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