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cricket20 [7]
2 years ago
5

Element X is a radioactive isotope such that every 9 years, its mass decreases by half. Given that the initial mass of a sample

of Element X is 70 grams, how long would it be until the mass of the sample reached 56 grams, to the nearest tenth of a year?
Mathematics
1 answer:
4vir4ik [10]2 years ago
5 0

Answer:

2.9 years

Step-by-step explanation:

Given that :

Half-life t1/2 = 9 years

Initial mass = I = 70 grams

A = final mass = 56 grams

t = time taken to reach final amount

Using the exponential half life relation :

A = I(0.5)^t/t1/2

56 = 70(0.5)^t/9

56/70 = 0.5^t/9

0.8 = 0.5^t/9

Log 0.8 = log 0.5^t/9

−0.096910 = −0.301029 * t/9

t/9 = 0.096910 / 0.301029

t/9 = 0.3219291

t = 0.3219291 * 9

t = 2.8973619

t = 2.9 years

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Answer:

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Step-by-step explanation:

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Montoya Construction needs to borrow $375,000 to build a road to install utilities in a small subdivision. It borrows the funds
MissTica

Step 1

Given;

\begin{gathered} \text{Principal(p)= \$375000} \\ \text{First rate = 8\%=}\frac{8}{100}=0.08 \\ \text{Second rate = 20\%= }\frac{20}{100}=0.2 \\ \text{Time}=\frac{90}{365}=\frac{18}{73} \end{gathered}

Required; To find the difference in interest between the two periods.

Step 2

State the formula for simple interest

A=P(1+rt)

Step 3

Find the interest when the rate is 8%

\begin{gathered} A=375000(1+(0.08\times\frac{18}{73}) \\ A=375000(1+\frac{36}{1825}) \\ A=\text{\$}382397.26 \end{gathered}

Therefore the interest is given as;

A-P=382397.26-375000=\text{\$}7397.26

Step 4

Find the interest in 1980 with a 20% rate

\begin{gathered} A=375000(1+(0.2\times\frac{18}{73}) \\ A=\text{\$}393493.15 \end{gathered}

The interest is given as;

A-p=393493.15-375000=\text{\$}18493.15\text{ }

Step 5

Find the difference in interest between the two rates.

\text{\$}18493.15-\text{\$}7397.26=\text{\$}11095.89

Hence, the difference in interest between the two rates = $11095.89

4 0
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Answer:

letter ( D ).

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7 0
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Degger [83]

Answer:

3,4,5

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Here, we want to select the set that could work and the sides of a right angled triangle

the correct answer is 3,4,5

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What we are saying here is that by adding the square of 3 and 4, we get the square of 5 which illustrates that Pythagoras theorem works for the set

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