Carbon-14 has a half-life of 5,730 years. If a sample contained 70 mg originally, how much is left after 17,190 years?

Mathematics

1 answer:

kow [346]3 years ago

7 0

17,190 / 5,730 = 3. This means we must divide 70mg in half 3 times

70 / 2 = 35

35 / 2 = 17.5

17.5 / 2 = 8.75

There will be 8.75 mg left after 17,190 years.

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Kevin and Marcus were comparing their ages. The same of their ages was at least 9 years but was at most 24 years. If Kevin is 6

saul85 [17]

The inequality that represent age of marcus is:

$24\geq 6+m\geq 9$

The possible values for age of Marcus is:

m = 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13, 14, 15, 16, 17, 18

Solution:

Given that,

The sum of ages of Kevin and Marcus was at least 9 years but was at most 24 years

Kevin is 6 years old

Let "m" be the age of Marcus

Thus we frame a inequality as:

$24\geq \text{ sum of ages } \geq 9$

at least 9 years means that "greater than or equal to 9"

at most 24 years means "less than or equal to" 24

From given,

Sum of ages = kevin age + marcus age

Sum of ages = 6 + m

Thus the inequality is:

$24\geq 6+m\geq 9$

Solve the inequality

$\mathrm{If}\:a\ge \:u\ge \:b\:\mathrm{then}\:a\ge \:u\quad \mathrm{and}\quad \:u\ge \:b\\\\24\ge \:6+m\quad \mathrm{and}\quad \:6+m\ge \:9\\\\Solve\ them\ separately$

$24\ge \:6+m\\\\\mathrm{Switch\:sides}\\\\6+m\le \:24\\\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\\\6+m-6\le \:24-6\\\\\mathrm{Simplify}\\\\m\le \:18$

Now solve another inequality

$6+m\ge \:9\\\\\mathrm{Subtract\:}6\mathrm{\:from\:both\:sides}\\\\6+m-6\ge \:9-6\\\\\mathrm{Simplify}\\\\m\ge \:3\\\\\mathrm{Combine\:the\:intervals}\\\\m\le \:18\quad \mathrm{and}\quad \:m\ge \:3\\\\\mathrm{Merge\:Overlapping\:Intervals}\\\\3\le \:m\le \:18$

$24\ge \:6+m\ge \:9\quad :\quad \begin{bmatrix}\mathrm{Solution:}\:&\:3\le \:m\le \:18\:\\ \:\mathrm{Interval\:Notation:}&\:\left[3,\:18\right]\end{bmatrix}$