Question

A new radioactive element is found on Mars. Every 10 years, the concentration of the element is a quarter of what it was at the start of the 10-year period. If the decay rate is given by Equation with left side as, C= Co (x)^t/10, where C0 is the initial concentration of the element, C is the final concentration of the element, and t is the time in years, what is the value of x?

Answer 1

`\boxed{(1)/(16)}`

Step-by-step explanation:

`C = C_(0)(x)^{(t)/(10)}`

Data:

t = 5 yr

C = ¼C₀

Calculation:

`\begin{array}{rcl}(1)/(4)C_(0) & = & C_(0)(x)^{(5)/(10)}\\(1)/(4) & = & x^{(1)/(2)}\\x & = & \mathbf{(1)/(16)}\\\end{array}`

`\text{The value of $x$ is $\boxed{\mathbf{(1)/(16)}}$}`

Answer 2

x=0.25

Step-by-step explanation:

The equation is

`C = Co*x^{(t)/(10) }`

For this problem, C would be equal to 0.25 * Co when t is equal to 10 years. If we put this information in the equation we're left with:

`0.25*Co=Co*x^{(10)/(10) }`

Now we solve for x:

`0.25*Co=Co*x^{(10)/(10) }\\0.25=x^(1 )\\0.25=x`

So the value of x is 0.25