Answer:
m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ
Half-Life = 46.21 years.
Step-by-step explanation:
Radioactive reactions always follow a first order reaction dynamic
Let the initial mass of radioactive substance be m₀ and the mass at any time be m
(dm/dt) = -Km (Minus sign because it's a rate of reduction)
The question provides K = 0.015 from the given differential equation
(dm/dt) = -0.015m
(dm/m) = -0.015dt
∫ (dm/m) = -0.015 ∫ dt
Solving the two sides as definite integrals by integrating the left hand side from m₀ to m and the Right hand side from 0 to t.
We obtain
In (m/m₀) = -0.015t
(m/m₀) = (e^(-0.015t))
m = m₀ e^(-0.015t)) = m₀ e⁻⁰•⁰¹⁵ᵗ
m(t) = m₀ e⁻⁰•⁰¹⁵ᵗ
At half life, m(t) = (m₀/2), t = T(1/2)
(m₀/2) = = m₀ e⁻⁰•⁰¹⁵ᵗ
e⁻⁰•⁰¹⁵ᵗ = (1/2)
In e⁻⁰•⁰¹⁵ᵗ = In (1/2)
-0.015t = - In 2
t = (In 2)/0.015
t = (0.693/0.015)
t = 46.21 years
Half life = T(1/2) = t = 46.21 years.
Hope this Helps!!!
X-intercept is (-3,0)
Y-intercept is (0,6)
The equation is y=2x+6
Replace y with 0 when solving for x-intercept and replace x with 0 when solving for y-intercept.
Answer:
B
Step-by-step explanation:
It isn't A. A has to be something like y = -x^2 That minus sign directs the parabola downward.
It could be B which is something like y = x^3 - 5
C has an odd number of real roots (5). In other words it crosses the x axis 5 times. So let's make one of these up and graph it
y = (x^2 - 4)(x^2 - 1)(x + 3) Study this for a minute. The leading coefficient of this polynomial is plus 1. There are no negatives in front of the x. The highest degree is x^5. Now here's the key question: on the left side where does the graph start? In Quad three right? It's coming from the lower left. That is exactly what graph C is NOT doing. So for C the graph has a leading coefficient that is less than 0. a<0
Not C.
D is not the answer either. It opens downward and looks like it is something like y = - x^4 + 3x^3 - 2x^2 + 5x - 2
Answer B: