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!!!
Subtract 10x from both sides
-5y=-10x+15
Divide all terms by -5
y=2x-3
Final answer: y=2x-3
1)first step
for the first inequation you need apply the following property, IXI < a
-a<x<a, them you must substitute 2x-3, of this form
-5<2x-3<5⇒ -5+3<2x-3+3<5+3⇒-2<2x<8⇒-2/2<2x/2<8/2⇒ -1<x<4,
the solution is
S1 =(-1,4)
2) For the second inequality apply the following property
IXI>a , x<-a or x >a , therefore
IX-2I>1 ⇒X-2 <-1 OR X-2> 1
X-2<-1⇒X-2+2<-1+2⇒ X< 1
OR
X-2>1⇒X>1+2⇒X>3
THE SOLUTION IS S2 = (-∞,1)∪(3,∞)