If 5000 gets increased by 200, then you end up with 5200.
if 900 gets increased by 200, then it'd be 1100.
so in short, if we take 5200 to be the 100%, what is 1100 in percentage from it?
Answer:
no solution
Step-by-step explanation:
3(x – 2) + 1 ≥ x + 2(x + 2).
Distribute
3x – 6 + 1 ≥ x + 2x + 4
Combine like terms
3x -5 ≥ 3x +4
Subtract 3x from each side
-5 ≥ 4
This is never true so there is no solution
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!!!
Part because the whole proportion would be 4/4 and 2/4 is part of it