We can use the Euclidean algorithm to find out:
46 = 1*33 + 13
33 = 2*13 + 7
13 = 1*7 + 6
7 = 1*6 + 1
The last remainder is 1, which means 33 and 46 are coprime and gcf(33, 46) = 1.
Another way to see this is to write out the prime factorizations of both numbers:
33 = 3*11
46 = 2*23
As you can see, there are no shared divisors, so the gcf is 1.
Answer:
D
Step-by-step explanation:
3 1/2 - 1 7/10
7/2 - 17/10 (Common denominator 10, multiply 7 and 2 by 5)
35/2 - 17/10 = 18/10
1 8/10 = 1 4/5
Answer:
y=3/5x-9
Step-by-step explanation:
9x – 15y = 135
Subtract 9x from each side
9x -9x – 15y =-9x+ 135
-15y = -9x +135
Divide each side by -15
-15y/-15 = -9x/-15 +135/-15
y=3/5x-9
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!!!
Answer:
400,000
Step-by-step explanation:
Ok so, su ecusión es 0.05p = 20,000 p = 400,000
lo siento si esto está mal no soy genial con el español.