We find the value of N₀ since we are provided with initial conditions.
The condition is that, at time t = 0, the amount of substance contains originally 10 grams.
We substitute:
10 = N₀ (e^(-0.1356)*0)
10 = N₀ (e^0)
N₀ = 10
When the substance is in half-life (meaning, the half of the original amount), it contains 5 grams. We solve t in this case.
5 = 10 e^(-0.1356*t)
0.5 = e^(-0.1356*t)
Multiply natural logarithms on both sides to bring down t.
ln(0.5) = -0.1356*t
Hence,
t = -(ln(0.5))/0.1356
t ≈ 5.11 days (ANSWER)
Answer:
5(6)+0.65y
Step-by-step explanation:
$5(6hrs)+65÷100=0.65
=5(6)+0.65y
Answer:
C
Step-by-step explanation:
We can use process of elimination
D is incorrect because the roots are 3 and -4 and there are no negative roots visible
B is wrong because the roots -3 and -6 are both negative
You can factor A into (x-2)(x-3) and the roots are 2 and 3 but the roots on the graph look closer to 3 and 6
For C it can be factored as (x-6)(x-3) so the roots are 3 and 6 which look accurate
C. The graph of G(x) is the graph of F(x) shifted 9 units to the left.
Inside changes do the opposite of what they seem like they'd do. "x+9" shifts it left 9 units.
