Since opening night, attendance at Play A has increased steadily, while attendance at Play B first rose and then fell. Equations
1 answer:
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Answer:
This system has no solution:
y = x²
y = 0 · x - 1
Step-by-step explanation:
Hi there!
I propose these two equations:
y = x²
y = 0 · x - 1
This system has no solution because in the quadratic function "y" is always positive for every value of "x", while in the linear function "y" is always -1 for every value of "x". Then, there will not be a pair (x,y) shared by both functions.
In other words, there is no value of "x" that satisfies this expression:
x² = 0 · x - 1
In the attached graphic, you can see that both curves never touch (in red is the quadratic function and in blue is the linear function), hence, the system has no solution.
Have a nice day!
Answer:
λN N(0) = 6
N(t) = N₀e^(λt)
Applying the inital value condition
N(t) = 6e^(λt)
Step-by-step explanation:
Summarizing the information briefly and stating the variables in the problem.
t = time elapsed during the decay
N(t) = the amount of the radioactive substance remaining after time t
λ= The constant of proportionality is called the decay constant or decay rate
Given the initial conditions
N(0) = N₀ = 6
The rate at which a quantity of a radioactive substance decays () is proportional to the quantity of the substance (N) and λ is the constant of proportionality is called the decay constant or decay rate :
λN
N(t) = N₀e^(λt) ......equ 1
substituting the value of N₀ = 6 into equation 1
N(t) = 6e^(λt)