Answer:
Step-by-step explanation:
Answer:
Lavania observed 39 fruit flies after 6 days of observation
Step-by-step explanation:
Let x be the number of fruit flies on the first day of Lavania's study.
After 6 days she had nine more than five times as many fruit flies as when she began the study.
Five times as many fruit flies as when she began the study = 5x
Nine more than five times as many fruit flies as when she began the study=5x+9
The expression to find the population of fruit flies Lavania observed after 6 days is 5x+9
If she observes 20 fruit flies on the first day of the study, then x=6, then
Answer:
t = ln(0.5)/-r
Step-by-step explanation:
The decay rate parameter is missing. I will assume a value of 4% per day.
The exponential decay is modeled by the following equation:
A = A0*e^(-r*t)
where A is the mass after t time (in days), A0 is the initial mass and r is the rate (as a decimal).
At half-life A = A0/2, then:
A0/2 = A0*e^(-0.04*t)
0.5 = e^(-0.04*t)
ln(0.5) = -0.04*t
t = ln(0.5)/-0.04
t = 17.33 days
In general the half-life time is:
t = ln(0.5)/-r
Answer:
y = 2x − 1
Step-by-step explanation:
By eliminating the parameter, first solve for t:
x = 4 + ln(t)
x − 4 = ln(t)
e^(x − 4) = t
Substitute:
y = t² + 6
y = (e^(x − 4))² + 6
y = e^(2x − 8) + 6
Taking derivative using chain rule:
dy/dx = e^(2x − 8) (2)
dy/dx = 2 e^(2x − 8)
Evaluating at x = 4:
dy/dx = 2 e^(8 − 8)
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
Now, without eliminating the parameter, take derivative with respect to t:
x = 4 + ln(t)
dx/dt = 1/t
y = t² + 6
dy/dt = 2t
Finding dy/dx:
dy/dx = (dy/dt) / (dx/dt)
dy/dx = (2t) / (1/t)
dy/dx = 2t²
At the point (4, 7), t = 1. Evaluating the derivative:
dy/dx = 2(1)²
dy/dx = 2
Writing equation of line using point-slope form:
y − 7 = 2 (x − 4)
y = 2x − 1
