Y’all I need help asap. I also need to show my work for this so if you can find the slope AND explain how that would be AMAZING
1 answer:
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I FOUND YOUR COMPLETE QUESTION IN OTHER SOURCES.
SEE ATTACHED IMAGE.
Part A:
The differential equation is:
y '= 0.08y
The initial condition is:
P (0) = 500
Part B:
The formula for this case is:
P (t) = 500exp (0.08 * t)
Part C:
After five days we have to evaluate t = 5 in the equation.
We have then:
P (5) = 500 * exp (0.08 * 5)
P (5) = 745.9123488
Rounding:
P (5) = 746
SEE ATTACHED IMAGE.
Part A:
The differential equation is:
y '= 0.08y
The initial condition is:
P (0) = 500
Part B:
The formula for this case is:
P (t) = 500exp (0.08 * t)
Part C:
After five days we have to evaluate t = 5 in the equation.
We have then:
P (5) = 500 * exp (0.08 * 5)
P (5) = 745.9123488
Rounding:
P (5) = 746