The independent variable i<span>n exponential growth and decay problems is the time because the value of the given population will depend on the how much time have elapsed from the star of the decay or growth. in the usual form of a growth function is A(t) = Ao(e)^(kt)</span>
First, to estimate we can round 53.82 up to 55.00. Then you can look at it like this: when you divide 100 by 20 you get 5, so 20% is one fifth of 100. Therefore, we want to find one fifth of $55. To do this, you multiply 55*1/5. When you multiply the numerators by each other and the denominators by each other, you get 55/5. When you divide the numerator and denominator by 5, you get 11/1 or $11.
Hope this helps
.53 of an hour because .53 (3.5) + .53 (4) = 3.975 which is the nearest estimate.
Answer:
Step-by-step explanation:
We are given that
y(0)=0
y'(0)=1
By comparing with
We get
q(x)=0
p(x),q(x) and g(x) are continuous for all real values of x except 3.
Interval on which p(x),q(x) and g(x) are continuous
and (3,
By unique existence theorem
Largest interval which contains 0=
Hence, the larges interval on which includes x=0 for which given initial value problem has unique solution=
