By letting
we get derivatives
a) Substitute these into the differential equation. After a lot of simplification, the equation reduces to
Examine the lowest degree term 
, which gives rise to the indicial equation,
with roots at r = 0 and r = 4/5.
b) The recurrence for the coefficients 
is
so that with r = 4/5, the coefficients are governed by
c) Starting with 
, we find
so that the first three terms of the solution are
Answer:
a₅=27.
Step-by-step explanation:
1) according to the condition the given subsequence is arithmetic sequence. It means, the a₁=7 and d=5, then
2) a₅=a₁+4*d; => a₅=7+4*5=27.
Answer:
i don't know nothing about that
The student will have $135 in her bank account at the end of the ninth week. You can fine this out by finding out the amount she deposits a week and to do this you would take the $30 and divide it by 2 because she had $30 at the end of the second week.
30/2=15
So you see that the student deposits $15 each week, so to find out how much money she will have in 9 weeks you will multiply her $15 by 9.
15x9=135
So the student will have $135 at the end of the ninth week.
Using the power of zero property, we find that:
a) The simplification of the given expression is 1.
b) Since , equivalent expressions are: and .
--------------------------------
The power of zero property states that any number that is not zero elevated to zero is 1, that is:
Thus, at item a, , thus the simplification is .
At item b, equivalent expressions are found elevating non-zero numbers to 0, thus and .
