Answer:
a) See the proof below.
b) 
Step-by-step explanation:
Part a
For this case we assume that we have the following series
and this series has a finite radius of convergence
and we assume that
for all n, this information is given by the problem.
We assume that the series converges at the point
since w eknwo that converges, and since converges we can conclude that:
For this case we need to show that converges also for 
So we need to proof that 
We can do some algebra and we can rewrite the following expression like this:
and we see that the last series is alternating.
Since we know that
converges then the sequence {
} must be positive and we need to have
And then by the alternating series test we can conclude that
also converges. And then we conclude that the power series
converges for
,and that complete the proof.
Part b
For this case we need to provide a series whose interval of convergence is exactly (-1,1]
And the best function for this 
Because the series
converges to
when
using the root test.
But by the properties of the natural log the series diverges at
because
and for
we know that converges since
is an alternating series that converges because the expression tends to 0.
Answer:
The butterfly travels 660 ft in 1.25 min.
Step-by-step explanation:
If the distance, d, a butterfly flies varies directly as the time,t in the air, then we can write the mathematical statement.

If we introduce the constant of proportionality, we obtain;

According to the question, when
,
.
We substitute these values to obtain;



The direct variation equation now becomes

When 


Therefore the butterfly will travel 660 ft in 1.25 min.
90%, as each section is 10 percent!
just a quick clarification, tis usually -4.9 and that's a rounded number to reflect earth's gravity on an object in motion, but -5 is close enough :)
![\bf ~~~~~~\textit{initial velocity in meters} \\\\ h(t) = -4.9t^2+v_ot+h_o \quad \begin{cases} v_o=\textit{initial velocity}\\ \qquad \textit{of the object}\\ h_o=\textit{initial height}\\ \qquad \textit{of the object}\\ h=\textit{object's height}\\ \qquad \textit{at "t" seconds} \end{cases} \\\\[-0.35em] \rule{34em}{0.25pt}](https://tex.z-dn.net/?f=%5Cbf%20~~~~~~%5Ctextit%7Binitial%20velocity%20in%20meters%7D%20%5C%5C%5C%5C%20h%28t%29%20%3D%20-4.9t%5E2%2Bv_ot%2Bh_o%20%5Cquad%20%5Cbegin%7Bcases%7D%20v_o%3D%5Ctextit%7Binitial%20velocity%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h_o%3D%5Ctextit%7Binitial%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bof%20the%20object%7D%5C%5C%20h%3D%5Ctextit%7Bobject%27s%20height%7D%5C%5C%20%5Cqquad%20%5Ctextit%7Bat%20%22t%22%20seconds%7D%20%5Cend%7Bcases%7D%20%5C%5C%5C%5C%5B-0.35em%5D%20%5Crule%7B34em%7D%7B0.25pt%7D)
![\bf h(x)=-5(\stackrel{\mathbb{F~O~I~L}}{x^2-8x+16})+180\implies h(x)=-5x^2+40x-80+180 \\\\[-0.35em] ~\dotfill\\\\ ~\hfill h(x)=-5x^2+\stackrel{\stackrel{v_o}{\downarrow }}{40} x+\stackrel{\stackrel{h_o}{\downarrow }}{\boxed{100}}~\hfill](https://tex.z-dn.net/?f=%5Cbf%20h%28x%29%3D-5%28%5Cstackrel%7B%5Cmathbb%7BF~O~I~L%7D%7D%7Bx%5E2-8x%2B16%7D%29%2B180%5Cimplies%20h%28x%29%3D-5x%5E2%2B40x-80%2B180%20%5C%5C%5C%5C%5B-0.35em%5D%20~%5Cdotfill%5C%5C%5C%5C%20~%5Chfill%20h%28x%29%3D-5x%5E2%2B%5Cstackrel%7B%5Cstackrel%7Bv_o%7D%7B%5Cdownarrow%20%7D%7D%7B40%7D%20x%2B%5Cstackrel%7B%5Cstackrel%7Bh_o%7D%7B%5Cdownarrow%20%7D%7D%7B%5Cboxed%7B100%7D%7D~%5Chfill)
Answer:
$5500 is the new value of the car.
Step-by-step explanation:
actual value of car=$7700
fraction of value decreased=2/7
so, amount of car's value decreased=
$(2/7)*7700.
=$2*(7700/7)
=$2*1100
=$2200
so, new value of car=$(7700-2200) =$5500