Answer:converge at 
Step-by-step explanation:
Given
Improper Integral I is given as
integration of 
is -
![I=\left [ -\frac{1}{x}\right ]^{\infty}_3](https://tex.z-dn.net/?f=I%3D%5Cleft%20%5B%20-%5Cfrac%7B1%7D%7Bx%7D%5Cright%20%5D%5E%7B%5Cinfty%7D_3)
substituting value
![I=-\left [ \frac{1}{\infty }-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%20%5Cfrac%7B1%7D%7B%5Cinfty%20%7D-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
![I=-\left [ 0-\frac{1}{3}\right ]](https://tex.z-dn.net/?f=I%3D-%5Cleft%20%5B%200-%5Cfrac%7B1%7D%7B3%7D%5Cright%20%5D)
so the value of integral converges at 
Greetings!We can factor this expression/ the GCF is:
Hope this helps.
-Benjamin
Answer:
They lose about 2.79% in purchasing power.
Step-by-step explanation:
Whenever you're dealing with purchasing power and inflation, you need to carefully define what the reference is for any changes you might be talking about. Here, we take <em>purchasing power at the beginning of the year</em> as the reference. Since we don't know when the 6% year occurred relative to the year in which the saving balance was $200,000, we choose to deal primarily with percentages, rather than dollar amounts.
Each day, the account value is multiplied by (1 + 0.03/365), so at the end of the year the value is multiplied by about
... (1 +0.03/365)^365 ≈ 1.03045326
Something that had a cost of 1 at the beginning of the year will have a cost of 1.06 at the end of the year. A savings account value of 1 at the beginning of the year would purchase one whole item. At the end of the year, the value of the savings account will purchase ...
... 1.03045326 / 1.06 ≈ 0.9721 . . . items
That is, the loss of purchasing power is about ...
... 1 - 0.9721 = 2.79%
_____
If the account value is $200,000 at the beginning of the year in question, then the purchasing power <em>normalized to what it was at the beginning of the year</em> is now $194,425.14, about $5,574.85 less.
30cm2 I believe this is the answer according to my calculations
