Answer: The only graph that shows a proportional relationship is the line that crosses the origin point (0,0).
Explanation
The other graphs are linear functions but not not proportional relationships.
The general form of a proportional relationship is y = kx, where k is the proportionality constant. So, for x = 0 you will always obtain y = 0.
The general form of a linear relationshio is y = kx + b, being b the y-intercept, so if the y-intercept is not 0, it is not a proportional relationship. That is what happens with the other three graphs.
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.
Answer:
Step-by-step explanation:
We have to diagonalize the matrix
![\left[\begin{array}{ccc}1&-1&0\\5&1&4\\0&1&1\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%26-1%260%5C%5C5%261%264%5C%5C0%261%261%5Cend%7Barray%7D%5Cright%5D)
we have to solve the expression
Thus, by applying the determinant we obtain the polynomial
and the eigenvector will be
HOPE THIS HELPS!!
