Answer: h = 9
Step-by-step explanation: A system of linear equations is consistent when it has at least one solution.
The matrix given is:
![\left[\begin{array}{ccc}-15&21&h\\5&-7&-3\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%2621%26h%5C%5C5%26-7%26-3%5Cend%7Barray%7D%5Cright%5D)
Transform this matrix in a row-echelon form:
![\left[\begin{array}{ccc}-15&21&h\\0&0&-9+h\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-15%2621%26h%5C%5C0%260%26-9%2Bh%5Cend%7Barray%7D%5Cright%5D)
For this row-echelon form to have solutions:
-9 + h = 0
h = 9
For this system to be consistent: h = 9.
Answer:
Step-by-step explanation:
Answer:
She runs 4 miles in one hour
Step-by-step explanation:
5 divided by 1.25=4
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.
