Answer:
If k = −1 then the system has no solutions.
If k = 2 then the system has infinitely many solutions.
The system cannot have unique solution.
Step-by-step explanation:
We have the following system of equations
The augmented matrix is
![\left[\begin{array}{cccc}1&-2&3&2\\1&1&1&k\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C1%261%261%26k%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
The reduction of this matrix to row-echelon form is outlined below.
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\2&-1&4&k^2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C2%26-1%264%26k%5E2%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&3&-2&k^2-4\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%263%26-2%26k%5E2-4%5Cend%7Barray%7D%5Cright%5D)
![\left[\begin{array}{cccc}1&-2&3&2\\0&3&-2&k-2\\0&0&0&k^2-k-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bcccc%7D1%26-2%263%262%5C%5C0%263%26-2%26k-2%5C%5C0%260%260%26k%5E2-k-2%5Cend%7Barray%7D%5Cright%5D)
The last row determines, if there are solutions or not. To be consistent, we must have k such that
Case k = −1:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-1-2\\0&0&0&(-1)^2-(-1)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&-3\\0&0&0&-2\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-1-2%5C%5C0%260%260%26%28-1%29%5E2-%28-1%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%26-3%5C%5C0%260%260%26-2%5Cend%7Barray%7D%5Cright%5D)
If k = −1 then the last equation becomes 0 = −2 which is impossible.Therefore, the system has no solutions.
Case k = 2:
![\left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&2-2\\0&0&0&(2)^2-(2)-2\end{array}\right] \rightarrow \left[\begin{array}{ccc|c}1&-2&3&2\\0&3&-2&0\\0&0&0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%262-2%5C%5C0%260%260%26%282%29%5E2-%282%29-2%5Cend%7Barray%7D%5Cright%5D%20%5Crightarrow%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Cc%7D1%26-2%263%262%5C%5C0%263%26-2%260%5C%5C0%260%260%260%5Cend%7Barray%7D%5Cright%5D)
This gives the infinite many solution.
Total weight=number of pennies times weight of 1 penny
total weight/weight of 1 penny=number of pennies
636.3/3.03=210
210 pennies
Answer: the interest on the loan is $39.38
Step-by-step explanation:
The formula for determining simple interest is expressed as
I = PRT/100
Where
I represents interest paid on the loan.
P represents the principal or amount taken as loan
R represents interest rate
T represents the duration of the loan in years.
From the information given,
P = $350
R = 4.5%
There are 12 months in a year. Converting 30 months into years, it becomes
30/12 = 2.5. so
T = 2.5 years
Therefore
I = (350 × 4.5 × 2.5)/100
I = $39.38
Answer:
30 hours is the right answer
Answer:
ok so the earth radius is 1534mi so if that is the radius than the area is 7.39×10^6so then you divide that into 3 pieces so each piece would be 2,463,333.3333 the nearest million would be 2,000,000
Step-by-step explanation:
