We have been given that on the day of his 18th birthday Harry decided to start saving money regularly
. Starting on that day, he could save 30.00 on the same date every month. We are asked to find the amount saved by the day before Harry's 60th birthday.
First of all, we will find years from 18 years to 60 years.
We know that 1 year equals 12 months.
To find total amount saved, we will multiply 504 months by amount saved per month.
Therefore, Harry would have saved 
by the day before his 60th birthday.
Answer:
thats alot of people don't you think
Step-by-step explanation:
Answer: The equation you will use would be 40x + 15x =115. you're welcome.
You find the eigenvalues of a matrix A by following these steps:
- Compute the matrix
, where I is the identity matrix (1s on the diagonal, 0s elsewhere) - Compute the determinant of A'
- Set the determinant of A' equal to zero and solve for lambda.
So, in this case, we have
![A = \left[\begin{array}{cc}1&-2\\-2&0\end{array}\right] \implies A'=\left[\begin{array}{cc}1&-2\\-2&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda\end{array}\right]=\left[\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right]](https://tex.z-dn.net/?f=A%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D%20%5Cimplies%20A%27%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%26-2%5C%5C-2%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%5Cend%7Barray%7D%5Cright%5D%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1-%5Clambda%26-2%5C%5C-2%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D)
The determinant of this matrix is
Finally, we have
So, the two eigenvalues are
First, rearrange the equation to standard line format
3x + 9y = 7
9y = -3x + 7
y = -3/9x + 7/9
y = -1/3x + 7/9
now we know the slope of the line (both the existing and the new parallel one, since they both have the same slope)
y = -1/3x + b
plug in the new point in and solve for b
4 = -1/3(6) + b
4 = -2 + b
b = 6
y = -1/3x + 6 is the equation for the parallel line
