Answer:
4 liters of 60% solution; 2 liters of 30% solution
Step-by-step explanation:
I like to use a simple, but effective, tool for most mixture problems. It is a kind of "X" diagram as in the attachment.
The ratios of solution concentrations are 3:6:5, so I've used those numbers in the diagram. The constituent solutions are on the left; the desired mixture is in the middle, and the numbers on the other legs of the X are the differences along the diagonals: 6 - 5 = 1; 5 - 3 = 2. This tells you the ratio of 60% solution to 30% solution is 2 : 1.
These ratio units (2, 1) add to 3. We want 6 liters of mixture, so we need to multiply these ratio units by 2 liters to get the amounts of constituents needed. The result is 4 liters of 60% solution and 2 liters of 30% solution.
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If you're writing equations, it often works well to let the variable represent the quantity of the greatest contributor—the 60% solution. Let the volume of that (in liters) be represented by v. Then the total volume of iodine in the mixture is ...
... 0.60·v + 0.30·(6 -v) = 0.50·6
... 0.30v = 0.20·6 . . . . subtract 0.30·6, collect terms
... v = 6·(0.20/0.30) = 4 . . . . divide by the coefficient of v
4 liters of 60% solution are needed. The other 2 liters are 30% solution.
Answer:
4
Step-by-step explanation:
Based on the information you have given me, p = 1/3.
All we need to do is substitute that value into the expression:
12p = 12(1/3) = 4
Answer:

Step-by-step explanation:
<u>Eigenvalues of a Matrix</u>
Given a matrix A, the eigenvalues of A, called
are scalars who comply with the relation:

Where I is the identity matrix
![I=\left[\begin{array}{cc}1&0\\0&1\end{array}\right]](https://tex.z-dn.net/?f=I%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D)
The matrix is given as
![A=\left[\begin{array}{cc}3&5\\8&0\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D)
Set up the equation to solve
![det\left(\left[\begin{array}{cc}3&5\\8&0\end{array}\right]-\left[\begin{array}{cc}\lambda&0\\0&\lambda \end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3%265%5C%5C8%260%5Cend%7Barray%7D%5Cright%5D-%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D%5Clambda%260%5C%5C0%26%5Clambda%20%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)
Expanding the determinant
![det\left(\left[\begin{array}{cc}3-\lambda&5\\8&-\lambda\end{array}\right]\right)=0](https://tex.z-dn.net/?f=det%5Cleft%28%5Cleft%5B%5Cbegin%7Barray%7D%7Bcc%7D3-%5Clambda%265%5C%5C8%26-%5Clambda%5Cend%7Barray%7D%5Cright%5D%5Cright%29%3D0)

Operating Rearranging

Factoring

Solving, we have the eigenvalues

If Lopez's mailing bill continues to deduct 19$ from his bank account every month for a year, then in total, the deductions total for the year would be $228.00.
Why?
$19.00 is being taken away each month. There are 12 months in a year.
$19.00 x 12 months = the total of $228.00 a year.