Answer:
Switch to the other streaming service
Step-by-step explanation:
So, 96$ with a 12% increase is $107.52.
But if you do 8.75×12 you get $105.
So Clare should switch to the other music stream.
Answer:
z = -1
Step-by-step explanation:
7 - z / -2 - (-1)
7 - z / -2+1
7 - z / -1
-7 +z = -8
z = -1
to check the work
(-1,-1) (-2,7)
7 + 1 / -2 + 1
8 / -1
= -8
The measure of Arc Q P is 96°. We also know that ∠QTP is central angle, then the measure of arc QP is 96°.
Step-by-step explanation:
<u>Step 1</u>
If QS is a circle diameter,
then m∠QTS=180°.
Let x be the measure of angle RTQ: ∠RTQ =x.
so, let ∠RTQ = x
<u />
<u>Step 2</u>
According to the question,
∠RTQ = ∠RTS - 12°
⇒ ∠RTS = x + 12°
∴ ∠QTS = ∠RTQ + ∠RTS
= x + x + 12° = 2x + 12° = 180°
⇒ 2x = 168°
⇒ x = 84°
⇒ ∠RTQ = 84°
<u></u>
<u>Step 3</u>
Now,
∵∠QTP and ∠RTS are vertical angles
∴ ∠QTP = 84° + 12° = 96°
As ∠QTP is the central angle, hence the measure of arc QP is 96°
<u></u>
<u>Step 4</u>
The Measure of arc QP = 96°
Answer:
the 2nd one
Step-by-step explanation:
I took the Quiz
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
