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taurus [48]
3 years ago
8

Need help ASAP pls due in 5 mins

Mathematics
2 answers:
castortr0y [4]3 years ago
8 0

Answer:

Jonathan Broke what is he doing buying cheap stuff

stealth61 [152]3 years ago
5 0
Well i cant do nothing about it
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zhannawk [14.2K]

Answer:

might wanna take a picture instead...

7 0
2 years ago
What is the missing value in the equation shown □ x1/10=0.034
grin007 [14]

Answer:

0.34

Step-by-step explanation:

Do the inverse operation so;

0.034 / 1/10 = 0.34

Now go back and subsitute,

0.34 * 1/10 = 0.034

4 0
3 years ago
For a field trip 11 students rode in cars and the rest filled eight buses. Which equation below will help you find how many stud
Helen [10]

Answer:

8x + 11= 315

Step-by-step explanation:

Let x = the number of students in a bus

There are 8 buses

8x is the number of students in buses

There are 11 students in cars

Add this together to get all the students, which is 315

8x + 11= 315

6 0
3 years ago
Read 2 more answers
What is the value of y?<br><br><br><br> Enter your answer in the box.<br> y =
Nataliya [291]
The answer to this question is y=28

6 0
3 years ago
Read 2 more answers
What eigen value for this matix <br> (1 -2)<br> (-2 0)
natali 33 [55]

You find the eigenvalues of a matrix A by following these steps:

  1. Compute the matrix A' = A-\lambda I, where I is the identity matrix (1s on the diagonal, 0s elsewhere)
  2. Compute the determinant of A'
  3. 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]

The determinant of this matrix is

\left|\begin{array}{cc}1-\lambda&-2\\-2&-\lambda\end{array}\right| = -\lambda(1-\lambda)-(-2)(-2) = \lambda^2-\lambda-4

Finally, we have

\lambda^2-\lambda-4=0 \iff \lambda = \dfrac{1\pm\sqrt{17}}{2}

So, the two eigenvalues are

\lambda_1 = \dfrac{1+\sqrt{17}}{2},\quad \lambda_2 = \dfrac{1-\sqrt{17}}{2}

5 0
2 years ago
Read 2 more answers
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