The answer is 1/16 i think. Hope this helps! :)
1) The set of words represented in the first task is definitely ''<span>member and class '' type of analogy. That's pretty easy, it reminds of definitions : simile is one of type of the figurative language and </span><span>isosceles is one of the types of triangles.
2) What about the next set, I am pretty sure that the gap should be filled with </span>jaywalking : offense :: bottle : container. I think that these words represent consequences, so jaywalking soneer or later will be spotted and it would have become an offence, and bottle will spent the rest of its life in the container.
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Hope that helps!</span>
Answer:
Jada should have multiplied both sides of the equation by 108.
Step-by-step explanation:
The question is incomplete. Find the complete question in the comment section.
Given the equation -4/9 = x/108, in order to determine Jada's error, we need to solve in our own way as shown:
Step 1: Multiply both sides of the equation by -9/4 as shown:
-4/9 × -9/4 = x/108 × -9/4
-36/-36 = -9x/432
1 = -9x/432
1 = -x/48
Cross multiplying
48 = -x
x = -48
It can also be solved like this:
Given -4/9 = x/108
Multiply both sides by 108 to have:
-4/9 * 108 = x/108 * 108
-4/9 * 108 = 108x/108
-432/9 = x
x = -48
Jada should have simply follow the second calculation by multiplying both sides of the equation by 108 as shown.
C)
Area of full triangle
1/2(8)(6)=24cm^2
Area of non shaded triangle
1/2(4)(3)=6cm^2
Area of full minus area of small triangle
24-6=18cm^2
D)
Area of Big rectangle
(4)(12)=48
Area of non shaded triangle
1/2(9)(4)=18
Area of Big rectangle minus non shaded triangle
48-18=30cm^2
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
