Answer:
120.0
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Step-by-step explanation:
Using translation concepts, the equation of F(x) is:
A. 
<h3>What is a translation?</h3>
A translation is represented by a change in the function graph, according to operations such as multiplication or sum/subtraction in it's definition.
The parent function is:

For function F(x), we have that:
- It was shifted down 2 units, hence F(x) = G(x) - 2.
- It was vertically compressed by a factor of 3, hence

Then the equation is:
A. 
More can be learned about translation concepts at brainly.com/question/28098112
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Answer:
The next step is to find the point on the circle which makes a tangent line that passes through the outside point.
Step-by-step explanation:
A tangent line to a circle is a line that touches the circle at exactly one point. You need two points to draw a line. You already have one point and the circle, then you need the other point, which lies on the circle. These two points have to lie on the same line. Notice that there are two possible tangent lines.
Answer:
19.2 centimeters
Step-by-step explanation:
Tia cut a 4 meter and 8 centimeter wire, converting this measure to centimeters

and she cut it into 10 equal pieces, thus each piece measures:
.
Martha has a wire of 540 centimeter wich she cutted into 9 equal pieces, so each piece measures:

The difference between the two wires is:

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
