Answer:
SAS
Step-by-step explanation:
From the image attached below, in triangle ABC, AB = 15, AC = 20 while in triangle ADE, AD = 12 and AE = 16
Two triangles are said to be similar if they have the same shape, that is either their corresponding angles are the same or their corresponding sides are proportional.
For both triangle ABC and triangle ADE, they have the same angle which is ∠A. Also:
Since two sides of triangle ABC are in the same proportion as two sides of triangle ADE and they have the same angle, therefore they are similar based on the SAS (side-angle-side) similarity
The answer of the question is B
Answer:
A² - 6A + 11 I = ![\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
Given the matrix
![A=\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right]](https://tex.z-dn.net/?f=A%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C-1%264%5Cend%7Barray%7D%5Cright%5D)
Calculate A² - 6A + 11 I
![A^2 = A*A= \left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] *\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] = \left[\begin{array}{ccc}2*2-3*1&2*3+3*4\\-1*2-4*1&-1*3+4*4\end{array}\right] =\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right]](https://tex.z-dn.net/?f=A%5E2%20%3D%20A%2AA%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C-1%264%5Cend%7Barray%7D%5Cright%5D%20%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C-1%264%5Cend%7Barray%7D%5Cright%5D%20%3D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%2A2-3%2A1%262%2A3%2B3%2A4%5C%5C-1%2A2-4%2A1%26-1%2A3%2B4%2A4%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2618%5C%5C-6%2613%5Cend%7Barray%7D%5Cright%5D)
![6A=6*\left[\begin{array}{ccc}2&3\\-1&4\end{array}\right] =\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right]](https://tex.z-dn.net/?f=6A%3D6%2A%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D2%263%5C%5C-1%264%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%2618%5C%5C-6%2624%5Cend%7Barray%7D%5Cright%5D)
![11 I = 11 * \left[\begin{array}{ccc}1&0\\0&1\end{array}\right] =\left[\begin{array}{ccc}11&0\\0&11\end{array}\right]](https://tex.z-dn.net/?f=11%20I%20%3D%2011%20%2A%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%260%5C%5C0%261%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%260%5C%5C0%2611%5Cend%7Barray%7D%5Cright%5D)
∴ A² - 6A + 11 I = ![\left[\begin{array}{ccc}1&18\\-6&13\end{array}\right] -\left[\begin{array}{ccc}12&18\\-6&24\end{array}\right] +\left[\begin{array}{ccc}11&0\\0&11\end{array}\right] =\left[\begin{array}{ccc}0&0\\0&0\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D1%2618%5C%5C-6%2613%5Cend%7Barray%7D%5Cright%5D%20-%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D12%2618%5C%5C-6%2624%5Cend%7Barray%7D%5Cright%5D%20%2B%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%260%5C%5C0%2611%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D0%260%5C%5C0%260%5Cend%7Barray%7D%5Cright%5D)
There is an indeterminate form: infinity/infinity. To solve it, you can use the infinite comparison method. First, you have to identify which has the highest degree term: whether the numerator or the denominator. In this case, it is the numerator, which has the negative term of degree 4: -7x ^ 4. Then, the result is -∞.
Therefore, the answer is the last option: Does not exist.
Answer:
<em><u>478</u></em>
Step-by-step explanation:
1*10+32(2+4+6)+76+2^3
First parenthesis
1*10+32(12)+76+2^3
Next exponents
1*10+32(12)+76+8
Next multiplication
10+384+76+8
Then add
394+84
Add
478
Therefore, your answer would be <u><em>478</em></u>
