the length of AB_ is 9 cm a dilation with a scale factor of 2 is applied to AB_ what is the length of the image of AB after the
dilation is applied
1 answer:
You might be interested in
We have the linear system:
which in Matrix format is
![\left[\begin{array}{ccc}a_1&b_1\\a_2&b_2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}c_1\\c_2\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26b_1%5C%5Ca_2%26b_2%5Cend%7Barray%7D%5Cright%5D%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc_1%5C%5Cc_2%5Cend%7Barray%7D%5Cright%5D%20)
![\left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] \left[\begin{array}{ccc}x\\y\end{array}\right] = \left[\begin{array}{ccc}-26\\13\end{array}\right]](https://tex.z-dn.net/?f=%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-6%5C%5C5%262%5Cend%7Barray%7D%5Cright%5D%20%20%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dx%5C%5Cy%5Cend%7Barray%7D%5Cright%5D%20%3D%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-26%5C%5C13%5Cend%7Barray%7D%5Cright%5D%20)
We then find the value of x and y use the Cranmer's Rule:
![x= \frac{ \left[\begin{array}{ccc}c_1&b_1\\c_2&b_2\end{array}\right] }{ \left[\begin{array}{ccc}a_1&a_2\\b_1&b_2\end{array}\right] } = \frac{c_1b_2-b_1c_2}{a_1b_2-a_2b_1}](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dc_1%26b_1%5C%5Cc_2%26b_2%5Cend%7Barray%7D%5Cright%5D%20%7D%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26a_2%5C%5Cb_1%26b_2%5Cend%7Barray%7D%5Cright%5D%20%7D%20%3D%20%5Cfrac%7Bc_1b_2-b_1c_2%7D%7Ba_1b_2-a_2b_1%7D%20)
![x= \frac{ \left[\begin{array}{ccc}-26&-6\\13&2\end{array}\right] }{ \left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] } = \frac{(-26)(2)-(-6)(13)}{-2)(2)-(-6)(5)} = \frac{26}{26}=1](https://tex.z-dn.net/?f=x%3D%20%5Cfrac%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-26%26-6%5C%5C13%262%5Cend%7Barray%7D%5Cright%5D%20%7D%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-6%5C%5C5%262%5Cend%7Barray%7D%5Cright%5D%20%7D%20%3D%20%5Cfrac%7B%28-26%29%282%29-%28-6%29%2813%29%7D%7B-2%29%282%29-%28-6%29%285%29%7D%20%3D%20%5Cfrac%7B26%7D%7B26%7D%3D1%20)
![y= \frac{ \left[\begin{array}{ccc}a_1&c_1\\a_2&c_2\end{array}\right] }{ \left[\begin{array}{ccc}a_1&b_1\\a_2&b_2\end{array}\right] } = \frac{a_1c_2-a_2c_1}{a_1b_2-a_2b_1}](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26c_1%5C%5Ca_2%26c_2%5Cend%7Barray%7D%5Cright%5D%20%7D%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da_1%26b_1%5C%5Ca_2%26b_2%5Cend%7Barray%7D%5Cright%5D%20%7D%20%3D%20%5Cfrac%7Ba_1c_2-a_2c_1%7D%7Ba_1b_2-a_2b_1%7D%20)
![y= \frac{ \left[\begin{array}{ccc}-2&-26\\5&13\end{array}\right] }{ \left[\begin{array}{ccc}-2&-6\\5&2\end{array}\right] }= \frac{(-2)(13)-(5)(-26)}{(-2)(2)-(-6)(5)}= \frac{104}{26}=4](https://tex.z-dn.net/?f=y%3D%20%5Cfrac%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-26%5C%5C5%2613%5Cend%7Barray%7D%5Cright%5D%20%7D%7B%20%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-2%26-6%5C%5C5%262%5Cend%7Barray%7D%5Cright%5D%20%7D%3D%20%5Cfrac%7B%28-2%29%2813%29-%285%29%28-26%29%7D%7B%28-2%29%282%29-%28-6%29%285%29%7D%3D%20%5Cfrac%7B104%7D%7B26%7D%3D4%20%20%20)
So we have the answers:
x = 1 and y = 4
Answer: Option A
which in Matrix format is
We then find the value of x and y use the Cranmer's Rule:
So we have the answers:
x = 1 and y = 4
Answer: Option A