It is okay to take the inverse of both side given that you remember to exclude the value that make the denominator zero which is in this case r=-1
For

Combine the fractions



Therefore our final answer is

Hope this helps!
When we partition a number line from 0 to 1 into six, each segment has a length of 1/6 and the segments are:
1/6. 2/6. 3/6, 4/6, 5/6 , 6/6
Breaking 2/6 into 4 equal parts, we get each part equal to 1/12
Fist you need to substitute the value of R into the equation
25-6(3)
Now you simply
First multiple 6(3)
25-6(3)
25-18
Now solve!
25-18
7
The answer is 7
Answer: A. ![\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Step-by-step explanation:
The question is asking us to find the product of the matrices. The key difference is the second A has a little <em>T</em> in the exponent. This <em>T</em> means transpose. You multiply A by the transpose of A. To find the transpose, you turn the rows into columns.
![A^T=\left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right]](https://tex.z-dn.net/?f=A%5ET%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Now that we have our transpose, we can multiply the matrices.
![\left[\begin{array}{ccc}5&2\\3&-1\\\end{array}\right] \left[\begin{array}{ccc}5&3\\2&-1\\\end{array}\right] =\left[\begin{array}{ccc}5*5+2*2&5*3+2(-1)\\3*5+2(-1)&3*3+(-1)(-1)\\\end{array}\right] =\left[\begin{array}{ccc}29&13\\13&10\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%262%5C%5C3%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%263%5C%5C2%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D5%2A5%2B2%2A2%265%2A3%2B2%28-1%29%5C%5C3%2A5%2B2%28-1%29%263%2A3%2B%28-1%29%28-1%29%5C%5C%5Cend%7Barray%7D%5Cright%5D%20%3D%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D29%2613%5C%5C13%2610%5C%5C%5Cend%7Barray%7D%5Cright%5D)