
The number "x" is given by:
Therefore, we have:
Answer:
Q.#1 is a
Q.#2 is d
Step-by-step explanation:
hope this helps:)
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)
Answer:
(-1,-7)
Step-by-step explanation:
Substitute 2x-5 (from the first equation) as the y-value for the second equation, so you have something that looks like this:
2x-5 = -2x-9
Add 2x to both sides: 4x-5 = -9
Add 5 to both sides 4x = -4
Divide both sides by 4: x = -1
Substitute -1 for x in either equation (you'll get the same y-value)