Answer:
A=63.6 approximately
Step-by-step explanation:
P=14.28
p=1/4(2pi*r)
p=1/2pi*r
14.28=1/2pi*r
28.56=3.14*r
r=9...... approximately
A=1/4pi*r^2
A=0.25*3.14*81
A=63.6......approximately
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)
-6(a+b+c) +6(x+y+z)
= -6(-10) +6(-10)
= 60 -60 = 0
I think you multiply 140 and 80% and then add thst to the 140