Here's what I got for the first question.
Did you want answers for the second one? I couldn't read that part of the image very well, if you did.
2x=-12 simplified, the expression or solving for X would be X=-6
Ok, so find circumference
arc/circumferece=xrad/2pirad
c=2pir
c=2pi6
c=12pi
arc=5.4cm
5.4/12pi=x/2pi
tims both sides by 2pi
5.4/6=x
0.9=x
0.9 radians
Answer:
see explanation
Step-by-step explanation:
The inverse of a matrix A = ![\left[\begin{array}{ccc}a&b\\c&d\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Da%26b%5C%5Cc%26d%5C%5C%5Cend%7Barray%7D%5Cright%5D)
is
= 
![\left[\begin{array}{ccc}d&-b\\-c&a\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7Dd%26-b%5C%5C-c%26a%5C%5C%5Cend%7Barray%7D%5Cright%5D)
If ad - bc = 0 then the matrix has no inverse
A = ![\left[\begin{array}{ccc}11&-5\\3&-1\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D11%26-5%5C%5C3%26-1%5C%5C%5Cend%7Barray%7D%5Cright%5D)
ad - bc = (11 × - 1) - (- 5 × 3) = - 11 - (- 15) = - 11 + 15 = 4 , then
= 
![\left[\begin{array}{ccc}-1&3\\-5&11\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-1%263%5C%5C-5%2611%5C%5C%5Cend%7Barray%7D%5Cright%5D)
= ![\left[\begin{array}{ccc}-0.25&0.75\\-1.25&2.75\\\end{array}\right]](https://tex.z-dn.net/?f=%5Cleft%5B%5Cbegin%7Barray%7D%7Bccc%7D-0.25%260.75%5C%5C-1.25%262.75%5C%5C%5Cend%7Barray%7D%5Cright%5D)
Answer:
Step-by-step explanation:
Given
Straight Line, PS
Such that
Required
Find PS
From the attachment above, it can be seen that RS is a continuation of PS;
This implies that
Substitute and
Collect like terms
Hence, the length of PS is
