Answer:
A and C
Step-by-step explanation:
To determine which events are equal, we explicitly define the elements in each set builder.
For event A
A={1.3}
for event B
B={x|x is a number on a die}
The possible numbers on a die are 1,2,3,4,5 and 6. Hence event B is computed as
B={1,2,3,4,5,6}
for event C
![C=[x|x^{2}-4x+3]\\solving x^{2}-4x+3\\x^{2}-4x+3=0\\x^{2}-3x-x+3=0\\x(x-3)-1(x-3)=0\\x=3 or x=1](https://tex.z-dn.net/?f=C%3D%5Bx%7Cx%5E%7B2%7D-4x%2B3%5D%5C%5Csolving%20%20x%5E%7B2%7D-4x%2B3%5C%5Cx%5E%7B2%7D-4x%2B3%3D0%5C%5Cx%5E%7B2%7D-3x-x%2B3%3D0%5C%5Cx%28x-3%29-1%28x-3%29%3D0%5C%5Cx%3D3%20or%20x%3D1)
Hence the set c is C={1,3}
and for the set D {x| x is the number of heads when six coins re tossed }
In the tossing a six coins it is possible not to have any head and it is possible to have head ranging from 1 to 6
Hence the set D can be expressed as
D={0,1,2,3,4,5,6}
In conclusion, when all the set are compared only set A and set C are equal
C:t=2:3
t:h=2:1
multiply first equation by 2 and 2nd by 3
c:t=4:6
t:h=6:3
so
c:t:h=4:6:3
if t is 4 more than c
t is 6 units and c is 4
6-4=2
so 4=2 units
2=1 unit
hollly is 3 units
3*2=6
holly has 6 cards
Using a graphing tool
Let's graph each of the cases to determine the solution of the problem
<u>case A)</u> 
see the attached figure N 
The range is the interval--------> (0,∞)
therefore
the function is not the solution
<u>case B)</u> 
see the attached figure N 
The range is the interval--------> (0,∞)
therefore
the function is not the solution
<u>case C)</u> 
see the attached figure N 
The range is the interval--------> (-∞,3)
therefore
the function is the solution
<u>case D)</u> 
see the attached figure N 
The range is the interval--------> (-3,∞)
therefore
the function is not the solution
<u>the answer is</u>
Hence the circumference of the circle is: 6π or 18.84 units
