Answer:
The fourth pair of statement is true.
9∈A, and 9∈B.
Step-by-step explanation:
Given that,
U={x| x is real number}
A={x| x∈ U and x+2>10}
B={x| x∈ U and 2x>10}
If 5∈ A, Then it will be satisfies x+2>10 , but 5+2<10.
Similarly, If 5∈ B, Then it will be satisfies 2x>10 , but 2.5=10.
So, 5∉A, and 5∉B.
If 6∈ A, Then it will be satisfies x+2>10 , but 6+2<10.
Similarly, If 6∈ B, Then it will be satisfies 2x>10 , and 2.6=12>10.
So, 6∉A, and 6∈B.
If 8∈ A, Then it will be satisfies x+2>10 , but 8+2=10.
Similarly, If 8∈ B, Then it will be satisfies 2x>10. 2.8=16>10.
So, 8∉A, and 8∈B.
If 9∈ A, Then it will be satisfies x+2>10 , but 9+2=11>10.
Similarly, If 9∈ B, Then it will be satisfies 2x>10. 2.9=18>10.
So, 9∈A, and 9∈B.
Answer: y = 2x+8
Check out this graph I supplied you with!
Answer: ![-\frac{1}{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D)
<u>Step-by-step explanation:</u>
![\frac{5x}{6x+3}](https://tex.z-dn.net/?f=%5Cfrac%7B5x%7D%7B6x%2B3%7D)
Restriction: 6x + 3 ≠ 0 <em>denominator cannot be zero</em>
<u> -3 </u> <u> -3 </u>
6x ≠ -3
<u> ÷6 </u> <u>÷6 </u>
x ≠ ![-\frac{1}{2}](https://tex.z-dn.net/?f=-%5Cfrac%7B1%7D%7B2%7D)
Answer: the odds would still be 1/11 or the percentage of drawing a "g" would be 9.1%
Step-by-step explanation: