x+2 > 10 solves to x > 8 after we subtract 2 from both sides
So set A is the set of real numbers that are larger than 8. The value 8 itself is not in set A. The same can be said about 5 as well.
Set B is the set of values that are larger than 5 since 2x > 10 turns into x > 5 after dividing both sides by 2. The value x = 5 is not in set B since x > 5 would turn into 5 > 5 which is false. The values x = 6, x = 8, and x = 9 are in set B.
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Summarizing everything, we can say...
5 is not in set A. True
5 is in set B. False
6 is in set A. False
6 is not in set B. False
8 is not in set A. True
8 is in set B. True
9 is in set A. True
9 is not in set B. False
12÷3=4
cause 1/3 in this situation is 3 cause 12÷4 is 3
Answer:
x>2 or x<-10
Step-by-step explanation:
9|x +4|>54
Divide each side by 9
9/9|x +4|>54/9
|x +4|>6
There are two solutions, one positive and one negative
x+4 >6 or x+4 < -6
Subtract 4 from each side
x+4-4 >6-4 or x+4-4 < -6-4
x>2 or x<-10
Answer: Choice C. p = 250(0.79)^t
Work Shown:
p = a*b^t
p = a*(1+r)^t
p = 250*(1+(-0.21))^t
p = 250(0.79)^t
Note that r = -0.21 is negative to indicate we have exponential decay.
