Hello I hope this helps. I'm pretty sure this is the answer.
Part A: C
Part B: A
Step-by-step explanation:
a). A = {x ∈ R I 5x-8 < 7}
5x - 8 < 7 <=> 5x < 8+7 <=> 5x < 15 =>
x < 3 => A = (-∞ ; 3)
A ∩ N = {0 ; 1 ; 2}
A - N* = (-∞ ; 3) - {1 ; 2}
b). A = { x ∈ R I 7x+2 ≤ 9}
7x+2 ≤ 9 <=> 7x ≤ 7 => x ≤ 1 => x ∈ (-∞ ; 1]
A ∩ N = {0 ; 1}
A-N* = (-∞ ; 1)
c). A = { x ∈ R I I 2x-1 I < 5}
I 2x-1 I < 5 <=> -5 ≤ 2x-1 ≤ 5 <=>
-4 ≤ 2x ≤ 6 <=> -2 ≤ x ≤ 3 => x ∈ [-2 ; 3]
A ∩ N = {0 ; 1 ; 2 ; 3}
A - N* = [-2 ; 3) - {1 ; 2}
d). A = {x ∈ R I I 6-3x I ≤ 9}
I 6-3x I ≤ 9 <=> -9 ≤ 6-3x ≤ 9 <=>
-15 ≤ -3x ≤ 3 <=> -5 ≤ -x ≤ 3 =>
-3 ≤ x ≤ 5 => x ∈ [-3 ; 5]
A ∩ N = {0 ; 1 ; 2 ; 3 ; 4 ; 5}
A - N* = [-3 ; 5) - {1 ; 2 ; 3 ; 4}
Answer:
after 75 minutes
Step-by-step explanation:
The least common multiple (LCM) of 15 and 25 is 75. It can be found a couple of ways:
1. List the factors of each number and find the product of the unique ones:
15 = 3·5
25 = 5²
The LCM is 3·5² = 75.
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2. Find the greatest common divisor (GCD) and divide the product of the numbers by that value. From the above list of factors, we see that 5 is the GCD of 15 and 25. Then the LCM is ...
15·25/5 = 75
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Or, you can simply list multiples of each number and see what the smallest number is that is in both lists:
15, 30, 45, 60, <em>75</em>, 90
25, 50, <em>75</em>, 100
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The two buses will appear together again after 75 minutes.
The answer is b and c
because they are both parallel to the y-axis
