Factors for 68: 1,2,4,17,34,68,<span>
Multiples of 6: </span>6<span>,12,18,24,30,36,42,48,54,60,66,72,78,84,90,96,102,108,114,120,126,132,138,144,150,156,162,168,174,180,186,192,198,204,210,216,222,228,234,240,246,252,258,264,270,276,282,288,294
Multiples for 68: </span>68,136,204,272,340,408,476,544,612,680,748,816,884,952,1020,1088,1156,1224,1292,1360,1428,1496,1564,1632,1700,1768,1836,1904,1972,2040,2108,2176,2244,2312,2380,2448,2516,2584,2652,2720,2788,2856,2924,2992,3060,3128,3196,3264,3332,
Factors of 6: <span>1,2,3,6
Happy studying ^_^</span>
The truth set for each predicate are:
a) d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}
b) d ∈ { 1, 2, 3, 6}
c) x ∈ { [-2, -1] U [1, 2]}
d) x ∈ {-2, -1, 1, 2}
<h3>
How to find the truth set of each predicate?</h3>
We only need to find the sets of values such that the statements are true.
a)
We want to find vales of d, such that d is an integer, and 6/d is an integer.
Here the possible values of d will be:
d ∈ { -6, -3, -2, -1, 1, 2, 3, 6}
Which are all the factors of 6, so all these integers divide 6.
b) Same as before, but this time the domain is a positive integer, so now the truth set will be:
d ∈ { 1, 2, 3, 6}
c) We want to find real values of x such that:
1 ≤ x² ≤ 4
If we apply the square root to the 3 sides, we get two inequalities:
-√1 ≥ x ≥ -√4
√1 ≤ x ≤ √4
Simplifying:
-1 ≥ x ≥ -2
1 ≤ x ≤ 2
So the truth set is:
x ∈ { [-2, -1] U [1, 2]}
c) Same as before, but now we only have integer solutions, so the truth set is:
x ∈ {-2, -1, 1, 2}
If you want to learn more about predicates:
brainly.com/question/18152046
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