The members of a possible set C are {6, 9, 12}
<h3>Set theory</h3>
The universal set is the set that contains all other sets. Given the following sets
ξ = {x:x is an integer, 2 ≤ x ≤ 14}
B = {x:x is a prime number}
A = {x:x is a multiple of 3}
The sets B and A are:
B = {2, 3, 5, 7, 11}
A = {3, 6, 9, 12}
If C is a subset of A with a cardinality of 3, hence the required set will be {6, 9, 12}. Note that 3 is excluded since B n C must be an empty set.
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Answer:
2.0 x 10^11
Step-by-step explanation:
2.6 divide it by 1.3 and get 2.0. 9+2 = 11. Together with the, it would be 10^11. That how I got 2.0 x 10^11
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5 x 4 x 2 = 40 would be the answer to how many different outfits.
There might be something missing in the given equation. Since C is a function of x, the equation must contain x. For sample purposes, let's just assume the equations are:
C(x) = 0.35x for x≤8000
C(x) = 0.75x for x≥8000
The domain is from 0 to 42000. This is the x-axis as shown in the picture. Sample calculations are shown on the left of the graph. The blue graph is for the function C(x) = 0.35x for x≤8000, while the orange graph is for the functionC(x) = 0.75x for x≥8000.The graph break is located when x = 8000.
2x3=6. So the Greatest Common Factor (GCF) is 6