All positive integers $t$ such that $1.2t \leq 9.6$?
1 answer:
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Answer:
(a)
(b) 10
(c)
(d) 0
(e) 1024
Step-by-step explanation:
(a)
A = {x ∈ Z | 0 < x, x² ≤ 100}
We need to find all the elements of given set.
The given conditions are
... (1)
Taking square root on both sides.
.... (2)
Using (1) and (2) we get
Since x ∈ Z,
(b)
We need to find the value of | {x ∈ Z | 0 < x, x² ≤ 100}| or |A|. It means have to find the number of elements in set A.
| {x ∈ Z | 0 < x, x² ≤ 100}| = 10
(c)
B = {x ∈ Z | x > 10, x² ≤ 100}
We need to find all the elements of given set.
The given conditions are
... (3)
It means
.... (4)
Inequality (3) and (4) have no common solution, so B is null set or empty set.
(d)
We need to find the value of |{x ∈ Z | x > 10, x² ≤ 100}| or |B|. It means have to find the number of elements in set B.
|{x ∈ Z | x > 10, x² ≤ 100}| = 0
(e)
We need to find the value of | P(A) |. P(A) is the power set of set A.
Number of elements of a power set is
where, n is the number of elements of set A.
We know that the number of elements of set is 10. So the value of |P(A)| is
Therefore |P(A)|=1024.