Question

01. Let π‘ˆ = {βˆ’2; 1; 0;1/2;3/4; 2;10/3; 4; βˆ’5}. Explain the elements of each of the following sets: a) {π‘₯ ∈ π‘ˆ/π‘₯ βˆ’ 4 < 0} b) {π‘₯ ∈ π‘ˆ/π‘₯ 2 βˆ’ 6π‘₯ + 8 < 0} c) {π‘₯ ∈ π‘ˆ/4 βˆ’ π‘₯ ≀ 0} d) {π‘₯ ∈ π‘ˆ/π‘₯ 2 + 1 ≀ 0}

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Answer to a math question 01. Let π‘ˆ = {βˆ’2; 1; 0;1/2;3/4; 2;10/3; 4; βˆ’5}. Explain the elements of each of the following sets: a) {π‘₯ ∈ π‘ˆ/π‘₯ βˆ’ 4 < 0} b) {π‘₯ ∈ π‘ˆ/π‘₯ 2 βˆ’ 6π‘₯ + 8 < 0} c) {π‘₯ ∈ π‘ˆ/4 βˆ’ π‘₯ ≀ 0} d) {π‘₯ ∈ π‘ˆ/π‘₯ 2 + 1 ≀ 0}

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Eliseo
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Given set π‘ˆ = {βˆ’2; 1; 0; 1/2; 3/4; 2; 10/3; 4; βˆ’5}

a) {π‘₯ ∈ π‘ˆ/π‘₯ - 4 < 0}
To solve this, first, we need to find the elements in π‘ˆ where π‘₯ - 4 < 0.
π‘₯ - 4 < 0
π‘₯ < 4

The elements in set π‘ˆ that satisfy π‘₯ < 4 are {-2; 1; 0; 1/2; 3/4; 2, 10/3}.

b) {π‘₯ ∈ π‘ˆ/π‘₯^2 - 6π‘₯ + 8 < 0}
To solve this, first, we need to find the elements in π‘ˆ where π‘₯^2 - 6π‘₯ + 8 < 0.
(π‘₯ - 4)(π‘₯ - 2) < 0
The solutions to this inequality are π‘₯ ∈ (2, 4).
Thus, the elements in set π‘ˆ that satisfy this inequality are {3/4}.

c) {π‘₯ ∈ π‘ˆ/4 - π‘₯ ≀ 0}
To solve this, first, we need to find the elements in π‘ˆ where 4 - π‘₯ ≀ 0.
4 - π‘₯ ≀ 0
π‘₯ β‰₯ 4
The only element in set π‘ˆ that satisfies this inequality is {4}.

d) {π‘₯ ∈ π‘ˆ/π‘₯^2 + 1 ≀ 0}
To solve this, first, we need to find the elements in π‘ˆ where π‘₯^2 + 1 ≀ 0.
This inequality has no real solutions because π‘₯^2 + 1 is always greater than 0 for all real values of π‘₯.
Therefore, there are no elements in π‘ˆ that satisfy this inequality.

\textbf{Answer:}
a) {-2; 1; 0; 1/2; 3/4; 2; 10/3}
b) {3/4}
c) {4}
d) No elements in set π‘ˆ satisfy the inequality.

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