Answer:
...let x be the number
x+12+12-9+7=24
x+31-9=24
x+22=24
x=24-22=2
i am 2
Step-by-step explanation:
are all fractions with a denominator of 9 recurring decimals investigate
b. { -2 ≤ x ≤ 2}
c. { -2 ≤ x ≤ 1 }
d. { -1 ≤ x ≤ 1 }
e. { 0 ≤ x ≤ 1 }
Tolong caranya...
The solution to the given inequality, |2x + 1| ≤ 3, is the set { -2 ≤ x ≤ 1 }, which corresponds to answer choice (c). This is achieved by solving two separate inequalities, 2x + 1 ≤ 3 and - (2x + 1) ≤ 3.
The subject of the question is Mathematics, specifically a High School algebra topic on solving absolute inequalities. The student's question is asking us to solve the inequality |2x + 1| ≤ 3. To do this, we need to create and solve two separate inequalities: 2x + 1 ≤ 3 and - (2x + 1) ≤ 3.
Solving 2x + 1 ≤ 3 gives us 2x ≤ 2 and x ≤ 1 . Solving - (2x + 1) ≤ 3 gives us -2x - 1 ≤ 3 , then -2x ≤ 4 , and finally x ≥ -2 . Combining these answers gives us the solution set { -2 ≤ x ≤ 1 }, which corresponds to answer choice (c).
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