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AnnZ [28]
2 years ago
14

Solve the inequality: |2x+5|>0.5

Mathematics
2 answers:
anyanavicka [17]2 years ago
8 0

Answer:

Step-by-step explanation:

Divide all three terms by 2 to isolate x:

|x + 5/2| > 0.25

Case 1:  x + 5/2 > 0.25 (no absolute value symbol needed), or x > -2.25

Case 2:   -(x + 5/2) > 0.25.  Multiplying both sides by -1 yields

                 x + 5/2 < -0.25.  Subtracting 5/2 from both sides:  x < 5.45

The solution set is    -2.25 < x < 5.45.

Yuliya22 [10]2 years ago
8 0

Answer:

x < - 2.75, and x > - 2.25

Step-by-step explanation:

Here we can apply the absolute value rule " If say | u | > a, and a > 0, then u < - a, and u > a. " If  a were to be less than 0 ( a < 0 ) then their would be infinitely many solutions as the absolute value will always be greater than or equal to 0.

In this case let us consider both options, u < - a, and u > a,

2x + 5 < - 0.5 or 2x + 5 > 0.5 - solving for each of these inequalities we can combine both intervals and receive the solution

2x + 5 < - 0.5 : 2x < - 5.5, x < - 5.5 / 2, x < - 2.75

2x + 5 > 0.5 : 2x > - 4.5, x > - 4.5 / 2, x > - 2.25

Our solution is hence x < - 2.75, and x > - 2.25. Combining each of the intervals we receive our solution. Note that we can't represent this as one compound inequality as their signs differ.

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Essentially, this is a problem of balls and sticks. The 8 identical blackboards can be represented as 8 balls, and you assign them to each school by using 3 sticks. Basically each school receives an amount of blackboards equivalent to the amount of balls between 2 sticks: The first school gets all the balls before the first stick, the second school gets all the balls between stick 1 and stick 2, the third school gets the balls between sticks 2 and 3 and the last school gets all remaining balls.

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If each school needs at least 1 blackboard you can give 1 blackbooard to each of them first and distribute the remaining 4 the same way we did before. This time there will be 4 balls and 3 sticks, so we have to put 3 sticks in 7 spaces (if a school takes what it is between 2 sticks that doesnt have balls between, then that school only gets the first blackboard we assigned to it previously). The amount of ways to localize the sticks is {7 \choose 3} = 35. Thus, there are only 35 ways to distribute the blackboards in this case.

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