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ValentinkaMS [17]
3 years ago
9

Solve the given inequality and graph the solution on a number line -x/2 + 3/2 + < 5/2 .

Mathematics
2 answers:
liq [111]3 years ago
5 0
Answers in Image attached

Hope this helps

irina [24]3 years ago
5 0
Well, in that case, here goes...

The first thing I'd do is clear the fractions- that is, I'd multiply each of them by 2 because 2 is the lowest common denominator. 

That leaves you with -x + 3 < 5
Now solving isn't too bad...

-x + 3 - 3 < 5 - 3                 (subtract 3 from both sides)
-x < 2

The x isn't alone! That - is hanging on...so multiply both sides by -1 to get rid of it! 

-x * -1 < 2 * -1          WAIT! We multiplied by a negative!!
-x * -1 > 2 * -1          Had to flip that inequality ofver!

x > -2


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Let D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36} Determine which of the following statements are true and which are false. a)
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Answer:

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is odd then x> 0 is true statement.

\forall x\in D if x is even then x≤0 is false statement.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

Step-by-step explanation:

Consider the provided information.

D = {-48, -14, -8, 0, 1, 3, 16, 23, 26, 32, 36}

Part (A) \forall x\in D if x is odd then x> 0

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (B) \forall x\in D if x is less than 0 then x is even.

Here only even numbers are less than 0 that means the statement is true.

\forall x\in D if x is odd then x> 0 is true statement.

Part (C) \forall x\in D if x is even then x≤0

Here we can see that 16, 26, 32, 36 are even number and also greater than 0. Thus the statement is false.

\forall x\in D if x is even then x≤0 is false statement.

Part (D) \forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4.

There is only one number whose ones digit is 2. i.e. 32 also the tens digit of the number 32 is 3. Which makes the above statement true.

\forall x\in D If the ones digit of x is 2, then the tens digit is 3 or 4 is true statement.

Part (E) \forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2.

Numbers having ones digit 6 are: 16, 26 and 36

Here, the tens digits are 1, 2 and 3 which is contradict to our statement. Hence the provided statement is false.

\forall x\in D if the ones digit of x is 6, then the tens digit is 1 or 2 is false statement.

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