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Sati [7]
3 years ago
14

Simplify. give your answer in lowest terms.

Mathematics
2 answers:
tamaranim1 [39]3 years ago
8 0

The one in the title is different than the picture so I'm going to solve the picture and hope that is what you are looking for.

First we must find the least common multiple of 2 and 4. It is 4.

To get the denominator of the first term, 1/2, to be 4 we must mulitply both top and bottom by 2.

(1/2)(2/2) = 2/4

The number 2 is the same as the fraction 2/1. To get it to have the denominator of 4, we multiply both top and bottom by 4

(2/1)(4/4) = 8/4

Now we solve.

2/4 - 3/4 + 1/4 + 8/4

2/4 + 1/4 - 3/4 + 8/4

3/4 - 3/4 + 8/4

0 + 8/4

8/4 = 2

The answer is 2

OR

(2 - 3 + 1 + 8) / 4

(- 1 + 1 + 8) / 4

(0 + 8) / 4

8 / 4

2

The answer is 2

nataly862011 [7]3 years ago
4 0
This is the answer simplified into the most specific terms

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Step-by-step explanation:

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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)
morpeh [17]

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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3 years ago
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Answer:

i cant see the image?

Step-by-step explanation:

if someone can please tell me what is shows

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