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balandron [24]
3 years ago
13

Three and two third divided by two

Mathematics
1 answer:
suter [353]3 years ago
8 0

Answer:

1⅚

Step-by-step explanation:

(3 + ⅔) ÷ 2

11/3 ÷ 2

11/3 × 1/2

11/6

1⅚

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Select the statements that are true based on the following given information. D = {x | x is a whole number} E = {x | x is a perf
likoan [24]
<h3>2 Answers: Choice C and Choice D</h3>

==========================================================

Explanation:

Let's write the roster notation of each set

D = {0, 1, 2, 3, ...} the dots indicate the pattern goes on forever

E = {0, 1, 4, 9, 16, 25} = list of perfect squares smaller than 36

F = {20, 22, 24, 26, 28, 30} = even numbers between 20 and 30

----------------

If we intersect sets D and E, we're looking for what numbers are in both sets at the same time. Therefore, D ∩ E = {0, 1, 4, 9, 16, 25} which is the exact same as set E. This is because all of set E is inside of set D. We say that set E is a subset of set D. So, D ∩ E = E.

Choice A is very close to being true. The problem is that 25 is missing from the set {1,4,9,16}. So this is why choice A is false.

----------------

Now let's intersect sets D and F. The numbers they have in common are {20, 22, 24, 26, 28, 30} which is exactly what set F is. So set F is a subset of set D. We can write D ∩ F = F in much the same way we can say D ∩ E = E.

D ∩ F = {12,14,16,18} is not true. A number like 12 is not between 20 and 30, so it cannot be in set F.

Choice B is false so we cross it off the list.

----------------

Choice C is true and here's why

D ∪ (E ∩ F) is the same as saying D ∪ G where G is the set of intersecting E and F together. In other words, G = E ∩ F

We don't really need to even worry about sets E, F or G. All that matters here is set D.

When we write D ∪ (E ∩ F) or D ∪ G, we're saying "a number is in set D, or it is in set G". If it is in D, then it's a whole number. Otherwise, it's in a subset of whole numbers.

Overall, D ∪ G and D ∪ (E ∩ F) form the entire set of whole numbers.

------------------

Choice D is true.

E and F have nothing in common

E = {0, 1, 4, 9, 16, 25}

F = {20, 22, 24, 26, 28, 30}

So intersecting them leads to the empty set. This is the set with nothing inside it, not even 0.

------------------

Choice F is false

These two sets below

E = {0, 1, 4, 9, 16, 25}

F = {20, 22, 24, 26, 28, 30}

union together to get

H = {0,1,4,9,16,20,22,24,25,26,28,30}

just toss all of the numbers together into one big set

Notice how each of these numbers are whole numbers, so they are part of set D. This means set H is also a subset of set D.

When we intersect sets D and H, we end up with set H. We do not simply end up with the set with 25 only inside it.

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I think the answer is 2 tyres but i am not sure but don’t blame me if it’s wrong
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