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Lesechka [4]
3 years ago
7

The inequality x + 2y ≥ 3 is satisfied by point (1, 1). True False

Mathematics
2 answers:
andre [41]3 years ago
7 0
In the given statement above, in this case, the answer would be TRUE. It is true that the  inequality x + 2y ≥ 3 is satisfied by point (1, 1). In order to prove this, we just have to plug in the values. 1 + 2(1) <span> ≥ 3 
So the result is 1 + 2 </span> ≥ 3. 3 <span> ≥ 3, which makes it true, because it states that it is "more than or equal to", therefore, our answer is true. Hope this answer helps.</span>
ki77a [65]3 years ago
6 0

Answer:

True

Step-by-step explanation:

We can check by substitute x=1 into the inequality to determine if the y will be equal to 1 or more:

1+2y≥3

solve for y:

2y≥3-1

y≥2/2

y≥1

y is equal to 1 or more. Therefore the inequality is satisfied for point (1,1)

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8 0
3 years ago
Given the following sets:
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<h3>Answer:  10</h3>

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

Explanation:

Even though your teacher doesn't want you to list the items of the set, it helps to do so.

We'll be working with these two sets

A = {b, d, f, h, j, I, n, p, r, t}

C =  {d, h, I, p, t}

When we union them together, we combine the two sets together. Think of it like throwing all the letters in one bin rather than two bins.

A u C = {b, d, f, h, j, I, n, p, r, t,   d, h, I, p, t  }

The stuff that isn't bolded is set A, while the stuff that is bolded is set C

After we toss out the duplicates, we end up with this

A u C = {b, d, f, h, j, I, n, p, r, t}

But wait, that's just set A. Notice how everything in set C can be found in set A. This indicates set C is a subset of set A.

That's why all of the stuff in bold was tossed out (because they were duplicates of stuff already mentioned).

Once we determine what set A u C looks like, we count out the number of items in that set to determine the final answer.

There are 10 items in {b, d, f, h, j, I, n, p, r, t} which means 10 is the final answer.

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

An alternative method is to use the formula below

n(A u C) = n(A) + n(C) - n(A and C)

n(A u C) = 10 + 5 - 5

n(A u C) = 10

The notation n(A and C) counts how many items are found in both sets A and C at the same time. But as mentioned earlier, this is identical to just counting how many items are in set C. So we'll have n(C) cancel out with itself.

In short, n(A u C) = n(A) = 10

8 0
3 years ago
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