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3 years ago

I don’t understand this problem

Mathematics

1 answer:

ikadub [295]3 years ago

4 0

Hello! the answer is c

First of all let's eliminate the answers that do not make sense. For A. numbers are NOT irrational because they are a fraction. Numbers are also not irrational because they are a repeating number. C and D are both correct as fractions and terminating decimals are both rational numbers.

34/3≈11.33333333

By using a certain formula you can convert repeating decimals to a fraction. But most people know that .3333333 is 1/3. This gives us 11 1/3 as our answer. So what does it match? It is not a terminating decimal. It repeats but can be written as a fraction. Therefore our answer is C) Rational, because it is a fraction.

I hope this helps!

bran-list please

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nikdorinn [45]

Answer:

When we know all 3 sides, we use Heron's Formula

area = sqrt [s *(s-a) * (s-b) * (s-c)]

where "s" is the semi-perimeter which in this case equals

s = (6 + 8 + 12) / 2 = 13

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Step-by-step explanation:

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3 years ago

What is the median for the set of data? 6, 7, 10, 12, 12, 13

elixir [45]

Answer:

Step-by-step explanation:

If the set of numbers are not in least to greatest first set them up into their positions. But since this set is already in least to greatest we can go to the next step.

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6, 7, 10, 12, 12, 13

Find the mean of the numbers in the middle:

10 + 12 = 22

22 / 2 = 11

The answer is 11

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2 years ago

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2 years ago

e) Given the following: Let X = {1, 2, 3, 4) and a relation R on X as R= {(1,2), (2,3), (3,4)}. Find the reflexive and transitiv

Naddika [18.5K]

Answer:

The answer is $\{(1,1),(2,2),(3,3),(4,4),(1,2)(2,3)(3,4),(1,3),(1,4)\}$

Step-by-step explanation:

Remember that a reflexive relation $R\subset \mathcal{P}(X)$ , where $\mathcal{P}(X)$ is the power set of $X$ , is one which conteins the ordered pairs of the form $(a,a)$ , for $a\in X$ .

So, As the reflexive and transitive closure of $R$ (that we will denote by $\overline{R}$ ) is in particular reflexive, we must add to $R$ the elements $\{(1,1) , (2,2),(3,3),(4,4) \}$

A transitive relation $R$ is one in which if the pair $(a,b)$ and the pair $(b,c)$ are in there, then the pair $(a,c)$ must be there too.

So, to complete the relation $R$ to be reflexive and transitive we must add the pair $(1,3)$ (because $(1,2),(2,3)$ are in $R$ ), the pair $(2,4)$ , and the pair $(1,4)$ because we added the pair $(2,4)$ .