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erica [24]
3 years ago
6

Numbers 11-14 pls I’m having a lot of trouble

Mathematics
1 answer:
Ksivusya [100]3 years ago
6 0

The term "closed" in math means that if you take two items from a set, do some operation, then you'll always get another value in the same set (sometimes you may get the same value as used before). For example, adding two whole numbers leads to another whole number. We therefore say "the set of whole numbers is closed under addition". This applies to integers as well because integers are positive and negative whole numbers. So we can say that integers are closed under addition.

Integers are not closed under division. Take two integers like 2 an 5 and divide: 2/5 = 0.4 which is not an integer. Integers don't have decimal parts.

The set of whole numbers is {0,1,2,3,...} and we can subtract the two values 1 and 2 to get 1-2 = -1. The order matters here. Subtracting a larger value from a smaller leads to a negative. The value -1 is not in the set of whole numbers. So we can say that whole numbers is not closed under subtraction

Finally, the set of irrational numbers is closed under addition. Adding any two irrational numbers leads to another irrational number. For instance, pi+sqrt(2) = 3.142 + 1.414 = 4.556; I'm using rounded decimals as approximate values. An irrational number is one where we cannot write it as a fraction of integers. Contrast that with a rational number in which we can write it as a fraction of integers. Example: 10 = 10/1 is a rational number.

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Im currently in the middle of taking the test :(
Ber [7]

Answer:

So..how was the test?

Step-by-step explanation:

Did you pass???

7 0
3 years ago
I know that real numbers consist of the natural or counting numbers, whole numbers, integers, rational numbers and irrational nu
ra1l [238]

The imaginary unit i belongs to the set of complex numbers, denoted by \mathbb C. These numbers take the form a+bi, where a,b are any real numbers.

The set of real numbers, \mathbb R, is a subset of \mathbb C, where each number in \mathbb R can be obtained by taking b=0 and letting a be any real number.

But any number in \mathbb C with non-zero imaginary part is not a real number. This includes i.

  • "is it possible that i can use an imaginary number for a real number"

I'm not sure what you mean by this part of your question. It is possible to represent any real number as a complex number, but not a purely imaginary one. All real numbers are complex, but not all complex numbers are real. For example, 2 is real and complex because 2=2+0i.

There are some operations that you can carry out on purely imaginary numbers to get a purely real number. A famous example is raising i to the i-th power. Since i=e^{i\pi/2}, we have

i^i=\left(e^{i\pi/2}\right)^i=e^{i^2\pi/2}=e^{-\pi/2}\approx0.2079

3 0
3 years ago
Express this in standard form 5.4×10^1​
Naily [24]
Well it would just be 54 since you just need to move the decimal over by 1.

So 5.4x10= 54
7 0
4 years ago
What is 824.653 rounded to hundredths place
iren2701 [21]

Answer:

824.65

Step-by-step explanation:

6 0
3 years ago
Read 2 more answers
Find the sum 2 + 4 + 6...+30
I am Lyosha [343]

Answer:

240

Step-by-step explanation:

If you are referring to:

2 + 4 + 6 + 8 + 10 + 12 + 14 + 16 + 18 + 20 + 22 + 24 + 26 + 28 + 30, then your answer would be 240

Hope this helps :)

6 0
3 years ago
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