Mathematics, 12.07.2019 23:40 EllaLovesAnime

List the members of these sets. a.{x | x is a real number such that x2 = 1} b.{x | x is a positive integer less than 12} c.{x | x is the square of an integer and x < 100} d.{x | x is an integer such that x2 = 2}

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Answer from: yoyo5752

The sets are:

a) {-1, 1}b) {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}c) {1, 4, 9, 16, 25, 36, 49, 64, 81}d) {∅}.How to find the elements of each set?

We need to find all the values of x that meet the given restrictions for each set.

a) Here we know that x is a real number and we must have:

x^2 = 1

Solving for x:

x = ±√1 = ±1

Then this set is:

{-1, 1}

b) Here x is a positive integer smaller than 12, this is just:

{1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11}

c) In this case x is a square number, and it must be smaller than 100, so let's find all the square values smaller than 100.

1^2 = 12^2 = 43^2 = 94^2 = 165^2 = 256^2 = 367^2 = 498^2 = 649*9 = 8110*10 = 100 (from this onwards, the squares don't meet the criteria).

Then this set is:

{1, 4, 9, 16, 25, 36, 49, 64, 81}

d) Here x must be an integer, such that x^2 = 2

Solving the equation we get:

x = ±√2

But √2 is an irrational number, so there is no integer number that meets this restriction, this means that we have an empty set, this is written as:

{∅}.

If you want to learn more about sets, you can read:

link

Answer from: babyskitt

a.{x | x is a real number such that x^2 = 1}

x^2 = 1 => x = +/- 1

=> {-1, 1} < answer

b.{x | x is a positive integer less than 12}

1 ≤ x < 12 => {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11} < answer

c.{x | x is the square of an integer and x < 100}

x = n^2 < 100 => n^2 - 100 < 0

=> (n - 10) (n + 10) < 0

=> a) n - 10 > 0 and n + 10 < 0 => n > 10 and n < - 10 which is not possible

b) n - 10 < 0 and n + 10 > 0 => n < 10 and n > - 10 => - 10 < n < 10

=> n = { - 9, - 8, - 7, - 6, - 5, -4, -3, -2, -1, 0, 1, 2, 3, 4, 5, 6, 7, 8, 9}

=> x = {0, 1, 4, 9, 16, 25, 36, 49, 64, 81} < answer

d.{x | x is an integer such that x^2 = 2}

x = {∅ } because x is √2 which is not an interger but an irrational number

=> { ∅ }

Answer from: Quest

answer: the answer is one

step-by-step explanation:

pemdas, in this situation there is no parenthesis or exponents so you go to multiplication, 1x1=1, then division 1/1= 1 then addition 1+1=2 then subtraction 2-1=1

Answer from: Quest

hence, the required result is $b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]$

step-by-step explanation:

suppose we have two matrices a and b

we will assume the two matrices

$a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]$

$b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]$

we have to compute b-a which means subtracting corresponding elements of a from b.

$b=\left[\begin{array}{ccc}2& 3& 1\\5& 6& 4\\8& 9& 7\end{array}\right]$ - $a=\left[\begin{array}{ccc}1& 2& 3\\4& 5& 6\\7& 8& 9\end{array}\right]$

$b-a=\left[\begin{array}{ccc}1& 1& -2\\1& 1& -2\\1& 1& -2\end{array}\right]$

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