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agasfer [191]
2 years ago
12

What is the solution set of {x | x < 2} {x | x ≥ 2}? all numbers less than -5 and greater than 5 the empty set the numbers be

tween -5 and 5 all real numbers
Mathematics
1 answer:
juin [17]2 years ago
3 0

Answer:

The solution set is the empty set.

Step-by-step explanation:

{x | x < 2} is the set of all numbers less than 2. This means x can take values such as 1.99, 0, -2000 and so on. That is, all values less than 2.

{x | x ≥ 2} is the set of all numbers equal to or greater than 2. This means x can take values such as 2, 2.1, 5000, and so on. That is, 2 or any value greater than 2.

Since there is no sign between the two sets, the question is asking for the intersection between these two sets. That is, what elements are common to these two sets? As we can see, the two sets don't have any common element. Hence, their intersection is the empty set.

(Note that the union of these two sets would be the set of all real numbers as that includes all elements from either set).

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Answer:

22

Step-by-step explanation:

40x.55=22

5 0
3 years ago
Simplify each exponential expression using the properties of exponents and match it to the correct answer.
saveliy_v [14]

1) (2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2 = 2

2) (3^3) (4^0)^2 (3\times2)^-3 (2^2) = 1/2

3) (3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4) = 2

4) (2.3)^-1 (2^0) / (2.3)^-1 = 1

<u>Step-by-step explanation</u>:

Step 1 :

(2\times3^-2)^3 (5\times3^2)^2 / (3^-2)(5\times2)^2

⇒ (2^3) (3^-6) (5^2) (3^4) / (3^-2) (10^2)

⇒ (2.2^2.5^2) (3^-6.3^4) / (3^-2) (10^2)

⇒ (2)(10^2) (3^-2) / (3^-2) (10^2)

⇒ 2

Step 2 :

(3^3) (4^0)^2 (3\times2)^-3 (2^2)

Any number with power 'zero' is 1.

⇒ (3^3) (1^2) (3^-3) (2^-3) (2^2)

⇒ (2^(-3+2))

⇒ 2^-1 = 1/2

Step 3 :

(3^7\times4^7) (2\times5)^-3 (5)^2 / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (12^7) (2^-3) (5^-3) (5^2) / (12^7) (5^-1) (2^-4)

⇒ (2^-3) (5^(-3+2)) / (5^-1) (2^-4)

⇒ (2^-3) (5^-1) / (5^-1) (2^-4)

⇒ 1 / (2^-1)

⇒ 2

Step 4 :

(2.3)^-1 (2^0) / (2.3)^-1

⇒ (2^0)

⇒ Any number with power 'zero' is 1

⇒ 1

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