Answer:
1. b ∈ B 2. ∀ a ∈ N; 2a ∈ Z 3. N ⊂ Z ⊂ Q ⊂ R 4. J ≤ J⁻¹ : J ∈ Z⁻
Step-by-step explanation:
1. Let b be the number and B be the set, so mathematically, it is written as
b ∈ B.
2. Let a be an element of natural number N and 2a be an even number. Since 2a is in the set of integers Z, we write
∀ a ∈ N; 2a ∈ Z
3. Let N represent the set of natural numbers, Z represent the set of integers, Q represent the set of rational numbers, and R represent the set of rational numbers.
Since each set is a subset of the latter set, we write
N ⊂ Z ⊂ Q ⊂ R .
4. Let J be the negative integer which is an element if negative integers. Let the set of negative integers be represented by Z⁻. Since J is less than or equal to its inverse, we write
J ≤ J⁻¹ : J ∈ Z⁻
Answer:
One solution: (2.5,0)
Step-by-step explanation:
We can use substitution for the solution. Substitute y=2x-5 to -8x-4y=-20 so that you end up with -8x-8x+20=-20. Next you want to add like terms which will be -16x+20=-20, next you want isolate x by subtracting 20 from both sides and leaves you with -16x=-40. Divide -16 on both sides to fully isolate x and will leave you with x=2.5. Now substiture in 2.5 for x in y=2x-5 to get y=2(2.5)-5 which will then lead to y=0.
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M=230
m-80=150
add 80 to each sides
