This system of equations has no solution.
Answer:
Step-by-step explanation:
a) Yes.
![\forall\ a,b\ \in\ B,\ a+b\in B\ (since\ 0+0=0)](https://tex.z-dn.net/?f=%5Cforall%5C%20a%2Cb%5C%20%5Cin%5C%20B%2C%5C%20a%2Bb%5Cin%20B%5C%20%28since%5C%200%2B0%3D0%29)
b) Yes
![W=\{0,1,2,3,..,\}\ (whole\ numbers)\\\forall\ a,b\ \in\ 3W,\ a+b\in 3W\\since:\\a\ \in\ 3W \Longrightarrow\ \exists\ k_1\in W | a=3*k_1\\\\b\ \in\ 3W \Longrightarrow\ \exists\ k_2\in W | b=3*k_2\\\\a+b=3*k_1+3*k_2=3*(k_1+k_2) \in 3W\\](https://tex.z-dn.net/?f=W%3D%5C%7B0%2C1%2C2%2C3%2C..%2C%5C%7D%5C%20%28whole%5C%20numbers%29%5C%5C%5Cforall%5C%20a%2Cb%5C%20%5Cin%5C%203W%2C%5C%20a%2Bb%5Cin%203W%5C%5Csince%3A%5C%5Ca%5C%20%5Cin%5C%203W%20%5CLongrightarrow%5C%20%5Cexists%5C%20k_1%5Cin%20W%20%7C%20a%3D3%2Ak_1%5C%5C%5C%5Cb%5C%20%5Cin%5C%203W%20%5CLongrightarrow%5C%20%5Cexists%5C%20k_2%5Cin%20W%20%7C%20b%3D3%2Ak_2%5C%5C%5C%5Ca%2Bb%3D3%2Ak_1%2B3%2Ak_2%3D3%2A%28k_1%2Bk_2%29%20%5Cin%203W%5C%5C)
c) Yes
![\forall n\ \in\ \mathbb{N}: \Longrightarrow\ n+1 \in\ \mathbb{N}\\](https://tex.z-dn.net/?f=%5Cforall%20n%5C%20%5Cin%5C%20%5Cmathbb%7BN%7D%3A%20%5CLongrightarrow%5C%20n%2B1%20%5Cin%5C%20%5Cmathbb%7BN%7D%5C%5C)
d) No
![3\in V , 3+3=6\notin V\\](https://tex.z-dn.net/?f=3%5Cin%20V%20%2C%203%2B3%3D6%5Cnotin%20V%5C%5C)
Answer:
F(x)= 4x-1 I think so hope it will help you
1. -6 ≤ x < -1 . . . . conjunction
2. x ≤ 6 . . or . . 10 ≤ x . . . . disjunction
3. 7 ≤ x < 12 . . . . conjunction
4. x < -9 . . or . . -3 ≤ x . . . . disjunction
5. 2 ≤ x ≤ 5 . . . . conjunction
6. x ≤ 54 . . or . . 66 ≤ x . . . . disjunction
7. 39 < x ≤ 43 . . . . conjunction
_____
Your problem statement provided no letters.
Answer:
4
Step-by-step explanation:
24÷6
6 x ? = 24
6 goes into 24 four times
24÷6=4