Answer:
[tex]A:\\R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z[/tex]
[tex]B:\\R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\neq G[/tex]
[tex]C:\\R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3[/tex]
D:
[tex]R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\neq8[/tex]
Step-by-step explanation:
Let:
[tex]R=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}red\hspace{3}dice\\G=The\hspace{3}number\hspace{3}shows\hspace{3}by\hspace{3}the\hspace{3}green\hspace{3}dice[/tex]
A:
This is easy, simply R=4, and we don't know the value of G, so:
[tex]R=4\hspace{8}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z[/tex]
B: Be careful here, we know the numbers add to 5, so we don't know the exact value of R or G, because R could be 4 and G could be 1 or R could be 2 and green could be 3. However we can be sure that they can´t have the same value, because 5 is an odd number, if we add the same number to a number the result is always even: n+n=2n. So:
[tex]R+G=5\hspace{8}R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{5}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{3}and\hspace{3}R\neq G[/tex]
C: R is 3 or G is 3, that's the only thing we know, hence:
[tex]R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5}R =3 \hspace{3}or\hspace{3}G=3[/tex]
D: Similar to B, we only know that the sum of R and G isn't 8:
[tex]R=\{x\hspace{3} ;\hspace{3}1\leq x \leq 6 \}\hspace{3}x\in Z\hspace{3}G=\{y\hspace{3} ;\hspace{3}1\leq y\leq 6 \}\hspace{3}y\in Z\hspace{5};\hspace{3}R+G\neq8[/tex]
