Let v ∈ Z, v > 5, Show that v ∈ im(f) if and only if v + 5 is even

the number of possible outcomes that do not show a 1 on the first spinner and show the number 4 on the second spinner is 2.

spinner 1 spinner 2

1                     1

1                     2

1                     3

1                     4

2                   1

2                   2

2                   3

2                   4   = not 1, and 4

3                   1

3                   2

3                   3

3                   4 = not 1, and 4

only 2 outcomes have passed the given condition.

step-by-step explanation:

6 and 24

step-by-step explanation:

6 x 4 = 24

12r - 8 - 12

you have to combine like terms. below i have bolded the like terms:

12r - 8 - 12

together it combines to:

12r - 20

hope this !

i think the answer is b x = 1, y = 35/2