Determine whether each of these functions from {a, b,c, d} to itself is one-to-one. a) f(a) = b, f(b) = a, f(c) = c, f(d) = d. b) f(a) = b, f(b) = b, f(c)=d, f(d) = c. c) f(a) = d, f(b) = b, f(c)=c, f(d) = d.

The first one is one-to-one.

The second one is not one-to-one.

Third one is not one-to-one.

The problem:

Are the following one-to-one from {a,b,c,d} to {a,b,c,d}:

a)

f(a)=b

f(b)=a

f(c)=c

f(d)=d

b)

f(a)=b

f(b)=b

f(c)=d

f(d)=c

c)

f(a)=d

f(b)=b

f(c)=c

f(d)=d

Step-by-step explanation:

One-to-one means that a y cannot be hit more than once, but all the y's from the range must be hit.

So the first one is one-to-one because:

f(a)=b

f(b)=a

f(c)=c

f(d)=d

All the elements that got hit are in {a,b,c,d} and all of them were hit.

The second one is not one-to-one.

The reason is because f(a) and f(b) both are b.

Third one is not one-to-one.

The reason is because f(a) and f(d) are both d.

= 5

step-by-step explanation:

to find the 15th term substitute n = 15 into the nth term formula

= (- 3 × 15 ) + 50 = - 45 + 50 = 5