Can you think of a number k for which k^2 < k is true?

Some k's that will work:

1/2

1/3

1/4

(There are infinitely many answers.)

Step-by-step explanation:

Obviously 5 won't work because 25<5 is not true.

What about a negative? -5?

25<-5 is not true.

How about 1/3? 1/9<1/3 is true so 1/3 works.

Note: Think of a number between 0 and 1.

1/4<1/2 is true and (1/2)^2=1/4.

1/16<1/4 is true and (1/4)^2=1/16.

You can solve k^2<k.

Subtract k on both sides.

k^2-k<0

Factor.

k(k-1)<0.

So k^2-k is a parabola with x-intercepts (k-intercepts) at k=0 and k=1.

The parabola is open up so any number between 0 and 1 will satisfy k^2<k since between 0 and 1 the parabola is less than y=0 ( the x-axis).

total = principal × ( 1 + rate ) ^ years

total = 600 * (1.04)^5

total = 729.99

answer is c

step-by-step explanation:

i would go with the first one

step-by-step explanation:

hope this