AvenJohn

yes, but just because 2xy is an integer, it doesn't say anything about x, y, or xy.

thats where im stumped

K=2/5*E

2/5*E*1.2-600=E*.45

.48*E-600=.45*E
600=.03*E

E=20000

You can work out Ken's savings in the end by yourself :)

If we were to randomly arrange the cards such that the first 3 cards would be for the first person and so on, there would be 9! ways to do so.

Now, if we were to distribute the 3 red cards such that each person gets one, there are 6 ways that could happen. Then, we can distribute the remaining 6 cards using the same above method to get 6! ways.

That means the final probability is 6*6!/9!=1/84

:D

x(x+1)=2550

x^2+x-2550=0

x=-51

x+1=-50

sum=-101

8/x=x/5 (similar triangles)

x^2=40

x=sqrt40=2sqrt10

Problem 1:

a) 20, 22, 24

b) 32

c) 16

d) Never. All the numbers will be even

Formula for b and c: If first day is a and common difference if r then formula for day n is a+r*(n-1) (Make sure you see why)

Problem 2:

a) 6, 10, 15

b) These are all triangle numbers. (1+2+3+4+5...+n)

c) 1+2+3+4...10=55

d) 1+2+3+4...30=465 

e) n(n-1)/2

Maybe pay attention next time :)

Smallest: 17*-2

Largest: 17*14

14-2+1=13

Let's see here...

The odd digits must be going 1,3,5,7,9

The even: 8,6,4,2,0

Normally, there would be 10! ways to arrange it, but we must divide by 5!^2 because there is only one way to arrange even and odds. Therefore there are 252 numbers.

1800/2^3/3^2=25=5^2