You pick a card at random. Without putting the first card back, you pick a second card at random. 4 5 6 What is the probability of picking a 6 and then picking a 5?

b

1st integer x

2nd integer x+1

3rd integer x+2

4th integer x+3

x+x+1 +x+2+x+3 =74

combine like terms

4x+6 = 74

subtract 6 from each side

4x = 68

x = 17

the integers are 17,18,19,20

x=-2

step-by-step explanation:

i began by flipping the problem to give me x= on the left (keeps it clearer for me).   so my problem simply became 10x-3=4x-6.  

1)   subtract 4x from the right side, which equals 0.   you do the same to the left of the equal as the right; therefore, subtract 4x from 10x, leaving 6x on the left.

              10x-3=4x-6   :   4x-4x on the right leaves only the -6        

2)             6x-3= 0-6 :    

                6x+3=   -6

3) divide

n+3

step-by-step explanation:

we are increasing our answer by 3 every time. we are moving from negatives to positives, which can make it look confusing, but we are indeed increasing by 3. our next to numbers would be 4, 7, and 10.