Let the smallest of 4 consecutive odd numbers be 2 n + 1 , where n is an integer. Show, using algebra, that the sum of any 4 consecutive odd numbers is always a multiple of 8. Give your answer as an expression that makes explicit that it is a multiple of 8.

Expression is 8(n+2)

Step-by-step explanation:

smallest of 4 consecutive odd numbers =2n + 1

consecutive odd integers are found by adding 2 to the any given odd numbers

Thus, 2nd consecutive odd integers  = 2n + 1 + 2 = 2n + 3

3rd consecutive odd integers  = 2n + 3 + 2 = 2n + 5

2nd consecutive odd integers  = 2n + 5 + 2 = 2n + 7

Thus, 4 consecutive odd integers are

2n + 1 ,2n + 3,2n + 5,2n + 7

sum of these numbers are = 2n + 1 +2n + 3 + 2n + 5+2n + 7 = 8n+16

sum of these numbers are = 8(n+2)

Thus, we see that the sum of numbers are 8(n+2)

As, 8 is common for n+2, whatever is value of n, the number will be multiple of 8 .

thus expression is 8(n+2)

< 4, -5>

step-by-step explanation:

starting from g. we move 4 right and then 5 down to get to h. that means the vector is < 4, -5>

about 168 miles

step-by-step explanation: