Ineed asap the sum of four consecutive odd integers is 400. write an equation that models this situation and find the value of the four odd integers.

97,99,101,103

Step-by-step explanation:

Let x = first odd integer

x+2 = 2nd odd integer

x+4 = 3rd odd integer

x+6 = 4th odd integer

Sum of 4 odd integers is 400

x+ (x+2) + (x+4)+(x+6) = 400

Combine like terms

4x +12 = 400

Subtract 12 from each side

4x+12-12 = 400-12

4x = 388

Divide by 4 on each side

4x/4 = 388/4

x=97

The first integer is 97

The 2nd is 97+2 =99

The third ix 97+4 = 101

The 4th is 97+6 = 103

Step-by-step explanation:

Here is another method. It might be shorter. You can decide. Make sure that the numbers are odd by doing this

2k + 1  is one of the odd numbers. 2k is even. When you add 1 to 2k + 1 is odd.

2k - 1 is another member of this family

2k + 3 is a third member

2k - 3 is the fourth member.

When you add them, all the coefficients will cancel out.

2k + 1 + 2k - 1 + 2k - 3 + 2k + 3 = 400   Add

8k = 400                                                 Divide by 8

k = 400/8

k = 50

2k + 1 = 2*50 + 1 = 100 + 1 = 101

2k - 1 =  2*50 - 1 = 100 - 1 = 99

2k + 3 = 2*50 + 3 = 103

2k - 3 = 2*50 - 3 = 97

You don't really need to use 2k. You can do the same thing with n - 1 and n + 1 and n - 3 and n + 3. The same thing will happen: all the coefficients will cancel.

i found two solutions,   n = 2 • ± √3 = ± 3.4641

step-by-step explanation: