Answer:
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
It will always be on the first or third quartile of the graph, or the point will always be on the “left side” of the graph. Brainliest answer?;)
3a+b+c
The perimeter is two times one side (to account for the opposite) and two times the adjacent side. So if the sides would be x and y, the perimeter would be 2*x + 2*y.
So, knowing that the sum is 16a+8b-6c, if we subtract the given side 5a+3b-4c from this, what remains is two times the "other" side:
16a+8b-6c - 2*(5a+3b-4c) =
16a+8b-6c -10a-6b+8c =
6a+2b+2c
half of that is
(6a+2b+2c)/2 = 3a+b+c
Answer:
B. 22
Step-by-step explanation:
Quotient of 18 and 2 is 9, and just take the 9 out of the next series and you will be left with 22.