It might be B I might not be right
There are 2 choices for the first set, and 5 choices for the second set. Each of the 2 choices from the first set can be combined with each of the 5 choices from the second set. Therefore there are 2 times 5 combinations from the first and second sets. Continuing this reasoning, the total number of unique combinations of one object from each set is:
If we divide 56 by 4 we get 14. Why by4? So that one number is 3 times the other So he had 14 balls that went into the green bags and 42 (56-14) that went into the red bags. We could just answer the question and say 14 but I think they want to know how many in each green bag.
14 and 42 don't work because they are not the same number of balls. What number is a common factor? 7 is,
We could have 2 green bags and split the 14 balls into 2 groups of 7 and with the remaining 42 - put them into 6 red bags of 7 each.
And so the answer to your question is:
7 ball in each bag = 2 bags are green, and 6 bags are red
14 balls + 42 balls = 56 bouncy balls
Hj and jk are the same length line segments ( because the midpoint divides a line into two equal parts)
So hj = jk.
hk is the line segment which has the mid point j. It is the double of hj or jk. It can be the sum of hj and jk.
hj + jk = hk
or
2 * hj = hk
or
2 * jk = hk
Answer:
red: x
blue: 2x+3
Step-by-step explanation:
let the number of red chips= x
let the number of blue chips= 2x+3
