In how many ways can the numbers 3, 4, 5, 6, 7, 8, 9, and 10 be divided into two sets A and B such that the sum of the numbers in B is three times the sum of the numbers in A?

4 ways

Step-by-step explanation:

3+4+5+6+7+8+9+10=52

1/4×52=13

13×3=39

3+10=13

4+9=13

5+8=13

6+7=13

so there are 4 ways to divide in two parts.

1.

A={3,10}

B={4,5,6,7,8,9}

2.

A={4,9}

B={3,5,6,7,8,10}

3.

A={5,8}

B={3,4,6,7,9,10}

4.

A={6,7}

B={3,4,5,8,9,10}

22.83 repeating

step-by-step explanation:

137 ÷ 6 is 22.8333333 therefore we understand that 3 is repeating.

happy to

pls mark as brainliest