To use the partial quotients method for each of the division problems, follow the steps below. The method involves subtracting multiples of the divisor from the dividend until the remaining number is less than the divisor. Let's solve the problems:
1.) \( 210 \div 15 \)
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Step 1: Estimate a large multiple of 15 to subtract. Let's start with 10 (because \(15 \times 10 = 150\)).
\[ 210 - 150 = 60 \]
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Step 2: Again, subtract 15 from 60. We can take out 4 more (since \(15 \times 4 = 60\)).
\[ 60 - 60 = 0 \]
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Step 3: Add the partial quotients (10 + 4 = 14).
Answer: \( 210 \div 15 = 14 \)
2.) \( 286 \div 13 \)
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Step 1: Start with a large multiple of 13. Let's try 20 (since \(13 \times 20 = 260\)).
\[ 286 - 260 = 26 \]
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Step 2: Now, find how many times 13 can fit into 26. (Since \(13 \times 2 = 26\)).
\[ 26 - 26 = 0 \]
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Step 3: Add the partial quotients (20 + 2 = 22).
Answer: \( 286 \div 13 = 22 \)
3.) \( 575 \div 25 \)
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Step 1: Start with a large multiple of 25. Let's begin with 20 (since \(25 \times 20 = 500\)).
\[ 575 - 500 = 75 \]
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Step 2: Next, see how many times 25 fits into 75. (Since \(25 \times 3 = 75\)).
\[ 75 - 75 = 0 \]
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Step 3: Add the partial quotients (20 + 3 = 23).
Answer: \( 575 \div 25 = 23 \)
Final Answers:
- \( 210 \div 15 = 14 \)
- \( 286 \div 13 = 22 \)
- \( 575 \div 25 = 23 \)
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