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Number Theory and Fractions Unit Test 1 of 151 of 15 Items Question Determine the number greater than 70 but less than 85 that...Number Theory and Fractions Unit Test
1 of 151 of 15 Items
Question
Determine the number greater than 70 but less than 85 that is divisible by both 2 and 9.(1 point)
Responses
78
78
81
81
72
72
84
Answers
Answered by
Missyyyyyy Catieeeeeee🫶(Connex-Us Q&A 6th)
Jesse lists four factor pairs of 72 as 2 × 36
, 4 × 18
, 6 × 12
, and 8 × 9
. Which factor pair has a prime number?(1 point)
Responses
6 × 12
6 times 12
4 × 18
4 times 18
8 × 9
8 times 9
2 × 36
, 4 × 18
, 6 × 12
, and 8 × 9
. Which factor pair has a prime number?(1 point)
Responses
6 × 12
6 times 12
4 × 18
4 times 18
8 × 9
8 times 9
2 × 36
Answered by
Missyyyyyy Catieeeeeee🫶(Connex-Us Q&A 6th)
The composite number 50 is between what two prime numbers?(1 point)
The composite number 50 lies between
and
, both prime numbers.
The composite number 50 lies between
and
, both prime numbers.
Answered by
Missyyyyyy Catieeeeeee🫶(Connex-Us Q&A 6th)
Express the prime factorization of 1,200 as a product of prime numbers with exponents.(2 points)
4
⋅
⋅
2
= 1,200
4
⋅
⋅
2
= 1,200
Answered by
GPT-4o mini
To find a number greater than 70 but less than 85 that is divisible by both 2 and 9, we first identify the multiples of 2 and 9 within that range.
- The multiples of 2 between 70 and 85 are: 72, 74, 76, 78, 80, 82, 84.
- The multiples of 9 between 70 and 85 are: 72, 81, 90 (but 90 is out of the range).
Next, we find the common multiples from both lists:
- From the multiples of 2, we have: 72, 74, 76, 78, 80, 82, 84.
- From the multiples of 9, we have: 72, 81.
The only number that appears in both lists and is between 70 and 85 is **72**.
Thus, the correct answer is **72**.
- The multiples of 2 between 70 and 85 are: 72, 74, 76, 78, 80, 82, 84.
- The multiples of 9 between 70 and 85 are: 72, 81, 90 (but 90 is out of the range).
Next, we find the common multiples from both lists:
- From the multiples of 2, we have: 72, 74, 76, 78, 80, 82, 84.
- From the multiples of 9, we have: 72, 81.
The only number that appears in both lists and is between 70 and 85 is **72**.
Thus, the correct answer is **72**.
Answered by
GPT-4o mini
To identify which factor pair of 72 contains a prime number, we can analyze each factor pair:
1. **2 × 36**: The factors are 2 and 36. The number 2 is a prime number.
2. **4 × 18**: The factors are 4 and 18. Neither of these numbers is prime (4 is 2 × 2 and 18 is 2 × 9).
3. **6 × 12**: The factors are 6 and 12. Neither of these is prime (6 is 2 × 3 and 12 is 2 × 6).
4. **8 × 9**: The factors are 8 and 9. Neither of these is prime (8 is 2 × 4 and 9 is 3 × 3).
From the analysis, the factor pair **2 × 36** contains the prime number **2**.
Therefore, the correct answer is **2 × 36**.
1. **2 × 36**: The factors are 2 and 36. The number 2 is a prime number.
2. **4 × 18**: The factors are 4 and 18. Neither of these numbers is prime (4 is 2 × 2 and 18 is 2 × 9).
3. **6 × 12**: The factors are 6 and 12. Neither of these is prime (6 is 2 × 3 and 12 is 2 × 6).
4. **8 × 9**: The factors are 8 and 9. Neither of these is prime (8 is 2 × 4 and 9 is 3 × 3).
From the analysis, the factor pair **2 × 36** contains the prime number **2**.
Therefore, the correct answer is **2 × 36**.
Answered by
GPT-4o mini
The composite number 50 lies between the prime numbers **47** and **53**.
So, the answer is:
The composite number 50 lies between **47** and **53**, both prime numbers.
So, the answer is:
The composite number 50 lies between **47** and **53**, both prime numbers.
Answered by
GPT-4o mini
To express the prime factorization of 1,200, we can start by dividing it by the smallest prime number, 2, and continue factoring until we are left with only prime numbers.
1. Divide by 2:
\[
1200 \div 2 = 600 \\
600 \div 2 = 300 \\
300 \div 2 = 150 \\
150 \div 2 = 75
\]
Now, we can no longer divide by 2 since 75 is not even.
2. Next, we check the next prime number, which is 3:
\[
75 \div 3 = 25
\]
3. Now, we factor 25, which is \(5 \times 5\):
\[
25 \div 5 = 5 \\
5 \div 5 = 1
\]
Now, putting this all together, we have:
\[
1200 = 2^4 \times 3^1 \times 5^2
\]
Thus, the prime factorization of 1,200 as a product of prime numbers with exponents is:
\[
2^4 \cdot 3^1 \cdot 5^2 = 1200
\]
1. Divide by 2:
\[
1200 \div 2 = 600 \\
600 \div 2 = 300 \\
300 \div 2 = 150 \\
150 \div 2 = 75
\]
Now, we can no longer divide by 2 since 75 is not even.
2. Next, we check the next prime number, which is 3:
\[
75 \div 3 = 25
\]
3. Now, we factor 25, which is \(5 \times 5\):
\[
25 \div 5 = 5 \\
5 \div 5 = 1
\]
Now, putting this all together, we have:
\[
1200 = 2^4 \times 3^1 \times 5^2
\]
Thus, the prime factorization of 1,200 as a product of prime numbers with exponents is:
\[
2^4 \cdot 3^1 \cdot 5^2 = 1200
\]
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