Question
Given the following information: a is 20 more than 3 times the value of b . b is an even integer. ...which is true, regarding operations on the two numbers?A a + b and a × b will always be an odd integer.
B a + b and a × b will always be an even integer.
C a + b will always be an odd integer, and a × b will always be an even integer.
D a + b will always be an even integer, and a × b will always be an odd integer.
E a + b and a × b will be rational numbers that are not in the set of integers.
B a + b and a × b will always be an even integer.
C a + b will always be an odd integer, and a × b will always be an even integer.
D a + b will always be an even integer, and a × b will always be an odd integer.
E a + b and a × b will be rational numbers that are not in the set of integers.
Answers
GPT-4o mini
To analyze the problem, we first express \( a \) in terms of \( b \):
Given:
\[ a = 3b + 20 \]
Since \( b \) is an even integer, we can denote:
\[ b = 2k \]
for some integer \( k \).
Substituting into the equation for \( a \):
\[
a = 3(2k) + 20 = 6k + 20
\]
Notice that \( a \) is expressed as:
\[
a = 6k + 20 = 2(3k + 10)
\]
This shows that \( a \) is also an even integer since it is divisible by 2.
Now we will analyze the operations \( a + b \) and \( a \times b \).
1. **Calculate \( a + b \)**:
\[
a + b = (6k + 20) + (2k) = 8k + 20
\]
Because \( 8k + 20 \) is the sum of two even numbers (both \( 8k \) and \( 20 \) are even), \( a + b \) is even.
2. **Calculate \( a \times b \)**:
\[
a \times b = (6k + 20) \times (2k) = 12k^2 + 40k
\]
Here, both terms \( 12k^2 \) and \( 40k \) are clearly even, so \( a \times b \) is also even.
Conclusively, we find that:
- \( a + b \) is always even.
- \( a \times b \) is always even.
Thus, the correct answer is:
**B**: \( a + b \) and \( a \times b \) will always be an even integer.
Given:
\[ a = 3b + 20 \]
Since \( b \) is an even integer, we can denote:
\[ b = 2k \]
for some integer \( k \).
Substituting into the equation for \( a \):
\[
a = 3(2k) + 20 = 6k + 20
\]
Notice that \( a \) is expressed as:
\[
a = 6k + 20 = 2(3k + 10)
\]
This shows that \( a \) is also an even integer since it is divisible by 2.
Now we will analyze the operations \( a + b \) and \( a \times b \).
1. **Calculate \( a + b \)**:
\[
a + b = (6k + 20) + (2k) = 8k + 20
\]
Because \( 8k + 20 \) is the sum of two even numbers (both \( 8k \) and \( 20 \) are even), \( a + b \) is even.
2. **Calculate \( a \times b \)**:
\[
a \times b = (6k + 20) \times (2k) = 12k^2 + 40k
\]
Here, both terms \( 12k^2 \) and \( 40k \) are clearly even, so \( a \times b \) is also even.
Conclusively, we find that:
- \( a + b \) is always even.
- \( a \times b \) is always even.
Thus, the correct answer is:
**B**: \( a + b \) and \( a \times b \) will always be an even integer.
Related Questions
Does anyone know where I can find information about the average industry ratios of financial informa...
In 1000 samples, assuming that the null hypothesis is true, how many times would you expect to commi...
Select the letter of the choice that identifies the source in which you
would most likely find...
How can the whole number 378 be expressed as a product of prime numbers with exponents?(1 point)Resp...