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f(x) = 2x – 3 f(m) = –11
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30 Jul 2018, 10:25
30 Jul 2018, 10:25
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Question Stats:
70% (00:47) correct
29% (00:43) wrong based on 98 sessions
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\(f(x) = 2x - 3\)
\(f(m) = -11\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(f(m) = -11\)
Quantity A |
Quantity B |
m |
\(\frac{1}{2} f(m)\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
Re: f(x) = 2x – 3 f(m) = –11
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11 Aug 2018, 16:40
11 Aug 2018, 16:40
1
Expert Reply
Explanation
The problem gives a function, \(f(x) = 2x – 3\), and then indicates that, when m is plugged in to the function, the answer is –11. Therefore:
\(2m - 3 = -11\)
\(2m = -8\)
\(m = -4\)
Quantity A is equal to –4. The problem indicates that f(m) = –11, so Quantity B is equal to = –5.5. Quantity A is greater.
The problem gives a function, \(f(x) = 2x – 3\), and then indicates that, when m is plugged in to the function, the answer is –11. Therefore:
\(2m - 3 = -11\)
\(2m = -8\)
\(m = -4\)
Quantity A is equal to –4. The problem indicates that f(m) = –11, so Quantity B is equal to = –5.5. Quantity A is greater.
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f(x) = 2x – 3 f(m) = –11
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10 Sep 2021, 08:31
10 Sep 2021, 08:31
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sandy wrote:
\(f(x) = 2x - 3\)
\(f(m) = -11\)
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
\(f(m) = -11\)
Quantity A |
Quantity B |
m |
\(\frac{1}{2} f(m)\) |
A) Quantity A is greater.
B) Quantity B is greater.
C) The two quantities are equal.
D) The relationship cannot be determined from the information given.
If \(f(m) = -11\), then we are saying that: \(f(m) = 2m - 3 = -11\)
So, now we can solve the following equation \(m\): \(2m - 3 = -11\)
Add \(3\) to both sides: \(2m = -8\)
Divide both sides by \(2\): \(m = -4\)
So, we have:
QUANTITY A: \(-4\)
QUANTITY B: \(\frac{1}{2} f(m) = (\frac{1}{2})(-11) = -5.5\)
Answer: A
