Get expert advice and community support on IDNLearn.com. Get prompt and accurate answers to your questions from our community of experts who are always ready to help.
Sagot :
Let's analyze the given information and mathematical conditions step-by-step:
1. Given conditions:
- [tex]\(ac = 1\)[/tex]
- [tex]\(\frac{b+c}{d}\)[/tex] is undefined
- [tex]\(abc = d\)[/tex]
2. Condition for [tex]\(\frac{b+c}{d}\)[/tex] to be undefined:
[tex]\(\frac{b+c}{d}\)[/tex] is undefined when the denominator [tex]\(d = 0\)[/tex].
3. First, from [tex]\(ac = 1\)[/tex]:
Since [tex]\(a\)[/tex] and [tex]\(c\)[/tex] are real numbers, neither [tex]\(a\)[/tex] nor [tex]\(c\)[/tex] can be zero because their product is 1. Therefore, we can eliminate option (A).
4. Given [tex]\(d = abc\)[/tex]:
Since we established that [tex]\(d = 0\)[/tex] to make [tex]\(\frac{b+c}{d}\)[/tex] undefined, substitute [tex]\(d\)[/tex] with [tex]\(abc\)[/tex]:
[tex]\[ abc = 0 \][/tex]
Since neither [tex]\(a\)[/tex] nor [tex]\(c\)[/tex] can be zero (from [tex]\(ac = 1\)[/tex]), it follows that [tex]\(b\)[/tex] must be 0.
This confirms that option (D) must be true:
[tex]\[ b = 0 \][/tex]
5. Checking other options:
- Option (B) [tex]\(a = 1\)[/tex] and [tex]\(c = 1\)[/tex]: If this were true, [tex]\(ac = 1 \cdot 1 = 1\)[/tex], which holds true, but this specific combination is not necessary for the given conditions to hold.
- Option (C) [tex]\(a = -c\)[/tex]: Substituting in [tex]\(ac = 1\)[/tex], we get [tex]\(a \cdot (-a) = -a^2 = 1\)[/tex], which implies [tex]\(a^2 = -1\)[/tex]. This is not possible for real number [tex]\(a\)[/tex].
- Option (E) [tex]\(b + c = 0\)[/tex]: Since [tex]\(b = 0\)[/tex], we get [tex]\(0 + c = 0\)[/tex], which implies [tex]\(c = 0\)[/tex]. This contradicts our given fact that [tex]\(c\)[/tex] cannot be zero. Therefore, this is not a valid option.
Given the conditions for the expressions to be defined and what we derived from the given equations, the condition that must be true is:
D. [tex]\(b = 0\)[/tex]
1. Given conditions:
- [tex]\(ac = 1\)[/tex]
- [tex]\(\frac{b+c}{d}\)[/tex] is undefined
- [tex]\(abc = d\)[/tex]
2. Condition for [tex]\(\frac{b+c}{d}\)[/tex] to be undefined:
[tex]\(\frac{b+c}{d}\)[/tex] is undefined when the denominator [tex]\(d = 0\)[/tex].
3. First, from [tex]\(ac = 1\)[/tex]:
Since [tex]\(a\)[/tex] and [tex]\(c\)[/tex] are real numbers, neither [tex]\(a\)[/tex] nor [tex]\(c\)[/tex] can be zero because their product is 1. Therefore, we can eliminate option (A).
4. Given [tex]\(d = abc\)[/tex]:
Since we established that [tex]\(d = 0\)[/tex] to make [tex]\(\frac{b+c}{d}\)[/tex] undefined, substitute [tex]\(d\)[/tex] with [tex]\(abc\)[/tex]:
[tex]\[ abc = 0 \][/tex]
Since neither [tex]\(a\)[/tex] nor [tex]\(c\)[/tex] can be zero (from [tex]\(ac = 1\)[/tex]), it follows that [tex]\(b\)[/tex] must be 0.
This confirms that option (D) must be true:
[tex]\[ b = 0 \][/tex]
5. Checking other options:
- Option (B) [tex]\(a = 1\)[/tex] and [tex]\(c = 1\)[/tex]: If this were true, [tex]\(ac = 1 \cdot 1 = 1\)[/tex], which holds true, but this specific combination is not necessary for the given conditions to hold.
- Option (C) [tex]\(a = -c\)[/tex]: Substituting in [tex]\(ac = 1\)[/tex], we get [tex]\(a \cdot (-a) = -a^2 = 1\)[/tex], which implies [tex]\(a^2 = -1\)[/tex]. This is not possible for real number [tex]\(a\)[/tex].
- Option (E) [tex]\(b + c = 0\)[/tex]: Since [tex]\(b = 0\)[/tex], we get [tex]\(0 + c = 0\)[/tex], which implies [tex]\(c = 0\)[/tex]. This contradicts our given fact that [tex]\(c\)[/tex] cannot be zero. Therefore, this is not a valid option.
Given the conditions for the expressions to be defined and what we derived from the given equations, the condition that must be true is:
D. [tex]\(b = 0\)[/tex]
We value your participation in this forum. Keep exploring, asking questions, and sharing your insights with the community. Together, we can find the best solutions. Thank you for choosing IDNLearn.com. We’re committed to providing accurate answers, so visit us again soon.