Westonci.ca is your trusted source for accurate answers to all your questions. Join our community and start learning today! Our platform provides a seamless experience for finding precise answers from a network of experienced professionals. Connect with a community of professionals ready to provide precise solutions to your questions quickly and accurately.
Sagot :
Let's analyze each set of ordered pairs to determine if any of them could be generated by an exponential function. An exponential function typically takes the form [tex]\( f(x) = a \cdot b^x \)[/tex], where [tex]\(a\)[/tex] and [tex]\(b\)[/tex] are constants, and [tex]\(b\)[/tex] is the base of the exponential.
1. [tex]\((1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
For this set, observe the pairs:
- [tex]\((1, 1)\)[/tex]
- [tex]\((2, \frac{1}{2})\)[/tex]
- [tex]\((3, \frac{1}{3})\)[/tex]
- [tex]\((4, \frac{1}{4})\)[/tex]
The values [tex]\(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)[/tex] do not follow a multiplicative pattern that is typical for exponential functions. For instance, [tex]\(\frac{1}{2} \cdot b \neq \frac{1}{3}\)[/tex]. Thus, this set cannot be generated by an exponential function.
2. [tex]\((1,1),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{9}\right),\left(4, \frac{1}{16}\right)\)[/tex]
For this set, let's check if each term can be expressed as [tex]\( a \cdot b^x \)[/tex]:
- [tex]\((1, 1) = 1 \cdot 1^1\)[/tex]
- [tex]\((2, \frac{1}{4}) = 1 \cdot \left(\frac{1}{2}\right)^2\)[/tex]
- [tex]\((3, \frac{1}{9}) = 1 \cdot \left(\frac{1}{3}\right)^2 \, or \, 1 \cdot \left(\frac{1}{2}\right)^3\)[/tex]
- [tex]\((4, \frac{1}{16}) = 1 \cdot \left(\frac{1}{2}\right)^4\)[/tex]
The base doesn't remain consistent; it varies from [tex]\(\left(\frac{1}{3}\right)\)[/tex] to [tex]\(\left(\frac{1}{2}\right)\)[/tex]. So, this set cannot be generated by an exponential function.
3. [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex]
Let's check this set:
- [tex]\((1, \frac{1}{2}) = \frac{1}{2}^1\)[/tex]
- [tex]\((2, \frac{1}{4}) = \frac{1}{2}^2\)[/tex]
- [tex]\((3, \frac{1}{8}) = \frac{1}{2}^3\)[/tex]
- [tex]\((4, \frac{1}{16}) = \frac{1}{2}^4\)[/tex]
Here, the form [tex]\(a \cdot b^x\)[/tex] holds where [tex]\(a = 1\)[/tex] and [tex]\(b = \frac{1}{2}\)[/tex]. Thus, this set [tex]\((1, \frac{1}{2}), (2, \frac{1}{4}), (3, \frac{1}{8}), (4, \frac{1}{16})\)[/tex] can be generated by an exponential function [tex]\(f(x) = \left(\frac{1}{2}\right)^x\)[/tex].
4. [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{6}\right),\left(4, \frac{1}{8}\right)\)[/tex]
Finally, consider this set:
- [tex]\((1, \frac{1}{2})\)[/tex]
- [tex]\((2, \frac{1}{4})\)[/tex]
- [tex]\((3, \frac{1}{6})\)[/tex]
- [tex]\((4, \frac{1}{8})\)[/tex]
These values also do not follow a pattern consistent with an exponential function, as the ratios and pattern of growth do not match.
Therefore, among the given sets of ordered pairs, the set that could be generated by an exponential function is the third one:
[tex]\[\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\][/tex]
1. [tex]\((1,1),\left(2, \frac{1}{2}\right),\left(3, \frac{1}{3}\right),\left(4, \frac{1}{4}\right)\)[/tex]
For this set, observe the pairs:
- [tex]\((1, 1)\)[/tex]
- [tex]\((2, \frac{1}{2})\)[/tex]
- [tex]\((3, \frac{1}{3})\)[/tex]
- [tex]\((4, \frac{1}{4})\)[/tex]
The values [tex]\(1, \frac{1}{2}, \frac{1}{3}, \frac{1}{4}\)[/tex] do not follow a multiplicative pattern that is typical for exponential functions. For instance, [tex]\(\frac{1}{2} \cdot b \neq \frac{1}{3}\)[/tex]. Thus, this set cannot be generated by an exponential function.
2. [tex]\((1,1),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{9}\right),\left(4, \frac{1}{16}\right)\)[/tex]
For this set, let's check if each term can be expressed as [tex]\( a \cdot b^x \)[/tex]:
- [tex]\((1, 1) = 1 \cdot 1^1\)[/tex]
- [tex]\((2, \frac{1}{4}) = 1 \cdot \left(\frac{1}{2}\right)^2\)[/tex]
- [tex]\((3, \frac{1}{9}) = 1 \cdot \left(\frac{1}{3}\right)^2 \, or \, 1 \cdot \left(\frac{1}{2}\right)^3\)[/tex]
- [tex]\((4, \frac{1}{16}) = 1 \cdot \left(\frac{1}{2}\right)^4\)[/tex]
The base doesn't remain consistent; it varies from [tex]\(\left(\frac{1}{3}\right)\)[/tex] to [tex]\(\left(\frac{1}{2}\right)\)[/tex]. So, this set cannot be generated by an exponential function.
3. [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\)[/tex]
Let's check this set:
- [tex]\((1, \frac{1}{2}) = \frac{1}{2}^1\)[/tex]
- [tex]\((2, \frac{1}{4}) = \frac{1}{2}^2\)[/tex]
- [tex]\((3, \frac{1}{8}) = \frac{1}{2}^3\)[/tex]
- [tex]\((4, \frac{1}{16}) = \frac{1}{2}^4\)[/tex]
Here, the form [tex]\(a \cdot b^x\)[/tex] holds where [tex]\(a = 1\)[/tex] and [tex]\(b = \frac{1}{2}\)[/tex]. Thus, this set [tex]\((1, \frac{1}{2}), (2, \frac{1}{4}), (3, \frac{1}{8}), (4, \frac{1}{16})\)[/tex] can be generated by an exponential function [tex]\(f(x) = \left(\frac{1}{2}\right)^x\)[/tex].
4. [tex]\(\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{6}\right),\left(4, \frac{1}{8}\right)\)[/tex]
Finally, consider this set:
- [tex]\((1, \frac{1}{2})\)[/tex]
- [tex]\((2, \frac{1}{4})\)[/tex]
- [tex]\((3, \frac{1}{6})\)[/tex]
- [tex]\((4, \frac{1}{8})\)[/tex]
These values also do not follow a pattern consistent with an exponential function, as the ratios and pattern of growth do not match.
Therefore, among the given sets of ordered pairs, the set that could be generated by an exponential function is the third one:
[tex]\[\left(1, \frac{1}{2}\right),\left(2, \frac{1}{4}\right),\left(3, \frac{1}{8}\right),\left(4, \frac{1}{16}\right)\][/tex]
Thank you for trusting us with your questions. We're here to help you find accurate answers quickly and efficiently. Your visit means a lot to us. Don't hesitate to return for more reliable answers to any questions you may have. Discover more at Westonci.ca. Return for the latest expert answers and updates on various topics.
