Westonci.ca is your trusted source for finding answers to a wide range of questions, backed by a knowledgeable community. Join our platform to connect with experts ready to provide accurate answers to your questions in various fields. Discover detailed answers to your questions from a wide network of experts on our comprehensive Q&A platform.
Sagot :
To determine whether an equation can be used to show that Vladas' hypothesis is incorrect, we need to check if any of the given equations represents a function of [tex]\( x \)[/tex]. A function of [tex]\( x \)[/tex] means for every value of [tex]\( x \)[/tex], there should be exactly one value of [tex]\( y \)[/tex].
Let's go through each equation step by step:
1. Equation: [tex]\( x + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x \][/tex]
[tex]\[ y = \pm \sqrt{25 - x} \][/tex]
For a given [tex]\( x \)[/tex], [tex]\( y \)[/tex] can be either [tex]\( \sqrt{25 - x} \)[/tex] or [tex]\( -\sqrt{25 - x} \)[/tex]. This implies that for each [tex]\( x \)[/tex] there are two possible values of [tex]\( y \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
2. Equation: [tex]\( x^2 - y = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y = x^2 - 25 \][/tex]
For any given [tex]\( x \)[/tex], there is exactly one corresponding value of [tex]\( y \)[/tex], which is [tex]\( x^2 - 25 \)[/tex]. Therefore, this equation is indeed a function because it satisfies the condition of providing exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
3. Equation: [tex]\( x^2 + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x^2 \][/tex]
[tex]\[ y = \pm \sqrt{25 - x^2} \][/tex]
For any given [tex]\( x \)[/tex] within [tex]\( -5 \leq x \leq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{25 - x^2} \)[/tex] and [tex]\( -\sqrt{25 - x^2} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it can provide more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
4. Equation: [tex]\( x^2 - y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = x^2 - 25 \][/tex]
[tex]\[ y = \pm \sqrt{x^2 - 25} \][/tex]
For [tex]\( |x| \geq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{x^2 - 25} \)[/tex] and [tex]\( -\sqrt{x^2 - 25} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
From the above examination, we observe that only the equation [tex]\( x^2 - y = 25 \)[/tex] (i.e., the second equation) is a function of [tex]\( x \)[/tex]. This disproves Vladas' hypothesis that an equation with a squared term can never be a function of [tex]\( x \)[/tex].
Therefore, the equation [tex]\( x^2 - y = 25 \)[/tex] can be used to show Vladas that his hypothesis is incorrect.
Let's go through each equation step by step:
1. Equation: [tex]\( x + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x \][/tex]
[tex]\[ y = \pm \sqrt{25 - x} \][/tex]
For a given [tex]\( x \)[/tex], [tex]\( y \)[/tex] can be either [tex]\( \sqrt{25 - x} \)[/tex] or [tex]\( -\sqrt{25 - x} \)[/tex]. This implies that for each [tex]\( x \)[/tex] there are two possible values of [tex]\( y \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
2. Equation: [tex]\( x^2 - y = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y = x^2 - 25 \][/tex]
For any given [tex]\( x \)[/tex], there is exactly one corresponding value of [tex]\( y \)[/tex], which is [tex]\( x^2 - 25 \)[/tex]. Therefore, this equation is indeed a function because it satisfies the condition of providing exactly one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
3. Equation: [tex]\( x^2 + y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = 25 - x^2 \][/tex]
[tex]\[ y = \pm \sqrt{25 - x^2} \][/tex]
For any given [tex]\( x \)[/tex] within [tex]\( -5 \leq x \leq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{25 - x^2} \)[/tex] and [tex]\( -\sqrt{25 - x^2} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it can provide more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
4. Equation: [tex]\( x^2 - y^2 = 25 \)[/tex]
Solving for [tex]\( y \)[/tex] in terms of [tex]\( x \)[/tex],
[tex]\[ y^2 = x^2 - 25 \][/tex]
[tex]\[ y = \pm \sqrt{x^2 - 25} \][/tex]
For [tex]\( |x| \geq 5 \)[/tex], there are two possible values of [tex]\( y \)[/tex]: [tex]\( \sqrt{x^2 - 25} \)[/tex] and [tex]\( -\sqrt{x^2 - 25} \)[/tex]. Therefore, it does not meet the criteria of a function of [tex]\( x \)[/tex] because it provides more than one value of [tex]\( y \)[/tex] for each [tex]\( x \)[/tex].
From the above examination, we observe that only the equation [tex]\( x^2 - y = 25 \)[/tex] (i.e., the second equation) is a function of [tex]\( x \)[/tex]. This disproves Vladas' hypothesis that an equation with a squared term can never be a function of [tex]\( x \)[/tex].
Therefore, the equation [tex]\( x^2 - y = 25 \)[/tex] can be used to show Vladas that his hypothesis is incorrect.
Thank you for your visit. We're committed to providing you with the best information available. Return anytime for more. We hope you found what you were looking for. Feel free to revisit us for more answers and updated information. Westonci.ca is your go-to source for reliable answers. Return soon for more expert insights.
