At Westonci.ca, we provide clear, reliable answers to all your questions. Join our vibrant community and get the solutions you need. Join our Q&A platform and connect with professionals ready to provide precise answers to your questions in various areas. Explore comprehensive solutions to your questions from knowledgeable professionals across various fields on our platform.

Vladas believes that an equation with a squared term is never a function of [tex]\( x \)[/tex]. Which equation can be used to show Vladas that his hypothesis is incorrect?

A. [tex]\( x + y^2 = 25 \)[/tex]
B. [tex]\( x^2 - y = 25 \)[/tex]
C. [tex]\( x^2 + y^2 = 25 \)[/tex]
D. [tex]\( x^2 - y^2 = 25 \)[/tex]


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.