Math

QuestionDetermine if the following relationships express yy as a function of xx:
1. x2+y2=25x^{2}+y^{2}=25
2. 2x+3y=102 x+3 y=10
3. x=y225x=y^{2}-25

Studdy Solution

STEP 1

Assumptions1. A function is a relation between a set of inputs and a set of permissible outputs with the property that each input is related to exactly one output. . We are examining whether yy is a function of xx in each of the given relationships.

STEP 2

The first relationship is x2+y2=25x^{2}+y^{2}=25. This is the equation of a circle centered at the origin with a radius of5. We can solve for yy to see if it is a function of xx.
y2=25x2y^{2} =25 - x^{2}

STEP 3

Taking the square root of both sides to solve for yy givesy=±25x2y = \pm \sqrt{25 - x^{2}}

STEP 4

Because we have a positive and a negative solution for yy, this means that for a given xx, we can have two possible values of yy. Therefore, yy is not a function of xx in this relationship.

STEP 5

The second relationship is 2x+3y=102x +3y =10. We can solve for yy to see if it is a function of xx.
3y=102x3y =10 -2x

STEP 6

Dividing both sides by3 to solve for yy givesy=102x3y = \frac{10 -2x}{3}

STEP 7

Because we have a single solution for yy for a given xx, this means that for a given xx, we have exactly one value of yy. Therefore, yy is a function of xx in this relationship.

STEP 8

The third relationship is x=y225x = y^{2} -25. We can solve for yy to see if it is a function of xx.
y2=x+25y^{2} = x +25

STEP 9

Taking the square root of both sides to solve for yy givesy=±x+25y = \pm \sqrt{x +25}

STEP 10

Because we have a positive and a negative solution for yy, this means that for a given xx, we can have two possible values of yy. Therefore, yy is not a function of xx in this relationship.
So, the relationship x2+y2=25x^{2}+y^{2}=25 does not represent yy as a function of xx, the relationship 2x+3y=102x +3y =10 does represent yy as a function of xx, and the relationship x=y225x = y^{2} -25 does not represent yy as a function of xx.

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