d. both a relation and a function:
Given:
Mark records his science scores in each monthly assessment over a period of 5 months. In the first assessment he scores 76%. In the second assessment he scores 73%. After that, his scores keep increasing by 2% in every assessment.
x represents the number of assessments since he starts recording and y represents the scores in each assessment.
In order for a relation to be a function the association has to be unambiguous that means that for a given input only one output can exist.If an input can have two or more outputs then you cannot determine which is the correct output for that input.
In the given situation:
x is the input that is number of assessments since mark starts recording the scores so there is only one assessment no repeating.so there is only one output.
Hence the relation is a function.
Learn more about the function here:
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Indefinitely many
The lines are on each other so every solution is correct
Answer:
<h2>B. 243</h2><h2 />
Step-by-step explanation:
F(x) = 3^x
then
F(5) = 3^5 = 3×3×3×3×3 =243
The standard form of the equation of a circle is (x-h)^2 + (y-k)^2 = r^2, where (h,k) is the center of the circle, (x,y) is a point of the circle, and r is the length of the radius of the circle. When the equation of a circle is written, h,k, and r are numbers, while x and y are still variables. (x-2)^2 + (y-k)^2 = 16 is an example of a circle. The problem gives us two of the three things that a circle has, a point (5,9) and the center (-2,3). We need to find the radius in order to write the equation. We substitute -2 for h, 3 for k, 5 for x, and 9 for y to get (5 - (-2))^2 + (9 - 3)^2 = r^2 We simplify: 49 + 36 = r^2, r^2 = 85. We only need to know r^2 because the equation of a circle has r^2. We now have all the information to write the equation of a circle. (x + 2)^2 + (y - 3)^2 = 85.
