The linear function f(x)=0.2x+79 represents the average test score in your math class, where x is the number of the test taken. The linear function g(x) represents the average test score in your science class, where x is the number of the test taken. X and g(x) 1,86 2,84 3,82. Part a determine the test average for your math class after completing test 2. Part b determine the test average for your science class after completing test 2. Part c: which class had a higher average after completing test 4? Show your work to support your answer.

The Science class had a higher average test score after completing test 4.

Step-by-step explanation:

Average test score in your math class  is denoted by function: f(x)=0.2x+79

x is the number of the test taken.

The linear function g(x) represents the average test score in your science class, where x is the number of the test taken.

X    g(x)

1     86

2     84

3     82

Now to find function we will use two point slope form

Part A: Substitute x = 2 in f(x)

f(2) = 0.2(2) + 79 = 79.4

The test average for your math class after completing test 2 = 79.4

Part B

y=-2x+88

y=-2(2)+88=84

The test average for your science class after completing test 2 = 84

Part C:

Substitute x=4 in f(x) to get the test average for Math class  

f(4) = 0.2(4) +79 = 79.8

y=-2x+88

To get the test average for science class

y=-2(4)+88

y = 80

After completing test 4, the test average for Math class is 79.8, and the test average for Science class is 80.

80>79.8

Hence the answer is  The Science class had a higher average test score after completing test 4.

step-by-step explanation:

we can use this expression to calculate the energy gain as a particle moves with a field or the energy loss as it moves against it. (note that it is also used to define the electron-volt, a unit of energy equal to that gained by an electron as it falls through a pd of 1 v, i.e. w = 1 ev = 1.6 10-19 c 1 v = 1.6 10-19j).in fact this expression is really nothing more than the statement that potential is the potential energy per unit charge. in this case, v= w/qis simply stating that the change in potential energy(or work done) per unit charge is equal to the change in potential (or potential difference), v.student questions: uniform electric fields2electron beamcathode+5 kvanode0 v