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.

geometric series converge the common ratio r has magnitude less than one.

which is equivalent to

for the example we need

|2-x| < 1

-1 < 2-x   and 2-x < 1

x < 3 and x > 1

we can write that

answer: 1 < x < 3

1a. common ratio r=250/25=10 > 1     diverges

1b.   r=50/100=1/2 converges

when these converge they converge to

where a is the first term.

answer: 200

1c. r=-1   so it's not the case that |r| < 1.     diverges

2a. x is the common ratio so we converge when |x|< 1

2b.   r=-10x / -2 = 5x.

we converge when

|5x| < 1

|x| < 1/5

2c.   r=3-2x

-1 < 3-2x and 3-2x < 1

-4 < -2x and 2 < 2x

x < 2 and 1 < x

1 < x < 2

2d. [/tex] r=-48x^3/12 = 4x^3[/tex]

|4x^3| < 1

|x^3| < \frac 1 4

|x| < 4^{-1/3}