PLZZZ HELP FAST FIRST ONE GETS THE BRAINIEST Gina has borrowed 100 songs from her friend. She plans to download an equal number of songs on her music player each week for 5 weeks. The graph shows the number of songs left to download, y, for a certain number of weeks, x: A graph titled Song Downloading shows Number of Weeks on x-axis and Number of Songs Left to Download on y-axis. The x-axis scale is shown from 0 to 5 at increments of 1, and the y-axis scale is shown from 0 to 140 at increments of 20. A straight line joins the ordered pairs 0, 100 and 1, 80 and 2, 60 and 3, 40 and 4, 20 and 5, 0. Part A: What is the rate of change and initial value of the function represented by the graph, and what do they represent in this scenario? Show your work to find the rate of change and initial value. (6 points) Part B: Write an equation in slope-intercept form to model the relationship between x and y. (4 points)

Step-by-step explanation:

The rate of change is equal to (y2-y1)/(x2-x1) where (x1,y1)=((0,150) and (x2,y2)=(5,0). Therefore the rate of change equals (0-150)/((5-0)=-30 songs/week. This is equal to the slope of the line that goes through these two points. The initial value is the y coordinate of the point when x=0 which is 150. The equation of the line is y=-30x+150, and this is the slope-intercept form (y=my+b) where m is the slope of the line and b is the y-intercept.

phosphate, calcium ion, chloride ion, neon atom

step-by-step explanation:

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}