In 1-4, write an equation in slope-intercept form to represent each table (just the first 4)

Q1 y = 17/2x - 2

Q2 y = -3x + 14

Q3 y = 1/4x

Q4 y = 1/8x + 10

Step-by-step explanation:

See, the problem here is, you did everything to NOT convert the decimals into fractions. You would´ve had the right answers if you just learned to deal with the fractions. I´ll only going to be explaining 2 questions since there´s four of them.

Question Two

You take two points from the table, then you do y2 - y1 over x2 - x1.  -1 - 5 / 6 - 3 = -6/3 and 3 divided by -6 is -3. Which should give you "m" the slope. Which the slope for this question is -3. To find "b" our y-intercept, we will need to plug in the points from one of the two points we took in the chart into the format of y = mx + b. I chose to do, 5 = -3(3) + b. Multiply m times the x, 5 = - 9 + b. So now, we add 9 to itself to cancel itself out and add it to 5. Which gives us, 14 = b. So our equation is, y = -3x + 14

Question 3

This one is relatively easy, since looking at the table we know theres no b or m. Just a constant rate times x. The same thing we did the last question, take two points from the equation. y2 - y1 over x2 - x1. 1 - 0 over 4 - 0. Which gives us 1/4. 1/4 is our constant rate of change. y = 1/4x

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}