What is the equation of the line that passes through the points (-3,0.5) and (3,-0.5)

y = -1/6x + 0

Step-by-step explanation:

Using the equation of line

y_2 - y_1 / x_2 - x_1 = y - y_1 / x_1

Using the information we are provided with

( -3, 0.5)(3 , -0.5)

y_1 = 0.5

x_1 = -3

x_2 = 3

y_2 = -0.5

-0.5 - 0.5 / 3- (-3) = y - 0.5 / x - (-3)

-0.5 - 0.5 / 3 +3 = y - 0.5 / x + 3

-1 / 6 = y - 0.5 / x + 3

Cross multiply

-1(x + 3) = 6(y - 0.5)

Open the brackets

-x - 3 = 6y - 3

-x - 3 + 3 = 6y

-x + 0 = 6y

Divide through by 6 to get y

Following the eqn y = mx + C

-x + 0 / 6 = 6y / 6

y = -x + 0 / 6

We can separate it

y = -x / 6 + 0/6

y = -1/6x + 0

Therefore, the equation of the line is

y = -1/6x + 0

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}

step-by-step explanation: