4. find the equations of the following lines:
a. having slope of -2 and passing through the point (-4, 2)
b. parallel to y = 6x + 3 and passing through the point (-1, 13)
c. perpendicular to 3x – 5y = 5 and having an x-intercept of 3
d. passing through (-2, -6) and (4, 2)
e. having an x-intercept of -3 and a y-intercept of 6
this is frustrating me to no ! !

Step-by-step explanation:

The general equation for a line is given by: y = mx + n (m: slope, n: y-intercept)

a)

given: m = -2, x = -4, y = 2

equation becomes: 2 = -2 * (-4) + n

solve for n: n = -6

solution: y = -2x - 6

b)

given: m = 6, x = -1, y = 13

equation becomes: 13 = 6 * (-1) + n

solve for n: n = 19

solution: y = 6x + 19

c)

given: perpendicular to: y = 3/5x - 1 => m = -5/3, x = 3, y = 0

equation becomes: 0 = (-5/3) * 3 + n

solve for n: n = 5

solution: y = -5/3x + 5

d) get the slope m from the two points.

given: m = Δy/Δx = 8/6 = 4/3, x = 4, y = 2

equation becomes: 2 = (4/3) * 4 + n

solve for n: n = -10/3

solution: y = (4/3)x - (10/3)

e)

given: point1 (-3,0), point2(0,6)

m = Δy/Δx = 6/3 = 2, x = 0, y = 6

equation becomes: 6 = 0 + n

solve for n: n = 6

solution: y = 2x + 6

good luck!

cutie~

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}