Consider the line - 3x + 7y=4. Find the equation of the line that is parallel to this line and passes through the point (-3, -4).
Find the equation of the line that is perpendicular to this line and passes through the point (-3, -4).

Parallel:  y = 3/7x - 17/7

Perpendicular:  y = -7/3x - 11

Step-by-step explanation:

Rearrange the given equation so that it is in slope-intercept form.

-3x + 7y = 4

7y = 3x + 4

y = 3/7x + 4/7

The slope of the line is 3/7.  A parallel line will have the same slope.  Using this slope and the given point, you can find an equation using point-slope form.

y - y₁ = m(x - x₁)

y - (-4) = 3/7(x - (-3))

y + 4 = 3/7(x + 3)

y + 4 = 3/7x + 9/7

y = 3/7x - 17/7

A perpendicular line will have a slope that is the negative inverse of the original slope.  This means that the slope will be -7/3.  Repeat above steps.

y - y₁ = m(x - x₁)

y - (-4) = -7/3(x - (-3))

y + 4 = -7/3(x + 3)

y + 4 = -7/3x - 7

y = -7/3x - 11

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}

answerthe vertex form

step-by-step explanation: