Identify an equation in point-slope form for the line parallel to y = 1*-7 that
passes through (-3,-2).

y + 2 = 1 (x+3) > y + 2 = x + 3

Explanation:

The general form of the equation of a line in point-slope form is:

y - y₁ = m (x - x₁)

where:

m is the slope of the line

(x₁ , y₁) is a point that belongs to the line

The general form of the equation of a line in slope-intercept form is:

y = mx + c

where:

m is the slope

c is the y-intercept

1- Getting the point:

We are already given the point (-3 , -2) so we'll directly use it

2- Getting the slope

We are given that our line is parallel to the line:

y = 1x - 7

We know that parallel lines have equal slopes

This means that:

slope of the line we want = 1

3- Getting the equation:

Substitute with the point and slope in the above form to get the equation

This is done as follows:

y - y₁ = m (x - x₁)

y - (-2) = 1 (x - (-3))

y + 2 = 1 (x + 3)

y + 2 = x + 3

Hope this helps :)

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}

i think the answer is seven, for if i'm being useless °ω°

step-by-step explanation: