Need help.

1.) Write the equation of the line that is parallel to y=3x+5 and passes through the point
(-1, 4).

2.) Write the equation of the line that is perpendicular to y=−2x+8 and passes through the point (-4, -3).

1. y= 3x+7

2.y=1/2 x -1

Step-by-step explanation:

1.

Parallel lines have the same gradient.In the equation given,

y= 3x+5 , the slope is 3

So now, you find the equation of a line passing through point (-1,4) with m=3

m=Δ y/Δx

3= y-4/ x--1

3= y-4 /x+1

3(x+1) = y-4

3x+3 =y-4

3x+3+4=y

⇒⇒ y= 3x+7   is the equation.

2.

Two lines that are perpendicular to each other have the product of their slopes to be -1

Given equation to be y= -2x +8 then, m₁ = -2

Finding m₂ will be;

m₁ *m₂ = -1

-2 * m₂ = -1

-2 m₂ = -1

m₂ = -1/-2 = 1/2

So writing the equation of the perpendicular line with m₂ =1/2 , point (-4,-3)

will be;

m=Δy/Δx

1/2=y--3/x--4

1/2 = y+3/x+4

x+4 = 2( y+3)

x+4 = 2(y+3)

x+4 =2y+6

x+4-6=2y

x-2=2y

x/2 -2/2=y

y=1/2 x -1   is the equation.

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 1: 2

step-by-step explanation: