Find the equation of the line
passing through the points (5, 2)
and (10, 6).
y =
Xt

y=4/5x-2

Step-by-step explanation:

here are the given points: (5,2) and (10,6)

First, let's find the slope, which is found with the equation (y2-y1)/(x2-x1)

so let's label the points

x1=5

y1=2

x2=10

y2=6

now we can substitute into the equation

m=(6-2)/(10-5)

subtract

m=4/5

so that means the slope is 4/5

Now, let's use point slope form, which is y-y1=m(x-x1) (m is the slope)

so let's substitute the numbers we know from earlier into the equation

y-2=4/5(x-5)

do distributive prop.

y-2=4/5x-4

add 2 to both sides

y=4/5x-2

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}