Plz help me.
how to solve | k-10 | = 3
will give brainliest!!!

Fast and loose:

If you stare at it for a while, you can see that the LHS is the distance between k and 10. Basically, you've asked to pick k so you get two numbers that are 3 apart from 10. And these two numbers are 7 to the left, and 13 to the right.

Rigorously

Split it in 2 situations, depending if the quantity inside the absolute value is greater or smaller than zero.

If it's greater or equal than zero:

And since 13 is indeed greater than 10, it's a valid solution, so you get the first one. For the second, you assume negative values for the quantity inside, and change sign:

In this case, 7 is a valid solution too since it's less than 10

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}

step-by-step explanation:

what are the statements? and what do you mean by "216 2 1 6"?