X^2 plus 2y divideed by 2x plus 3z equal

(x^2 plus 2y")

    2x(2x + 3)

Step-by-step explanation:

Doing this problem correctly requires the application of "order of operations" rules.  "x^2 plus 2y" actually means "(x^2 plus 2y"), and so your

x^2 plus 2y divided by 2x

comes out to:

(x^2 plus 2y")

         2x

If we now divide this result by "2x plus 3z," we get:

  (x^2 plus 2y")

            2x

      2x + 3

or:

(x^2 plus 2y")

    2x(2x + 3)

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:

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