Find two posiitive even consecutive integers such that the square of the smaller integer is 10 more than the larger integer.

The two numbers are 4 and 6

Step-by-step explanation:

Suppose we have:

x=The smallest even integer

x+2=Next even integer

Recall even numbers occur every other integer.

The condition of the problem states the square of the smaller integer is 10 more than the longer integer:

Rearranging and simplifying:

Factoring:

We have two solutions:

x=4, x=-3

Since the integers must be positive:

x=4

Next even integer= x+2=6

The two numbers are 4 and 6

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 would pic false cause it goes down towards negative

step-by-step explanation: