Álgebra/help There are 3 consecutive integers that add up to 15. What is the least of these integers?

The answers are 4,5,6,

the answer you are looking for is 4

Step-by-step explanation:

the first thing you have to do is define a variable for stuff that you don't know. In this problem, you don't know any of the integers, so you should assign a variable to them.

will put  the first integer as n. The second integer, since it is right after the first, will be n+1. The third integer, since it is right after the second, will be (n+1)+1=n+2. consider the integers 1, 2, and 3. 2 is one more than 1, or in other words, 2=1+1. same for 3, except 3 is two more than 1, so 3=1+2. Since the integers are consecutive, each is one more than the last.We're told that the sum of our three integers is 15. Therefore, n+(n+1)+(n+2)=15 Solving this equation is pretty straightforward:

3n+3=15

3n=12

n=4 That means our first integer is 4. Our second integer is 4+1, or 5, and our third integer is 5+1, or 6. Our answer is confirmed because 4+5+6=15

i hope I helped!!

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}