There are three numbers. The first number is twice the second number. The third number is twice the first number. The sum of the three numbers is 112.

x = 32 y = 16 z = 64

Step-by-step explanation:

x is the 1st number

y is the 2nd number

z is the 3rd number

The first number is twice the second number so

x = 2 y

The third number is twice the first number.

z = 2 x

Their sum is 112

x + y + z = 112

Plus in what we know:

x + y + z = 112

2 y + y + 2x = 112

3y + 2x = 112 Let's solve for y and subtract 2x from each side

3y = 112 - 2x  Divide both sides by 3

y =

Now plug our answer back in to solve for x.

x = 2y

x = 2 ( )

x = (224 - 4x) / 3    Multiply each side by 3.

3x = 224 - 4x         Add 4x to each side

3x + 4x = 224

7x = 224              Divide each side by 7

7x / 7 = 224 / 7

x = 32

Now we can solve for z.

z = 2x

z = 2 ( 32 )

z = 64

Now we can solve for the numerical value of y.

x + y + z = 112

32 + y + 64 = 112

96 + y = 112       Subtract 96 from each side.

y = 112 - 96

y = 16

1st number = 36

Second = 16

Third = 64

Step-by-step explanation:

We have 3 numbers

Let's call the second number g

Then the first would be , 2x g = 2g

Then the third would be twice the first = 2 x 2g = 4g

We add up all numbers and equate it to 112

2g + g +4g = 112

7g = 112

g = 112/7

g = 16

First number = 2g = 2x16 = 32

Second number = g = 16

Third number = 4g = 4x16 = 64

Therefore the numbers are 32, 16, 64.

I 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}

26?

step-by-step explanation: