Divine has ten less dollars than Teresa. Cherish has two times as much money than Divine. Altogether they have fifty dollars. How much money does each cadet have?

Teresa has $20,Divine has $10,Cherish has $20

Step-by-step explanation:

let the amount Teresa has=x

Amount Divine has=x-10

Amount Cherish has=2(x-10)

Since they have $50 altogether;

x+(x-10)+2(x-10)=50

x+x-10+2x-20=50

4x-30=50

4x=50+30

4x=80

x=20

Therefore,

Teresa has $20,Divine has $10,Cherish has $20

Teresa: $20

Divine: $10

Cherish: $20

Step-by-step explanation:

Teresa has x.

Divine has $10 less than Teresa, so Divine has x - 10.

Cherish has two times as much as Divine, so Cherish has 2(x - 10) = 2x - 20.

In summary:

Teresa: x

Divine: x - 10

Cherish: 2x - 20

The sum of their amounts is equal to $50:

x + x - 10 + 2x - 20 = 50

4x - 30 = 50

4x = 80

x = 20

Teresa: 20

Divine: 20 - 10 = 10

Cherish: 2(20) - 20 = 20

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}

$113.42

i hope that is right.

step-by-step explanation: