Which division problems have quotients of 682? check all that apply.

(600 + 80 + 2) 10
(6,000 + 800 + 20) 10
(60,000 + 8,000 + 200) 100
682,000 1,000
6,820,000 1,000

The Options which has quotient equivalent to 682 are given below,

(6,000 + 800 + 20) 10

(60,000 + 8,000 + 200) 100

(682,000) 1,000

Step-by-step explanation:

Given :  Quotient is 682

Given Options are for which we need to find Quotients:

(600 + 80 + 2) 10

(6,000 + 800 + 20) 10

(60,000 + 8,000 + 200) 100

682,000 1,000

6,820,000 1,000

We will check for for each:

1) (600 + 80 + 2) 10

Solving Above we get;

Hence for this Option the quotient is not equal to 682.

2) (6,000 + 800 + 20) 10

Solving Above we get;

Hence for this Option the quotient is equal to 682.

3) (60,000 + 8,000 + 200) 100

Solving Above we get;

Hence for this Option the quotient is equal to 682.

4) (682,000) 1,000

Solving Above we get;

Hence for this Option the quotient is equal to 682.

5) (6820,000) 1,000

Solving Above we get;

Hence for this Option the quotient is not equal to 682.

Hence the Options which has quotient equivalent to 682 are given below,

(6,000 + 800 + 20) 10

(60,000 + 8,000 + 200) 100

(682,000) 1,000

answer: c, the green line

step-by-step explanation:

there is no black line, eliminate b.

simple interest implies periodic payments, but the red line shows no payments, eliminate d.

simple interest has equal payments. if paid say every day, the graph is about a straight line (actually a stairstep).

the yellow line is not straight, eliminate a.

check: the green line is straight, answer is c.

simple interest with positive interest rate gives a straight line sloping up to the right. green line:

g(t) = $100 + ($200-$100) ÷ (20 years) × (t years)

= $100 + (4%/year) × ($100)(t years)

incidentally, the yellow line curving upward represents continuously compounded interest

y = $100 × 2^(t years / (14 assuming the yellow line hits $200 at 14 years. 2^0 is 1, so y(0) = $100; 2^(14/14) is 2, so y(14) = $200.