the prices of stock A at 9am was $12.18 since then the price has been increasing at the rate of $0.06 each hour. At noon the price of stock B was $12.68. It begins to descrease at the rate of $0.09 per hour. If the 2 rates continue in how many hours will the prices of the two stocks be the same?

3.33hours

Step-by-step explanation:

Where the number of hours = h

the prices of stock A at 9am was $12.18 since then the price has been increasing at the rate of $0.06 each hour.

The Algebraic expression =

$12.18 + $0.06 × h

12.18 + 0.06h

At noon the price of stock B was $12.68. It begins to descrease at the rate of $0.09 per hour.

$12.68 - $0.09×h

12.68 - 0.09h

If the 2 rates continue in how many hours will the prices of the two stocks be the same?

Stock A = Stock B

12.18 + 0.06h = 12.68 - 0.09h

Collect like terms

0.06h + 0.09h = 12.68 - 12.18

0.15h = 0.5

h = 0.5/0.15

h = 3.3333333333 hours

h = 3.33hours

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.