Question

 The price of Stock A at 9 A.M. was ​$13.56. Since​ then, the price has been increasing at the rate of ​$0.12 each hour. At noon the price of Stock B was ​$14.31. It begins to decrease at the rate of ​$0.14 each hour. If the two rates​ continue, in how many hours will the prices of the two stocks be the​ same?

Answer 1

Answer:

The prices of the two stocks be the​ same after 2.88 hours after 12 pm

Step-by-step explanation:

Price of stock A at 9 a.m.= $13.56

The price has been increasing at the rate of ​$0.12 each hour.

Increase in price after 3 hours =$0.12 \times 3 =0.36$

Price of Stock A at noon= 13.56+0.36 = $13.92

Let n be the no. of hours after which the prices of both the stocks will be same .

So, Increase in price after n hours = 0.12 n

Price of Stock A after n hours = 13.92+0.12 n

Price of stock B = $14.31

The price has been decreasing at the rate of ​$0.14 each hour.

Let n be the no. of hours

So, Increase in price after n hours = 0.14 n

Price of Stock B after n hours =14.31-0.14 n

ATQ

$13.56+0.12 n= 14.31-0.14 n$

$-13.56+14.31=0.14 n + 0.12 n$

$0.75=0.26 n$

$\frac{0.75}{0.26}=n$

$2.88=n$

Hence the prices of the two stocks be the​ same after 2.88 hours after 12 pm .

Answer 2

$6.36 would be your answer!