the price of stock a at 9 a.m. was $15.75. since then, the price has been increasing at the rate of $0.05…

the price of stock a at 9 a.m. was $15.75. since then, the price has been increasing at the rate of $0.05 per hour. at noon, the price of stock b was $16.53. it begins to decrease at a rate of $0.13 per hour. if the stocks continue to increase and decrease at the same rates, in how many hours will the prices of the stocks be the same?\n\nthe prices of the stocks will be the same (square) hour(s) after noon.\n(type an integer or a decimal)

the price of stock a at 9 a.m. was $15.75. since then, the price has been increasing at the rate of $0.05 per hour. at noon, the price of stock b was $16.53. it begins to decrease at a rate of $0.13 per hour. if the stocks continue to increase and decrease at the same rates, in how many hours will the prices of the stocks be the same?\n\nthe prices of the stocks will be the same (square) hour(s) after noon.\n(type an integer or a decimal)

Answer

Explanation:

Step1: Calculate Stock A's price at noon

From 9 A.M. to noon (3 hours), Stock A increases at a rate of $0.05$ per hour. Its price at noon is $15.75 + 0.05\times3=15.75 + 0.15 = 15.9$.

Step2: Set up an equation

Let $t$ be the number of hours after noon when the prices are the same. The price of Stock A after $t$ hours is $15.9+0.05t$, and the price of Stock B after $t$ hours is $16.53 - 0.13t$. Set the two - expressions equal: $15.9+0.05t=16.53 - 0.13t$.

Step3: Solve the equation for $t$

Add $0.13t$ to both sides: $15.9+0.05t + 0.13t=16.53-0.13t + 0.13t$, which simplifies to $15.9 + 0.18t=16.53$. Then subtract 15.9 from both sides: $0.18t=16.53 - 15.9$, so $0.18t=0.63$. Divide both sides by 0.18: $t=\frac{0.63}{0.18}=3.5$.

Answer:

$3.5$