<u>ANSWER:</u>
<span>The initial value of the truck is $22,200
</span>
<u>EXPLANATION:</u>
For the given function
<span>
f(t) = 22,200(0.92)^t
We know that at t = 0, we have
</span> f(0) = 22,200(0.92)^0 = 22,200
<span>
which is the value of the truck initially. So, the correct answer is
</span>The initial value of the truck is $22,200
I think it’s 180 sorry if it’s wrong
Answer:
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Step-by-step explanation:
Given the data in the question;
If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is?
dA/dt = rate in - rate out
first we determine the rate in and rate out;
rate in = 3pound/gallon × 5gallons/min = 15 pound/min
rate out = A pounds/1000gallons × 5gallons/min = 5Ag/1000pounds/min
= 0.005A pounds/min
so we substitute
dA/dt = rate in - rate out
dA/dt = 15 - 0.005A
Therefore, If A(t) represents the amount of salt in the tank at time t, the correct differential equation for A is is dA/dt = 15 - 0.005A
Option C) dA/dt = 15 - 0.005A is the correction Answer
Answer:
An equivalent ratio in simplest term is: 3 : 5
