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
For example if there was 5.73810 You would round up to 6.
if the decimal place is .5 or higher you round up but if it is .49 and lower you round down!
So, the area of a circle is a=pi r^2
so do that and get a=pi(4)
and since the circle touches the edge of the rectangle, the rectangle is 4 cm tall, so 4*8 is 32 and pi(4) is about 12.56, so 32-12.56 is 19.44, and I believe that is the answer. (19.4)
X = (1/2)y - 3 for smaller number
y = 10x - 1 for larger number
Answer:
250 cm.
Step-by-step explanation:
Half = 50% so by proportion:
Half length of the rope = 75 *50/15
= 250