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:
y = 3x + 8
Step-by-step explanation:
y = mx + b
m is the slope
b is the y intercept
9514 1404 393
Answer:
77
Step-by-step explanation:
The sum of the measures of the two given arcs is 180°, the measure of semicircle ADC. Using this fact, we can find x and the measure of DC.
AD +DC = 180
(19x -49) +(5x +37) = 180
24x = 192 . . . . . . . . . . . add 12, simplify
x = 8 . . . . . . . . . . divide by 24
Then the measure of DC is ...
DC = 5x +37 = 5(8) +37 = 40 +37
DC = 77
