(-4,4) (2,1)
gradient = (1-4)/(2--4) = -1/2
y = mx + c
y = -1/2x + c
Replace point (2,1) in the equation
1=-1/2(2) +c
c = 2
Equation : y = -1/2x + 2
y-2 = -1/2x
Answer is C.
Hope it helped!
Since TSQ and QSR are supplementary, they give 180 when summed. TSQ is 150, so QSR must be 30.
Therefore, you have
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:
-2
Step-by-step explanation:
You can see that the line is passing through the negative point, which is -2.
I hope this helped!
~natasha
