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:
8x²- x -8
Step-by-step explanation:
The question is not well written. I guess what you mean is:
Subtract -2x^2+4x-1 ( minus 2x squared plus 4x minus 1) from 6x^2+3x-9 (6x squared plus 3x minus 9).
To subtract -2x^2+4x-1 from 6x^2+3x-9, that is
6x^2+3x-9 - (-2x^2+4x-1). This can be written as
6x²+3x-9 - (-2x²+4x-1)
Now, opening the bracket, we will get
6x²+3x-9 +2x²-4x+1
Then, collecting like terms, we get
6x²+2x²+3x-4x-9+1
6x²+2x² = 8x²
3x-4x = -x
-9+1 = -8
∴6x²+2x²+3x-4x-9+1 becomes
8x²-x -8
The answer is 8x²-x -8.
Since the angles add to 180°, angle F is 72°.
Using the law of sines,
DF/sin72° = 24/sin45°
x = 32.28
So, did you just guess at A?
This is just an answer from another website, but it should still work.
I'm sorry I can't answer this but the triangle looks like it's wearing a thong
Y=-x+8
gradient of the parallel line is the same as the gradient of the line given, which is -1
to find the y-intercept, sub the values (9,-1) into the equation we are finding which is y=-x+c
