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:
bet
Step-by-step explanation:
I believe the correct answer from the choices listed above is option D. The graph <span>G(x) as compared to the graph of F(x) would be that the </span><span>graph of G(x) is the graph of F(x) stretched vertically and shifted 5 units down. 2 is a stretch factor and -5 is the shift downwards of the graph. Hope this answers the question.</span>
answer
10
step-by-step explanation
the equation given is sin(x) = cos(y) with x = 2k + 3 and y = 6k + 7
substitute in 2k+ 3 for x in sin(x) and substitute in 6k + 7 for y in cos(y)
sin(x) = cos(y)
sin(2k + 3) = cos(6k + 7)
we know that sin(x) = cos(90 -x)
sin(2k + 3)
= cos(90 - (2k + 3) )
= cos(90 - 2k - 3)
= cos(87 - 2k)
substitute this into the equation sin(2k + 3) = cos(6k + 7)
sin(2k + 3) = cos(6k + 7)
cos(87 - 2k) = cos(6k + 7)
87 - 2k = 6k +7
80 = 8k
k = 10
Commutative property of addition
