Answer:
Time,t that it will elapse before the concerntration of the dye in the tank reaches1% of its original value is 460.52 seconds
Step-by-step explanation:
Let Q(t) = amount of dye for all time,t.
Let Q'= rate in - rate out.
But Q/200 = concerntration of dye
Therefore rate out= Q/200 ×2.
Q'/Q = - 1/100
Dividing by Q gives
Ln/Q/ + c = -1/100t + c1
Integrating both sides with respect to t
Ln/Q/ = -1/100t + c2.
c1 and c1 are just another constants
Q= c3e^-t/100
Exponentiating both sides will cause the absolute values to get absorbed into the constants giving rise to a new constant c3
200= c3-0/100
c3= 200
2= 200e^-t/100
Ln1/100= - t/100
Ln0.01 = - t/ 100
Cross multiply
t= Ln0.01 × 100
t = 4.6052 ×100
t = 460.52seconds
Answer:
I think it is 12 and 1/3 and 121 over 3
Is this for a quiz? It looks familiar
Answer:
C None of the above
Step-by-step explanation:
Here are the required options
Choose all answers that apply
A (5g+3h).8
B (5g+3h).6
C None of the above
Given the expression;
(5g+3h+4).2
Expand
(5g+3h+4).2
= 2(5g+3h+4)
= 2(5g) + 2(3h) + 2(4)
= 10g+6h+8
Checking option A;
(5g+3h).8
Expand
= 8(5g+3h)
= 8(5g) + 8(3h)
= 40g+24h
Checking option B:
(5g+3h).6
Expand
= 6(5g+3h)
Using the distributive law;
= 6(5g) + 6(3h)
= 30g+18h
Since none of the expressions for A and B is equivalent to 10g+6h+8, hence <u>NONE of the options are correct</u>
All you have to do is draw a net in the box.
can you help me with something since i helped you?
