Answer:
A(t) = 300 -260e^(-t/50)
Step-by-step explanation:
The rate of change of A(t) is ...
A'(t) = 6 -6/300·A(t)
Rewriting, we have ...
A'(t) +(1/50)A(t) = 6
This has solution ...
A(t) = p + qe^-(t/50)
We need to find the values of p and q. Using the differential equation, we ahve ...
A'(t) = -q/50e^-(t/50) = 6 - (p +qe^-(t/50))/50
0 = 6 -p/50
p = 300
From the initial condition, ...
A(0) = 300 +q = 40
q = -260
So, the complete solution is ...
A(t) = 300 -260e^(-t/50)
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The salt in the tank increases in exponentially decaying fashion from 40 grams to 300 grams with a time constant of 50 minutes.
40*9.2=360.8
10.7*2.25=24.075
360.8
+24.075
384.875. I disagree he actually made 384.875 dallors
Answer:
9. 
11. 
Step-by-step explanation:
9)
11)
Hope this helps!
Answer:
4x2−25
Step-by-step explanation:
(2x+5)(2x−5)
=(2x+5)(2x+−5)
=(2x)(2x)+(2x)(−5)+(5)(2x)+(5)(−5)
=4x2−10x+10x−25
4x2−25
