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.
Step-by-step explanation:
Answer:
$1.75
Step-by-step explanation:
The selling for each candy bar may be determined by a set of linear equations. This pair of linear equations may be solved simultaneously by using the elimination method. This will involve ensuring that the coefficient of one of the unknown variables is the same in both equations.
It may be solved by substitution in that one of the variable is made the subject of the equation and the result is substituted into the second equation
.
Let the cost of a snack bag be s and that of a candy bar be c, then if on Wednesday the students or 23 snack bags and 36 candy bars that raised $114.75 on Thursday the seventh so 37 snack bags and 36 candy bars that raised $146.25
23s + 36c = 114.75
37s + 36c = 146.25
14s = 31.5
s = $2.25
23(2.25) + 36c = 114.75
36c = 114.75 - 51.75
36c = 63
c = 63/36
= $1.75
Answer:
See below
Step-by-step explanation:
Solve x-5y=6 for x
First, we need to isolate the x by moving -5y to the other side.
To do this, we need to add 5y to both sides
x - 5y= 6
+5y +5y
x= 5y+6
So, your third answer is correct
0.59 increase because it went up
