Answer:
Step-by-step explanation:
Volume of tank is 3000L.
Mass of salt is 15kg
Input rate of water is 30L/min
dV/dt=30L/min
Let y(t) be the amount of salt at any time
Then,
dy/dt = input rate - output rate.
The input rate is zero since only water is added and not salt solution
Now, output rate.
Concentrate on of the salt in the tank at any time (t) is given as
Since it holds initially holds 3000L of brine then the mass to volume rate is y(t)/3000
dy/dt= dV/dt × dM/dV
dy/dt=30×y/3000
dy/dt=y/100
Applying variable separation to solve the ODE
1/y dy=0.01dt
Integrate both side
∫ 1/y dy = ∫ 0.01dt
In(y)= 0.01t + A, .A is constant
Take exponential of both side
y=exp(0.01t+A)
y=exp(0.01t)exp(A)
exp(A) is another constant let say C
y(t)=Cexp(0.01t)
The initial condition given
At t=0 y=15kg
15=Cexp(0)
Therefore, C=15
Then, the solution becomes
y(t) = 15exp(0.01t)
At any time that is the mass.
Answer:
z=6
Step-by-step explanation:
987/z=164.5
multiply both sides by z
Answer:
Jelly beans: $6 for 1 kg
Gummy worms: $7 for 1 kg
Step-by-step explanation:
In this question, we need to find the cost of 1kg of jelly beans and gummy worms.
To do this, make equations for both scenarios and solve:
2j + 2g = 26
2j + 3g = 33
Turn the bottom equation negative, as we will first solve for gummy worms (g).
2j + 2g = 26
-2j - 3g = -33
Solve:
-g = -7
Divide both sides by -1.
g = 7
We now know that gummy worms cost $7 for 1 kg
To solve for jellybeans (j), plug in 7 to g in one of the equations and solve:
2j + 2(7) = 26
2j + 14 = 26
Subtract 14 from both sides.
2j = 12
Divide both sides by 2.
j = 6
Jellybeans cost $6 for 1 kg.
Check answer by plugging in values to one of the equations:
2(6) + 2(7) = 26
12 + 14 = 26
26 = 26
239.99 x 75%(75/100) ~ $180
$180 x 90%(90/100) = $162
As we know that the standard equation of circle is
, where <em>r</em> is the radius of circle and centre at <em>(h,k) </em>
Now , as the circle passes through <em>(2,9)</em> so it must satisfy the above equation after putting the values of <em>h</em> and <em>k</em> respectively

After raising ½ power to both sides , we will get <em>r = +5 , -5</em> , but as radius can never be -<em>ve</em> . So <em>r = +</em><em>5</em><em> </em>
Now , putting values in our standard equation ;
<em>This is the required equation of </em><em>Circle</em>
Refer to the attachment as well !