Area of the square: A s = s².
Area of the circle: A c = ( 1/2 s )² π = 1/4 s² · 3.14 = 0.785 s²
Area inside the square but outside the circle: s² - 0.785 s² = 0.215 s²
The ratio:
0.215 s² : s² = 0.215 : 1 = 215 : 1000 = 43 : 200 .
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:
144m
Step-by-step explanation:
we would used tan 30 to determine the height of the building
tan 30 = opposite / adjacent
0.5774 x 250 = 144m
Answer:
-2
Step-by-step explanation:
(-3x/2) - 3 = 3
add 3 to both sides
(-3x/2) = 3
multiply both sides by 2
-3x = 6
divide both sides by -3
x= -2
