Answer:
t = 460.52 min
Step-by-step explanation:
Here is the complete question
Consider a tank used in certain hydrodynamic experiments. After one experiment the tank contains 200 liters of a dye solution with a concentration of 1 g/liter. To prepare for the next experiment, the tank is to be rinsed with fresh water flowing in at a rate of 2 liters/min, the well-stirred solution flowing out at the same rate.Find the time that will elapse before the concentration of dye in the tank reaches 1% of its original value.
Solution
Let Q(t) represent the amount of dye at any time t. Q' represent the net rate of change of amount of dye in the tank. Q' = inflow - outflow.
inflow = 0 (since the incoming water contains no dye)
outflow = concentration × rate of water inflow
Concentration = Quantity/volume = Q/200
outflow = concentration × rate of water inflow = Q/200 g/liter × 2 liters/min = Q/100 g/min.
So, Q' = inflow - outflow = 0 - Q/100
Q' = -Q/100 This is our differential equation. We solve it as follows
Q'/Q = -1/100
∫Q'/Q = ∫-1/100
㏑Q = -t/100 + c

when t = 0, Q = 200 L × 1 g/L = 200 g

We are to find t when Q = 1% of its original value. 1% of 200 g = 0.01 × 200 = 2

㏑0.01 = -t/100
t = -100㏑0.01
t = 460.52 min
Answer:
Area covered by the fences will be 16.1 unit²
Step-by-step explanation:
Let the first parabola is represented by the function f(x) = 6x²
and second parabola by g(x) = x² + 9
point of intersection of the graphs will be determined when f(x) = g(x)
6x² = x² + 9
5x² = 9
x² = 1.8
x = ± 1.34
Now we will find the area between these curves drawn on the graph.
Area = ![\int_{-1.34}^{1.34}[f(x)-g(x)]dx=\int_{-1.34}^{1.34}[6x^{2}-(x^{2}+9)]dx](https://tex.z-dn.net/?f=%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5Bf%28x%29-g%28x%29%5Ddx%3D%5Cint_%7B-1.34%7D%5E%7B1.34%7D%5B6x%5E%7B2%7D-%28x%5E%7B2%7D%2B9%29%5Ddx)
= 
= ![[\frac{5}{3}x^{3}-9x]_{-1.34}^{1.34}](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7Dx%5E%7B3%7D-9x%5D_%7B-1.34%7D%5E%7B1.34%7D)
= ![[\frac{5}{3}(-1.34)^{3}-9(-1.34)-\frac{5}{3}(1.34)^{3}+9(1.34)]](https://tex.z-dn.net/?f=%5B%5Cfrac%7B5%7D%7B3%7D%28-1.34%29%5E%7B3%7D-9%28-1.34%29-%5Cfrac%7B5%7D%7B3%7D%281.34%29%5E%7B3%7D%2B9%281.34%29%5D)
= ![[-4.01+12.06-4.01+12.06]](https://tex.z-dn.net/?f=%5B-4.01%2B12.06-4.01%2B12.06%5D)
= 16.1 unit²
Answer:
13 units^2
Step-by-step explanation:
one side of the square =



So area of square = Length * Length
=
*
= 13
is the answer.
Step-by-step explanation:
For this question,
•<u> </u><u>First</u>, find the slope of the line parallel to the line 5/8x−7/6y=2 :-

•<u>Second</u>, find the equation of the line,

For f(x), which has a vertex at (2,0), the y-intercept at (0,4) is above this vertex, so the parabola opens upward. This means that the vertex is the only point that touches the x-axis, so there is only 1 x-intercept.
For h(x), the graph does not have any x-intercepts.
For g(x) = x^2 + x - 2 = (x+2)(x-1), this intersects the x-axis at x = -2 and x = 1, so there are 2 x-intercepts.
From least to greatest: h(x), f(x), g(x).