Answer: The linear equation could be: r = 800,000 - 175,000h, where r represents the amount of water in the reservoir and h represents the number of hours since 4:00 am.
To find this equation, you must first find the slope. Divide the total change in water by the total number hours of the change.
We have: 800000/4 = 175,000 This means that the water is going down by 175,000 gallons per hour.
Now, just add in the starting amount of 800,000 gallons and you have your equation.
Area. The amount of space covered by a two-dimensional object; the amount of carpet you'd need to cover a floor would depend on the floor's area.
Perimeter. The distance around a two-dimensional object; the amount of fence you'd need to surround your yard equals the perimeter of your yard. (The perimeter of a circular object is called the circumference.
Volume. The amount of three-dimensional space inside an object; the amount of liquid inside a soda can represents the can's volume.
Surface area. Measures the amount of "skin" needed to cover a three-dimensional object, neglecting its thickness; the amount of siding you'd need to completely cover a house represents the surface area of that house's walls.
Description Formula Variables
Area of a rectangle A = l · w l = length, w = width
Perimeter of a rectangle P = 2(l + w) l = length, w = width
Area of a circle A = r2 r = radius of circle
Circumference of a circle C = 2r r = radius of circle
Volume of a rectangular solid V = l · w · h l = length, w = width, h = height
Volume of a cylinder V = r2h r = radius; h = height
Surface area of a cube SA = 6l2 l = length of side
X>Y
X=7Y
3X=7+4Y
X=?
Y=?
----------------------------
3*7=7+4Y
21Y=7+4Y
21Y-4Y=7
17Y=7
Y=7/17
----------------------
X=7*7/17
X=49/17
Range is the smallest value taken away from the biggest- so here it would be 100- 68, which you can easily work out!!
Answer:
X(s) = 2/[s³(s + 9)]
Step-by-step explanation:
Here is the complete question
Solve for the function X(s) in the Laplace domain by taking the Laplace transform of the following differential equations with given initial conditions.
dx/dt + 9x = 16t² x(0) = 0 and dx(0)/dt = 5
Solution
dx/dt + 9x = 16t²
Taking the Laplace transform of the differential equation, we have
L{dx/dt + 9x} = L{16t²}
L{dx/dt} + L{9x} = L{16t²}
L{dx/dt} + 9L{x} = 16L{t²}
sL{x} - x(0) + 9L{x} = 16L{t²}
L{x} = X(s)
L{t²} = 2!/s³
Substituting X(s) and L{t²} into the equation above, we have
sX(s) - x(0) + 9X(s) = 2!/s³
Substituting x(0) = 0 into the equation above, we have
sX(s) - x(0) + 9X(s) = 2!/s³
sX(s) - 0 + 9X(s) = 2!/s³
sX(s) + 9X(s) = 2!/s³
Factorizing X(s), we have
(s + 9)X(s) = 2!/s³
X(s) = 2/[s³(s + 9)]
So X(s) in the Laplace domain is
X(s) = 2/[s³(s + 9)]
