Let t = number of hours
The first candle starts at 8 inches.
It burns at 7/10 inch per hour, so in t hours it burns (7/10)t inches.
After t hours, its length is 8 - (7/10)t
The second candle starts at 6 inches.
It burns at 1/5 inch per hour, so in t hours it burns (1/5)t inches.
After t hours, its length is 6 - (1/5)t
You want the lengths to be equal, so the equation is
8 - (7/10)t = 6 - (1/5)t
You add them together to get 5 and 3/4 feet or 5.75 feet.
Answer:
radius r = 3 cm
height h = 10 cm
volume V = 282.743339 cm^3
lateral surface area L = 188.495559 cm^2
top surface area T = 28.2743339 cm^2
base surface area B = 28.2743339 cm^2
total surface area A = 245.044227 cm^2
In Terms of Pi π
volume V = 90 π cm3
lateral surface area L = 60 π cm^2
top surface area T = 9 π cm^2
base surface area B = 9 π cm^2
total surface area A = 78 π cm^2
Step-by-step explanation:
Cylinder Formulas in terms of r and h:
Calculate volume of a cylinder:
V = πr2h
Calculate the lateral surface area of a cylinder (just the curved outside)**:
L = 2πrh
Calculate the top and bottom surface area of a cylinder (2 circles):
T = B = πr2
Total surface area of a closed cylinder is:
A = L + T + B = 2πrh + 2(πr2) = 2πr(h+r)
Agenda: r = radius
h = height
V = volume
L = lateral surface area
T = top surface area
B = base surface area
A = total surface area
π = pi = 3.1415926535898
√ = square root
The question is:
Check whether the function:
y = [cos(2x)]/x
is a solution of
xy' + y = -2sin(2x)
with the initial condition y(π/4) = 0
Answer:
To check if the function y = [cos(2x)]/x is a solution of the differential equation xy' + y = -2sin(2x), we need to substitute the value of y and the value of the derivative of y on the left hand side of the differential equation and see if we obtain the right hand side of the equation.
Let us do that.
y = [cos(2x)]/x
y' = (-1/x²) [cos(2x)] - (2/x) [sin(2x)]
Now,
xy' + y = x{(-1/x²) [cos(2x)] - (2/x) [sin(2x)]} + ([cos(2x)]/x
= (-1/x)cos(2x) - 2sin(2x) + (1/x)cos(2x)
= -2sin(2x)
Which is the right hand side of the differential equation.
Hence, y is a solution to the differential equation.
