The volume is twice that of the cube of width, so the width of the original prism is
.. ∛(16 cm³/2) = 2 cm
Length = 4 cm
Width = 2 cm
Height = 2 cm
Answer: Identify the shapes you will need to determine the area of the figure.
Calculate and add the areas of the unshaded triangle and two circles.
Step-by-step explanation:
Answer:
y =
+ 
Step-by-step explanation:
y''- 9 y' + 18 y = t²
solution of ordinary differential equation
using characteristics equation
m² - 9 m + 18 = 0
m² - 3 m - 6 m+ 18 = 0
(m-3)(m-6) = 0
m = 3,6
C.F. = 
now calculating P.I.


hence the complete solution
y = C.F. + P.I.
y =
+ 
The volume of the first cube is (5h^2)^3, while the volume of the second cube is (3k)^3, so their total volume is (5h^2)^3 + (3k)^3. We can use the special formula for factoring a sum of two cubes:
x^3 + y^3 = (x + y)(x^2 - xy + y^2)
(5h^2)^3 + (3k)^3 = (5h^2 + 3k)((5h^2)^2 - (5h^2)(3k) + (3k)^2)
= (5h^2 + 3k)(25h^4 - 15(h^2)(k) + 9k^2)
This is the second of the given choices.