Consider
.. X = {1, 2, 3, 4}, x = 4
.. Y = {2, 3, 4, 5, 6}, y = 5, w = 3
The elements in X or Y (X ∪ Y) are {1, 2, 3, 4, 5, 6}, n = 6.
.. n = 6 = 4 + 5 - 3
Note that if we just add x and y, we count the common elements twice. In order to just count the common elements once, we need to subtract that count from the total of x and y.
selection B is appropriate.
Answer:
The area of the paralellogram is 24 square units.
Step-by-step explanation:
Geometrically speaking, the area of the parallelogram has an equation equivalent to the area formula for a rectangle, that is:
(1)
Where:
- Area.
- Base.
- Height.
The base and height of the parallelogram are, respectively:
Base
Height
Then, the area of the parallelogram is:
The area of the paralellogram is 24 square units.
I think it is 24...Im not sure tho
The answer to this is letter D
MrBillDoesMath!
Answer to #4: 81/256 * s^8 * t^ 12
Comments:
(7x^3) ^ (1/2) = 7 ^ (1/2) * x^(3/2) where ^(1/2) means the square root of a quantity. The answer written (7x^3) is NOT correct.
---------------------
(1) (27s^7t^11)^ (4/3)
= 27^(4/3) * (s^7)^(4/3) * (t^11)^ (4/3)
As 27 = 3^3, 27 ^(4/3) = 3^4 = 81
(2) (-64st^2)^ (4/3) = (-64)^(4/3) * (s^4/3) * t(^8/3)
As 64 = (-4)^3, (-64)^(4/3) = (-4)^4 = +256
So (1)/(2) =
81 * s^(28/3)* t^(44/3)
------------------------------- =
256 s^(4/3) * t^((8/3)
81/256 * s ^ (28/3 - 4/3) * t^(44/3 - 8/3) =
81/256 * s^(24/3) * t (36/3) =
81/256 * s^8 * t^ 12
MrB
