Let x, y, z ∈ C, the set of complex numbers.
By definition of ○,
(x ○ y) ○ z = (x + y - 2i) ○ z
… = (x + y - 2i) + z - 2i
… = x + y + z - 4i
and
x ○ (y ○ z) = x ○ (y + z - 2i)
… = x + (y + z - 2i) - 2i
… = x + y + z - 4i
so (x ○ y) ○ z = x ○ (y ○ z), which means ○ is indeed associative under C.
The answer to 0.6c+2.5=3.22 solving for C Answer:
c=1.2
First Photo - This is the problem drawn into the paper. If you have two vectors and they form a right angle (90°) between them, the Resultant Vector will appear like this, from the right angle.
Obs.:
1 - VR = Resultant Vector;
2 - The angle formed by the VR and the two vectors ISN'T a bisector.
Second Photo - You can move the bottom extremity of the Y vector to the arrow of the X vector and the tips from the Y and VR vectors will meet, forming a Right Triangle, whose the VR is the Hypotenuse (Opposes the right angle / Larger Side).
Now you just put the values in the Pythagorean Theorem:
Where VR is the "a".
Answer:
Terrence's
Step-by-step explanation:
The length of the square that will be cut out is the height of the box.
1. a
Anya's method: 8.5 -1.5 =7, 11- 1.5 =9.5, the height is 1.5, so the volume is height x length x width which is 1.5 x 9.5 x 7 =99.75 squared inches.
Terrence's method: 8.5-3 = 5.5, 11-3 = 8. Vol= 5.5 x 8 x 3 =132 squared inches. 99.75 < 132 squared inches, Terrence's idea would create larger volume.
1. b
The box's size depends on the length/width/height of the cardboard being cut, which is why different measurements / cutting methods for the same size cardboard can result in different box sizes.
2. The square would be cut from all four corners, therefore the sum of the 2 squares on the cardboard cannot exceed the short side of the cardboard. The shorter side of the cardboard is 8.5 inches, divided by 2 = 4.25 inches, hence the squares cannot be larger than 4.25 inches. Keep in mind that if you cut exactly 4.25 inches you will have a strip of 2.5 inches width that cannot be turned into a box.
If you want to cut 5 inches squares out, depending on how you draw it, it would either overlap or go outside of the paper because 5+5 is ten, surely on the 11 inches side that would still be perfectly fine but for the 8.5 inches side, there isn't any room for the 10 inches.
