Diagonal of a Rhombus are perpendicular & intersects in their middle point: Assume the diagonals intersects in H
A(0,-8), B(1,-0), C(8,-4) & D(x, y) are the vertices of the rhombus and we have to calculate D(x, y)
Consider the diagonal AC. Find the coordinate (x₁, y₁) H, the middle of AC Coordinate (x₁, y₁) of H, middle of A(0,-8), C(8,-4) x₁ (0+8)/2 & y₁=(-8-4)/2 ==> H(4, -6) Now let's calculate again the coordinate of H, middle of the diagonal BD B(1,-0), D(x, y) x value = (1+x)/2 & y value=(y+0)/2 ==> x= (1+x)/2 & y=y/2 (1+x)/2 & y/2 are the coordinate of the center H, already calculated, then: H(4, -6) = [(1+x)/2 , y/2]==>(1+x)/2 =4 ==> x=7 & y/2 = -6 ==> y= -12
Every rational number has a base-2 representation, but only the ones with denominators that are powers of 2 will require a finite number of bits to fully represent it.
For example,
whereas a number whose denominator contains anything else like 1/3 will need an infinite number of bits to represent it exactly.
and so on, so that it has a repeating but non-terminating base-2 representation
Im sorry i do not know the answer but i hope you have a wonderful day and never give up on things i almost gave up on a test i said i cant do it but i did i know this might sound pufft yeah sure its fake well its not