The answer would be
4
5
6
Yes
C would be the answer. When you simplify the equations C does not contain a radical.
A square is a figure with four equal sides and four right angles.In a square the diagonals bisect at right angles .
In the above given options option A is the right answer.
If the diagonals of any parallelogram are perpendicular then the figure is not necessarily a square.
Diagonals are perpendicular in a rhombus too.
A square and Rhombus both have diagonals that are perpendicular.
So it is not a necessary condition for a parallelogram to be a square .
The other options are right .
So option B is the right option that if diagonals are perpendicular is not sufficient to prove the figure to be a square.
4x - 2y - z = - 5 ______×2x - 3y + 2z = 3 _______×13x + y - 2z = - 5 ______×1
8x - 4y - 2z = - 10x - 3y + 2z = 33x + y - 2z = - 5
8x - 4y - 2z = - 10(+) x - 3y + 2z = 3_____________9x - 7y = - 77y = 9x + 7y = 9/7x + 1
x - 3y + 2z = 3(+) 3x + y - 2z = - 5_____________4x - 2y = - 24x - 2(9/7x + 1) = - 24x - 18/7x - 2 = - 210/7x = 0x = 0
y = 9/7(0) + 1y = 1
0 - 3(1) + 2z = 32z = 6z = 3
Step-by-step explanation:
A. x²-x-10=0
is part of the solution process for solving this rational equation using the least common denominator
