Answer:
a1 = 6 and an = an-1 - 5
a2 = a2-1 - 5 = a1 - 5 = 1
a3 = a3-1 - 5 = a2 - 5 = -4, etc
Step-by-step explanation:
Answer:
25/14
Step-by-step explanation:
Answer:
x° = ∠OBR = ∠ABC (base angles of a cyclic isosceles trapezoid)
Step-by-step explanation:
APRB form a cyclic trapezoid
∠APO = x° (Base angle of an isosceles triangle)
∠OPR = ∠ORP (Base angle of an isosceles triangle)
∠ORB = ∠OBR (Base angle of an isosceles triangle)
∠APO + ∠OPR + ∠OBR = 180° (Sum of opposite angles in a cyclic quadrilateral)
Similarly;
∠ORB + ∠ORP + x° = 180°
Since ∠APO = x° ∠ORB = ∠OBR and ∠OPR = ∠ORP we put
We also have;
∠OPR = ∠AOP = ∠BOR (Alternate interior angles of parallel lines)
Hence 2·x° + ∠AOP = 180° (Sum of angles in a triangle) = 2·∠OBR + ∠BOR
Therefore, 2·x° = 2·∠OBR, x° = ∠OBR = ∠ABC.
4x -3y + z = -10...............(1)
2x +y + 3z = 0...............(2)
-x +2y - 5z = 17...............(3)
First we multiply 3*(2) and add it to (1)
6x +3y +9z =0..................+(1)..................> 10x + 10z = -10......(4)
Then we multiply -2*(2) and add it to (3)
-4x -2y -6z =0 ...................+(3)................> -5x -11z = 17...........(5)
Multiply 2*(5) and add it to (4)
-10x -22z = 34...................+(4).................> -12 z = 24 ..............>>> z = -2
Substitute z in (4)............> 10x +10(-2) = -10.............................>>> x = 1
Substitute x and z in (2).....> 2(1) +y + 3(-2) = 0..................>>> y = 4
Solution (x,y,z) = (1,4,-2)
