Answer:
x² - 3x - 10 = (x - 5) (x + 2)
x² - 3x - 18 = (x + 3) (x - 6)
Step-by-step explanation:
<u>x² - 3x - 10</u>
x² + 2x - 5x - 10
x(x + 2) - 5(x + 2)
(x - 5) (x + 2)
<u>x² - 3x - 18</u>
x² - 6x + 3x - 18
x(x - 6) + 3(x - 6)
(x + 3) (x - 6)
<u>-TheUnknown</u><u>S</u><u>cientist</u>
The 6 is in the ones place the 1 is in the tens place and the 2 is in the hundreds place.
Step-by-step explanation:
Hey there!
By looking at the figure, the given "2x" and "6x + 28" are co-interior angle.
So,
2x + 6x + 28 = 180°. ( sum of co-interior angle is 180°)
8x + 28°=180°
8x = 180° - 28°
8x = 152°
X = 19°
<u>There</u><u>fore</u><u>,</u><u> </u><u>X </u><u>=</u><u> </u><u>1</u><u>9</u><u>°</u><u>.</u>
<em><u>Hope</u></em><em><u> </u></em><em><u>it</u></em><em><u> helps</u></em><em><u>.</u></em><em><u>.</u></em><em><u>.</u></em>
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.
Answer:
4 Is greater than 10. Hope this helped.
