
Let AB be a chord of the given circle with centre and radius 13 cm.
Then, OA = 13 cm and ab = 10 cm
From O, draw OL⊥ AB
We know that the perpendicular from the centre of a circle to a chord bisects the chord.
∴ AL = ½AB = (½ × 10)cm = 5 cm
From the right △OLA, we have
OA² = OL² + AL²
==> OL² = OA² – AL²
==> [(13)² – (5)²] cm² = 144cm²
==> OL = √144cm = 12 cm
Hence, the distance of the chord from the centre is 12 cm.
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Answer:
The Z is has no lines of symmetry. The W and Y have one line of symmetry. The X has two lines of symmetry.
Substitute the x with 5
-(5)2 + 4(5) - 10 = 0
Part A:5=2x^2-x^2+13
5=x^2+13
0=x^2+13-5
0=x^2+8
Part B:5=2x^3-x^3+13
5=x^3+13
0=x^3+13-5
0=x^3+8
Answer:
Subtract 4
Step-by-step explanation:
-3 - 4 = -7
-7 - 4 = -11
-11 - 4 = -15