
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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The answer is 10 because when you use the slope formula y^2 - y^1 divided by x^2 - x^1 and plug your points in you get 10/0 which makes the answer ten
The two fractions are divisible by 2, can be broken down
Answer:
Octave
Step-by-step explanation:
F to F is an Octave.
It would take the bud 15 minutes to travel at 125