
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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To be parallel, gradients are the same!
so the gradient is -5/2.
y=-5/2x +c
substitute the points youve been givrn into this to find c!
7=-5/2 (-6) +c
7=15 +c
c must equal -7
y=-5/2x -7
Answer:
103.95
Step-by-step explanation:
99×5/100=4.95
99+4.95=103.95
Answer:
-72
Step-by-step explanation:
-32 + (2-6)(10)
-32 + (-4)(10)
-32 + -40
-72
Answer:£149
Step-by-step explanation: