When two chords intersect in a cirlce, the ___ of the segments of one chord?

1 answer:

3 0

<span>Whenever two chords intersect in a circle, the products of their segments are equal.</span>

Source:
http://www.1728.org/circangl.htm

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Factorise<br> x^2 – 8x - 48

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Help please you don’t know how much this means to me

ICE Princess25 [194]

<h2>Problem 6:</h2><h2>a)</h2>

$a(0) = 1 \: \: \: \: \: \: b(0) = 2 \: \: \: \: \: c(0) = 3 \\ a(1) = b(0) + c(0) = 2 + 3 = 5 \\ b(1) = a(0) + c(0) = 1 + 3 = 4 \\ c(1) = a(0) + b(0) = 1 + 2 = 3 \\ \\ a(2) = b(1) + c(1) = 4 + 3 = 7 \\ b(2) = a(1) + c(1) = 5 + 3 = 8 \\ c(2) = a(1) + b(1) = 5 + 4 = 9$

$a(3) = b(2) + c(2) = 8 + 9 = 17\\ b(3) = a(2) + c(2) =7 + 9 = 16 \\ c(3) = a(2) + b(2) = 7 + 8 = 15 \\ \\ a(4) = b(3) + c(3) = 16 + 15 = 31 \\ b(4) = a(3) + c(3) = 17 + 15 = 32 \\ c(4) = a(3) + b(3) = 17 + 16 = 33$

$a(5) = 32 + 33 = 65 \\ b(5) = 31 + 33 = 64 \\ c(5) =31 + 32 = 63 \\ \\ a(6) = 64 + 63 = 127 \\ b(6) = 65 + 63 = 128 \\ c(6) = 65 + 64 = 129 \\$

$a(7) = 128 + 129 = 257 \\ b(7) = 127 + 129 = 256 \\c (7) = 127 + 128 = 255 \\ \\ a(8) = 256 + 255 = 511 \\ b(8) = 257 + 255 = 512 \\ c(8) = 257 + 256 = 513$

$a(9) = 512 + 513 = 1025 \\ b(9) = 511 + 513 = 1024 \\ c(9) = 511 + 512 = 1023 \\ \\ a(10) = 1024 + 1023 = 2047 \\ b(10) = 1025 + 1023 = 2048 \\ c(10) = 1025 + 1024 = 2049$