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LiRa [457]
3 years ago
6

Abs(sin(x) - sin(y)) ≤ abs(x-y)

Mathematics
1 answer:
barxatty [35]3 years ago
3 0
|sin(x)−sin(y)|=|2sin(x−y2)cos(x+y2)|≤|2sin(x−y2)|
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k:\ y=m_1x+b_1\ \ \ \ and\ \ \ \ l:\ y=m_2x+b_2\\ \\the\ line\k\ is\ perpendicular\ to\ the\ line\ l\ \ \ \Leftrightarrow\ \ \ m_1\cdot m_2=-1\\----------------------------\\\\A.\\2y=-3x+5\ \ \Rightarrow\ \ \ y=-  \frac{3}{2}  x+2.5\ \ \Rightarrow\ \ m_1=- \frac{3}{2} \\\\2x+3y=4\ \ \Rightarrow\ \ 3y=-2x+4\ \ \Rightarrow\ \ y=- \frac{2}{3}x+1 \frac{1}{3}  \ \ \Rightarrow\ \ m_2=- \frac{2}{3} \\\\m_1\cdot m_2=- \frac{3}{2} \cdot (- \frac{2}{3})=1 \neq -1

B.\\5x-8y=9\ \ \Rightarrow\ \ -8y=-5x+9\ \ \Rightarrow\ \ \ y=  \frac{5}{8}  x-1 \frac{1}{8} \ \Rightarrow\ \ m_1= \frac{5}{8} \\\\12x-5y=7\ \Rightarrow\ \ -5y=-12x+7\ \ \Rightarrow\ \ y= \frac{12}{5}x-1 \frac{2}{5}  \ \ \Rightarrow\ \ m_2= \frac{12}{5} \\\\m_1\cdot m_2= \frac{5}{8} \cdot  \frac{12}{5}= \frac{3}{2}  \neq -1

C.\\y=2x-7\ \ \Rightarrow\ \ m_1=2 \\\\x+2y=3\ \Rightarrow\ \ 2y=-x+3\ \ \Rightarrow\ \ y= -\frac{1}{2}x+1 \frac{1}{2}\ \ \Rightarrow\ \ m_2=- \frac{1}{2} \\\\m_1\cdot m_2= 2 \cdot (- \frac{1}{2})= -1\ \ \Rightarrow\ \ y=2x-7\ \ \ \ \perp\ \ \ \ x+2y=3

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