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Viktor [21]
3 years ago
5

Find the length of the segment. round to the nearest tenth of a unit P(1,2) Q(5,4)

Mathematics
2 answers:
Alborosie3 years ago
7 0
Hello, 

d= \sqrt{ (x_{2}-x_{1})^{2}+(y_{2}-y_{1})^{2}  }  \\  \\ d=\sqrt{ (5-1)^{2}+(4-2)^{2}  } \\  \\ d=\sqrt{ 4^{2}+2^{2}  }  \\  \\ d= \sqrt{16+4}  \\  \\ d= \sqrt{20}  \\  \\d= \sqrt{2^{2}*5 }\\  \\d= 2\sqrt{5 } \\ \\  d=4.5

Answer: lenght= 4.5
yuradex [85]3 years ago
7 0
P(1,2)\ \ \ Q(5,4)\\\\
PQ=\sqrt{(x_Q-x_P)^2+(y_Q-y_P)^2}\\\\
PQ=\sqrt{(5-1)^2+(4-2)^2}=\sqrt{16+4}=\sqrt{20}=2\sqrt{5}\approx4,5\\\\
Length\ of\ this\ segment\ is\ equal\ to\ 4,5.
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The solution for the given differential equation is lnlxl = \frac{-x}{y} +C

Given,

(y^{2} +y^{x} )dx - x^{2} dy =0

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y = ux, dy = udx + xdu

Then,

((ux)^{2} +(ux)x)dx-x^{2} (udx+xdu)=0\\=u^{2} x^{2} dx+ux^{2} dx-x^{2} udx-x^{3} du=0\\=u^{2} x^{2} dx=x^{3} du\\=\frac{x^{2} }{x^{3} } dx=\frac{1}{u^{2} } du

Now,

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That is lnlxl =\frac{-x}{y} +C

Learn more about differential equation here: brainly.com/question/21852102

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