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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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Step-by-step explanation:

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Answer:

Total surface area : 733

The shape of the base is a rectangle with sides 11 in. and 12 in.

Step-by-step explanation:

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Area of the base = 11*12 = 132

Area of the two triangles =  (11*16)/2 = 88

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2 years ago
How do I do 8b(ii) ? Please help me thank you!
Mila [183]
Step One
======
Find the length of FO (see below)

All of the triangles are equilateral triangles. Label the center as O
FO = FE = sqrt(5) + sqrt(2)

Step Two
======
Drop a perpendicular bisector from O to the midpoint of FE. Label the midpoint as J. Find OJ

Sure the Pythagorean Theorem. Remember that OJ is a perpendicular bisector.

FO^2 = FJ^2 + OJ^2
FO = sqrt(5) + sqrt(2)
FJ = 1/2 [(sqrt(5) + sqrt(2)]                                           \
OJ = ??

[Sqrt(5) + sqrt(2)]^2 = [1/2(sqrt(5) + sqrt(2) ] ^2 + OJ^2
5 + 2 + 2*sqrt(10) = [1/4 (5 + 2 + 2*sqrt(10) + OJ^2
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28 + 8* sqrt(10) = 7 + 2sqrt(10) + 4 OJ^2    Subtract 7 + 2sqrt From both sides
21 + 6 sqrt(10) = 4OJ^2   Divide both sides by 4
21/4 + 6/4* sqrt(10) = OJ^2
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sqrt OJ^2 = sqrt(21/4 + 3/2 sqrt(10) )
OJ = sqrt(21/4 + 3/2 sqrt(10) )

Step three
find h
h = 2 * OJ
h = 2* sqrt(21/4 + 3/2 sqrt(10) ) <<<<<< answer.


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