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 #1
avatar+26396 
+25

In a triangle ABC, AB = AC. D is the mid-point of BC, E is the foot of the perpendicular drawn from D to AC, and F is the mid-point of DE. Prove that AF is perpendicular to BE.

 

I. Definition:

 a=BC|a|=ab=AC|b|=bbb=b2ab=a2aab=a22d=ADd=b12a  f=AEf=(bd)b2bf=(b(b12a))b2bf=(b2(ab)2)b2bf=(b2a24)b2bf=(1a24b2)b 

 e=EDe=dfe=(b12a)fe2=(b12a)1212fAF=e2+f=(b12a)1212f+fAF=(b12a)12+12fAF=(b12a)12+12(1a24b2)bAF=b[12+12(1a24b2)]14aAF=b(12+12a28b2)14aAF=b(1a28b2)14a  BE=a(bf)BE=ab+fBE=ab+(1a24b2)bBE=ab+b(a24b2)bBE=a(a24b2)b 

 

II. Prove that AF is perpendicular to BE.

 AF=14a+b(1a28b2)BE=ab(a24b2)Perpendicular, if  ( AFBE )=0( AFBE )=(14a+b(1a28b2))(ab(a24b2))( AFBE )=14aa+14ab(a24b2)+ba(1a28b2)bb(1a28b2)(a24b2)aa=a2ab=ba=a22bb=b2( AFBE )=a24+14a22(a24b2)+a22(1a28b2)b2(1a28b2)(a24b2)( AFBE )=a24+a432b2+a22a416b2(1a28b2)(a24)( AFBE )=a24+a432b2+a22a416b2a24+a432b2( AFBE )=a24a24+a22+a432b2+a432b2a416b2( AFBE )=a22+a22+a416b2a416b2=0 

 

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25.11.2015
 #1
avatar+26396 
+15

sin(x+15 degree) = 3 cos( x-15 degree)

 

 Formula: sin(x+y)=sin(x)cos(y)+sin(y)cos(x)cos(xy)=cos(x)cos(y)+sin(x)sin(y) sin(x+15)=3cos(x15)sin(x)cos(15)+sin(15)cos(x)=3[ cos(x)cos(15)+sin(x)sin(15) ]sin(x)cos(15)+sin(15)cos(x)=3cos(x)cos(15)+3sin(x)sin(15)|:cos(x)x90tan(x)cos(15)+sin(15)=3cos(15)+3tan(x)sin(15)tan(x)cos(15)3tan(x)sin(15)=3cos(15)sin(15)|:cos(15)tan(x)3tan(x)tan(15)=3tan(15)tan(x)[ 13tan(15) ]=3tan(15)tan(x)=3tan(15)13tan(15)tan(x)=2.73205080760.1961524227tan(x)=13.9282032303x=85.8933946491±k180kZ

 

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25.11.2015
 #2
avatar+26396 
+23

Will can paint a house in 3 hours. Sam can paint a house in 5 hours. How long will it take for them to paint it together?

 

Will painting per hour13 house Sam painting per hour15 house  13househourt+15househourt=1  house  t(13househour+15househour)=1  house t(13+15)househour=1  house t(1355+1533)househour=1  house t(515+515)househour=1  house t(815)househour=1  house t=158  house hourhouset=158  hour t=1  hour 52.5  minutes 

 

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25.11.2015
 #8
avatar+26396 
+10

The sum is 1+11+111+1111+1111... find the sum of n

 

Sorry blush

New edit, without mistakelaugh:

 

 

a1=1a2=11a3=111a4=1111a1=1a2=1+101a3=1+101+102a4=1+101+102+103an=1+101+102+103+104++10n2+10n1Sn=(n0)100+(n1)101+(n2)102+(n3)103+(n4)104++[n(n2)]10n2+[n(n1)]10n1Sn=n1+n101+n102+n104++n10n2+n10n101001101210231034104(n2)10n2(n1)10n1Sn=n(1+101+102+104++10n2+10n1)geometric seriesnr=1(r1)10r1'Arithmetic-geometric' seriesSn=n(10n1101)nr=1(r1)10r1Sn=n(10n1101)snsn=nr=1(r1)10r1=0100+1101+2102+3103+4104++(n1)10n110sn=0101+1102+2103+3104++(n2)10n1+(n1)10nsn10sn=1101+1102+1103++110n2+110n1geometric series(n1)10n9sn=10(10n11101)(n1)10n9sn=10(10n119)(n1)10nsn=10(10n1199)(n1)10n9Sn=n(10n19)sn Sn=n(10n19)+10(10n1199)(n1)10n9or Sn=181[ 10(10n1)9n ]Example: S4=181[ 10(1041)94 ]S4=181[ 10(100001)36 ]S4=181[ 10(9999)36 ]S4=181[ 9999036 ]S4=181[ 99954 ]S4=1234 

 

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24.11.2015
 #7
avatar+26396 
+11

The sum is 1+11+111+1111+1111... find the sum of n

 

a1=1a2=11a3=111a4=1111a1=1a2=1+101a3=1+101+102a4=1+101+102+103an=1+101+102+103+104++10n2+10n1Sn=(n0)100+(n1)101+(n2)102+(n3)103+(n4)104++[n(n2)]10n2+[n(n1)]10n1Sn=n1+n101+n102+n104++n10n2+n10n101001101210231034104(n2)10n2(n1)10n1Sn=n(1+101+102+104++10n2+10n1)geometric seriesnr=1(r1)10r1'Arithmetic-geometric' seriesSn=n(10n11101)nr=1(r1)10r1Sn=n(10n11101)snsn=nr=1(r1)10r1=0100+1101+2102+3103+4104++(n1)10n110sn=0101+1102+2103+3104++(n2)10n1+(n1)10nsn10sn=1101+1102+1103++110n2+110n1geometric series(n1)10n9sn=10(10n11101)(n1)10n9sn=10(10n119)(n1)10nsn=10(10n1199)(n1)10n9Sn=n(10n119)sn Sn=n(10n119)+10(10n1199)(n1)10n9 

 

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24.11.2015