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Punkte9488
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 #5
avatar+9488 
+4
12.07.2019
 #2
avatar+9488 
+5

2)  This one is tricky but here's my attempt...

 

I started by making this list:

 

b( 1 ) _ = _ 1
b( 2 ) = 1
b( 3 ) = 2
b( 4 ) = 2
b( 5 ) = 2
b( 6 ) = 2
b( 7 ) = 3
b( 8 ) = 3
b( 9 ) = 3
b( 10 ) = 3
b( 11 ) = 3
b( 12 ) = 3
b( 13 ) = 4
b( 14 ) = 4
b( 15 ) = 4
b( -16- ) = 4

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b( 1980 ) = 44
b( 1981 ) = 45
b( 1982 ) = 45
b( 1983 ) = 45
b( 1984 ) = 45
b( 1985 ) = 45
b( 1986 ) = 45
b( 1987 ) = 45
b( 1988 ) = 45
b( 1989 ) = 45
b( 1990 ) = 45
b( 1991 ) = 45
b( 1992 ) = 45
b( 1993 ) = 45
b( 1994 ) = 45
b( 1995 ) = 45
b( 1996 ) = 45
b( 1997 ) = 45
b( 1998 ) = 45
b( 1999 ) = 45
b( 2000 ) = 45
b( 2001 ) = 45
b( 2002 ) = 45
b( 2003 ) = 45
b( 2004 ) = 45
b( 2005 ) = 45
b( 2006 ) = 45
b( 2007 ) = 45

 

By now we can notice something...

 

b( 7 )  =  3   because  7   is closer to  9  than  4,  and
b( 12 )  =  3   because  12   is closer to  9  than  16

 

Between  4  and  9 ,  there are  4  numbers and  4/2  =  2

 

Between  9  and  16  there are  6  numbers and  6/2  =  3

 

So the number of 3's   =  1 + 2 + 3   =   6

 

We can backtrack the previous steps to get...

 

the number of 3's   =   1+42+62 = 1+9412+16912 = 1+32(31)212+(3+1)23212

 

So in general,

 

the number of  n's   =   1+n2(n1)212+(n+1)2n212     which simplifies to...

 

the number of n's   =   2n

 

This agrees with the list from earlier!!! laughlaugh

 

Except the number of  45's  is only  27  because the list ends at  b(2007)

 

2007p=1b(p) = b(1)+b(2)+b(3)+b(4)+b(5)+b(6)+b(7)++b(2006)+b(2007) 2007p=1b(p) = 1+1+2+2+2+2+3++45+45 2007p=1b(p) = 2(1)+4(2)+6(3)++88(44)+27(45) 2007p=1b(p) = 44k=1(2k)(k)+ 27(45) 2007p=1b(p) = 44k=1(2k)(k)+ 27(45) 2007p=1b(p) = 244k=1k2+ 27(45) 2007p=1b(p) = 2(44(44+1)(2(44)+1)6)+ 27(45) 2007p=1b(p) = 58740+ 1215 2007p=1b(p) = 59955

 

 

P.S. There definitely might be a better way to do it... smiley

12.07.2019
 #1
avatar+9488 
+5
07.07.2019
 #1
avatar+9488 
+5

(a)  https://web2.0calc.com/questions/math-halp-plz#r1

 

(b)

 

 

 

 

Let     AB  =  c     (because it is the side across from angle C)

and    AC  =  b     (because it is the side across from angle B)

and    BC  =  a     (because it is the side across from angle A)

 

Draw a height from  M  which meets side  AC  at point  D

 

m∠BAC  =  m∠MAD     because they are the same angle

m∠ACB  =  m∠ADM     because they are both right angles

 

So by AA similarity,  △ABC ~ △AMD

 

And we know   AM  =  c / 2   because  M  is the midpoint of  AB

 

So the scale factor from  △ABC  to  △AMD  is  1/2   And so...

 

AM  =  c / 2

AD  =  b / 2

DM  =  a / 2

 

Then by SAS congruence we can determine that  △ADM  ≅  △CDM   and so...

 

CM  =  AM

CM  =  c / 2

CM  =  (1/2)(AB)

07.07.2019