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heureka

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 #1
avatar+26399 
+1

Find the size of the smallest angle between the two hands of a clock displaying these times.

10:20, 6:55, 4:35, 2:27 and 11:09

 

1. angular speed = ω

minutes hand: ωm=3601 h

hour hand: ωh=36012 h

 

2. angle between the two hands = α

α=(ωmωh)tα=(3601 h36012 h)tα=360(11 h112 h)tα=3601112 ht

 

3. Solution for 10:20, 6:55, 4:35, 2:27 and 11:09

t=10:20=10.ˉ3 hα=3601112 h10.ˉ3 hα=3410α=3410(mod360)α=170t=6:55=6.91ˉ6 hα=3601112 h6.91ˉ6 hα=2282.5α=2282.5(mod360)α=122.5t=4:35=4.58ˉ3 hα=3601112 h4.58ˉ3 hα=1512.5α=1512.5(mod360)α=72.5t=2:27=2.45 hα=3601112 h2.45 hα=808.5α=808.5(mod360)α=88.5t=11:09=11.15 hα=3601112 h11.15 hα=3679.5α=3679.5(mod360)α=79.5

 

 

laugh

15.06.2017
 #1
avatar+26399 
+2

What is the general form of the equation of a circle with its center at (-2, 1) and passing through (-4, 1)?

 

center at (xc=2, yc=1)

passing through point at (xp=4, yp=1)

radius2=r2=(xcxp)2+(ycyp)2

 

general form of the equation of a circle: (xxc)2+(yyc)2=r2

 

so we have:

(xxc)2+(yyc)2=r2|r2=(xcxp)2+(ycyp)2(xxc)2+(yyc)2=(xcxp)2+(ycyp)2x22xcx+x2c+y22ycy+y2c=x2c2xcxp+x2p+y2c2ycyp+y2px22xcx+y22ycy=2xcxp+x2p+2ycyp+y2px22xcx+y22ycy+2xcxpx2p+2ycypy2p=0x2+y22xcx2ycy+2xcxp+2ycypx2py2p=0x2+y22xcx2ycy+xp(2xcxp)+yp(2ycyp)=0

 

The general form of the equation of a circle with its center (xc,yc)and passing through point (xp,yp) is:

x2+y22xcx2ycy+xp(2xcxp)+yp(2ycyp)=0

 

xc=2, yc=1xp=4, yp=1x2+y22xcx2ycy+xp(2xcxp)+yp(2ycyp)=0x2+y22(2)x21y+(4)[2(2)(4)]+1(211)=0x2+y2+4x2y+(4)(4+4)+1(21)=0x2+y2+4x2y+0+11=0x2+y2+4x2y+1=0

 

The equation of the circle is x2+y2+4x2y+1=0

 

laugh

07.06.2017
 #1
avatar+26399 
+6

There is a row of Pascal's triangle that has three successive positive entries,"a" "b"  and "c"

such that "b" is double "c" 

and "a" is triple "c"

If this row begins "1,n,"  then find n.

 

Three successive positive entries:

a=(nk1)b=(nk)c=(nk+1)

 

"b" is double "c" and "a" is triple "c"

a=3c=(nk1)b=2c=(nk)c=(nk+1)

 

(1)2c=(nk)|c=(nk+1)2(nk+1)=(nk)|(nk+1)=(nkk+1)(nk)2(nkk+1)(nk)=(nk)2(nkk+1)=1nk=k+12(2)3c=(nk1)|c=(nk+1)3(nk+1)=(nk1)|(nk+1)=(nkk+1)(nk)3(nkk+1)(nk)=(nk1)|(nk1)=(knk+1)(nk)3(nkk+1)(nk)=(knk+1)(nk)3(nkk+1)=knk+13(nk)(nk+1)=k(k+1)|nk=k+123(k+12)(k+12+1)=k(k+1)3(k+12)(k+32)=k(k+1)34(k+1)(k+3)=k(k+1)34(k+3)=k34k+94=kk34k=9414k=94k=9nk=k+12n9=9+12n9=5n=14

 

The three successive positive entries are:

a=3003=(148)b=2002=(149)c=1001=(1410)

and n is 14.

 

laugh

22.05.2017
 #2
avatar+26399 
+1

The center of a circle is located at (−2, 7) . The radius of the circle is 2.

What is the equation of the circle in general form?

 

A circle can be defined as the locus of all points that satisfy the equation
(xh)2+(yk)2=r2  ( Standard Form )
where r is the radius of the circle,
and h,k are the coordinates of its center.

 

The general Form is:
x2+y2+ax+by+c=0

 

Standard Form to general Form:

(xh)2+(yk)2=r2x22xh+h2+y22yk+k2=r2x2+y2+x(2h)=a+y(2k)=b+h2+k2r2=c=0

 

a,b and c ?

x2+y2+x(2h)=a+y(2k)=b+h2+k2r2=c=0a=2hb=2kc=h2+k2r2

 

If we have h,k and r, we can calculate a,b and c:

x2+y2+ax+by+c=0a=2hb=2kc=h2+k2r2

 

h=2k=7r=2

a=2ha=2(2)a=4b=2kb=2(7)a=14c=h2+k2r2c=(2)2+7222c=4+494c=49x2+y2+ax+by+c=0x2+y2+4x14y+49=0

 

The equation of the circle in general form is: x2+y2+4x14y+49=0

 

laugh

16.05.2017