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heureka

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 #6
avatar+26400 
+3

I think it goes through the points (2.75,2.75) (1.75,5) and (5,1.75)

How do I find a graph that will go through these points ?? 

 

The formula of a hyperbola is given in general form: y=ax+c+d

 

Set P1(x1,y1)=(2.75,2.75)
Set P2(x2,y2)=(1.75,5)
Set P3(x3,y3)=(5,1.75)

 

(1)y1=ax1+c+d(y1d)(x1+c)=ay1x1+cy1dx1=a+cd(2)y2=ax2+c+d(y2d)(x2+c)=ay2x2+cy2dx2=a+cd(3)y3=ax3+c+d(y3d)(x3+c)=ay3x3+cy3dx3=a+cd

 

(1)y1x1+cy1dx1=a+cd(2)y2x2+cy2dx2=a+cd(3)y3x3+cy3dx3=a+cd(1)(2)y1x1+cy1dx1(y2x2+cy2dx2)=0c(y1y2)+d(x2x1)=y2x2y1x1(1)(3)y1x1+cy1dx1(y3x3+cy3dx3)=0c(y1y3)+d(x3x1)=y3x3y1x1

 

(1)c(y1y2)+d(x2x1)=y2x2y1x1(2)c(y1y3)+d(x3x1)=y3x3y1x1

 

c=|y2x2y1x1x2x1y3x3y1x1x3x1||y1y2x2x1y1y3x3x1|c=|51.752.752.751.752.751.7552.752.7552.75||2.7551.752.752.751.7552.75|c=0.95

 

d=|y1y2y2x2y1x1y1y3y3x3y1x1||y1y2x2x1y1y3x3x1|d=|2.75551.752.752.752.751.751.7552.752.75||2.7551.752.752.751.7552.75|d=0.95

 

a=y1x1+cy1dx1cda=2.752.750.952.750.952.75(0.95)0.95a=3.24

 

The formula of a hyperbola is given through  P1,P2,P3: y=3.24x0.95+0.95

 

The graph is:

 

laugh

27.06.2017
 #3
avatar+26400 
+1

Suppose that for some a,b,c we have

a+b+c = 6,

ab + ac + bc = 5, and

abc = -12.

What is a^3 + b^3 + c^3?

 

1. 

(a+b+c)×(ab+ac+bc)=65(a+b+c)×(ab+ac+bc)=30a2b+a2c+abc+ab2+abc+b2c+abc+ac2+bc2=30a2(b+c)+b2(a+c)+c2(a+b)+3abc=30a2(b+c)+b2(a+c)+c2(a+b)=303abc|abc=12a2(b+c)+b2(a+c)+c2(a+b)=303(12)a2(b+c)+b2(a+c)+c2(a+b)=30+36a2(b+c)+b2(a+c)+c2(a+b)=66

 

2.

(a+b+c)3=(a+b+c)2×(a+b+c)63=(a2+b2+c2+2(ab+ac+bc))×(a+b+c)|ab+ac+bc=5216=(a2+b2+c2+25)×(a+b+c)216=(a2+b2+c2+10)×(a+b+c)216=a3+a2(b+c)+b3+b2(a+c)+c3+c2(a+b)+10(a+b+c)216=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)+10(a+b+c)|a+b+c=6216=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)+106216=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)+6021660=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)156=a3+b3+c3+a2(b+c)+b2(a+c)+c2(a+b)=66156=a3+b3+c3+6615666=a3+b3+c390=a3+b3+c3a3+b3+c3=90

 

laugh

20.06.2017
 #2
avatar+26400 
+1

integral( (x+4) / (x^2-5x+6) ) dx

 

x+4x25x+6 dx|x25x+6=(x3)(x2)=x+4(x3)(x2) dx

 

Partial fraction decomposition:

x+4(x3)(x2)=Ax3+Bx2x+4=A(x2)+B(x3)x=2:2+4=A(22)+B(23)6=0+B(1)6=B(1)B=6x=3:3+4=A(32)+B(33)7=A(1)+B(0)7=A(1)A=7x+4(x3)(x2)=7x36x2

 

x+4x25x+6 dx=x+4(x3)(x2) dx=(7x36x2) dx=71x3 dx61x2 dx=7ln(|x3|)6ln(|x2|)+c

 

laugh

16.06.2017