Suggested: show that the circles x2+y2-6x-2y+1=0 and x2+y2+2x-8y+13=0 - x^3+6x^2y+11xy^2+6y^3 - solve 6x+2y-5z=13 3x+3y-2z=13 7x+5y-3z=26 - solve by cramer's rule 5x-7y+z=11 6x-8y-z=15 3x+2y-6z=7 - (6x-y 13)=(1 3x+2y) - solve by cross multiplication method 6x+7y-11=0 5x+2y=13 - dy/dx=2x+9y-20/6x+2y-10 - 6x-2y=10 - (6x-4y+1)dy-(3x-2y+1)dx=0 - find the value of 6x-5+4y when 3x+2y=1 - 2x+4y=3 12y+6x=6 - solve (2x+1)2y’’-2(2x+1)y’-12y=6x+5 - show that the line 5x+12y-4=0 touches the circle x2+y2-6x+4y+12=0 - 12y=6x Browse related:
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