Question 64227
1.) The graph of -4x+2y=6 crosses the x-axis when x = ?
Let y=0 and solve for x to find this:
-4x+2(0)=6
-4x+0=6
-4x/-4=6/-4
{{{highlight(x=-3/2)}}}
2.) Find the center and the radius from the equation of the circle in standard form (x-5)^2+(y+3)^2
Standard form of a circle is:{{{(x-h)^2+(y-k)^2=r^2}}}, where (h,k)=center, and {{{sqrt(r^2)}}}=radius.
You forgot to include the =r^2 part so I can't tell you what the radius is, but the center (h,k)=(5,-3)
:

3.) The formula for the area of a triangle is A=1/2bh. If A=20 and h=b-6, then what does b= ?
20=(1/2)b(b-6)
2(20)=2(1/2)b(b-6)
40=b(b-6)
40=b^2-6b
40-40=b^2-6b-40
0=b^2-6b-40
0=(b-10)(b+4)
b-10=0 or b+4=0
b-10+10=0+10  or b+4-4=0-4
b=10 or b=-4
Since you can't have a negative dimension, we ignore -4 and say that {{{highlight(b=10)}}}.
:
4. The solution set of the equation {{{4/(x+2)-2/(x-4)=-6/(x^2-2x-8)}}} is ?
{{{4/(x+2)-2/(x-4)=-6/((x-4)(x+2))}}}  The LCD is (x+2)(x-4), multiply evertyhing by the LCD to clear your fractions.
{{{4(x+2)(x-4)/(x+2)-2(x+2)(x-4)/(x-4)=-6(x+2)(x-4)/((x+2)(x-4))}}}
{{{4*cross((x+2))(x-4)/cross((x+2))-2(x+2)*cross((x-4))/cross((x-4))=-6*cross((x+2)(x-4))/cross((x+2)(x-4))}}}
{{{4(x-4)-2(x+2)=-6}}}
{{{4x-16-2x-4=-6}}}
{{{2x-20=-6}}}
{{{2x-20+20=-6+20}}}
{{{2x=14}}}
{{{2x/2=14/2}}}
{{{highlight(x=7)}}}
Double check to make sure that the equation is defined at x=7 (it is), because you can get extraneous solutions with rational equations.  If the solution had of been x=-2 or x=4, we would have had to reject it because it would have made one of the fractions undefined.
Happy Calculating!!!!