The slope of the tangent line to the curve x^3y+y^2-x^2=5 at the point (2,1) is Enter any fractions in reduced for as a/b.

-4/5

Step-by-step explanation:

To find the slope of the tangent to the equation at any point we must differentiate the equation.

x^3y+y^2-x^2=5

3x^2y+x^3y'+2yy'-2x=0

Gather terms with y' on one side and terms without on opposing side.

x^3y'+2yy'=2x-3x^2y

Factor left side

y'(x^3+2y)=2x-3x^2y

Divide both sides by (x^3+2y)

y'=(2x-3x^2y)/(x^3+2y)

y' is the slope any tangent to the given equation at point (x,y).

Plug in (2,1):

y'=(2(2)-3(2)^2(1))/((2)^3+2(1))

Simplify:

y'=(4-12)/(8+2)

y'=-8/10

y'=-4/5

point a is the center of this circle.

the ratio of the lengths of (ef/bc) and (ad/ih) is 2: 1.

step-by-step explanation:

c

step-by-step explanation: