Answer:
-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
Answer:
128/2=64/2=32
Step-by-step explanation:
you really need a help with that ? you don't know how to divide by 2 ? and then to divide the result another time by 2 ?
not to mention that you could achieve that with in division by 4 (as this is the save as dividing by 2 and then again by 2).
putting such questions in here is a farce and a waste of time for the helping people.
The answer for this problem is 4
