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

let's solve your equation step-by-step.

24−3x=−27

step 1: simplify both sides of the equation.

24−3x=−27

24+−3x=−27

−3x+24=−27

step 2: subtract 24 from both sides.

−3x+24−24=−27−24

−3x=−51

step 3: divide both sides by -3.

−3x

−3

=

−51

−3

x=17

answer:

x=17

step-by-step explanation:

mark brainliest and have a great day!

step-by-step explanation: