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
-4 for the first one and 2/3 for the second
Answer:
-x > -5
Step-by-step explanation:
Let's start with a positive number less than 5.
4<5
likewise when applied to -x> -5
-4>-5
Answer:
14
Step-by-step explanation:
THE ANSWER TO THE PROBLEM IS 14
Answer:
C
Step-by-step explanation:
If you add the first equation up you get 2z+6 by combining like terms
if you see A a is equal to 3z+6 so a doesn’t work
If you look at B b is equal to z+12 so b doesn’t work
finally C if you distibute it out you get 2(z)+2(3) which equal 2z+6 which works
C
