Answer:
3.
Step-by-step explanation:
Implicit differentiation:
x^2 y + (xy)^3 + 3x = 0
x^2 y + x^3y^3 + 3x = 0
Using the product rule:
2x* y + x^2*dy/dx + 3x^2 y^3 + x^3* (d(y^3)/dx) + 3 = 0
2xy + x^2 dy/dx + 3x^2 y^3 + x^3* 3y^2 dy/dx + 3 = 0
dy/dx(x^2 + 3y^2x^3) = (-2xy - 3x^2y^3 - 3)
dy/dx= (-2xy - 3x^2y^3 - 3) / (x^2 + 3y^2x^3)
At the point (-1, 3).
the derivative = (6 - 81 - 3)/(1 -27)
= -78/-26
= 3.
Yup just answered this question. Lol
Answer: y =
x + 8
Step-by-step explanation:
This is asking us to remove the parameter. In other words, we want an equation with only the relation between x and y, so we need to remove the t. There are a few ways to do this, but I am going to set one equation equal to t and then plug it into the next one.
Given:
x = 5t
Divide both sides of the equation by 5:
t =
x
-
Given:
y = t + 8
Plug in:
y = (
x) + 8
Answer:
y =
x + 8
Answer:
x = 4 or x = -1
Step-by-step explanation:
2x^2 - 8x + 6 =0
divide by 2 we get : x^2 -4x +3 = 0
(x -4) (x+1)
x = 4 or x = -1