Answer:
I love algebra anyways
the ans is in the picture with the steps
(hope it helps can i plz have brainlist :D hehe)
Step-by-step explanation:
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:
A. 5.8
Step-by-step explanation:
From the given graph d has two coordinates;
the first coordinates of d = (0, 0)
the second coordinate of d = (-3 , -5), that is x = -3 when traced up and y = -5 when traced horizontal
The distance between the two coordinates = distance of d;

Option A is correct
Answer:
√
15 +
√
10
Step-by-step explanation:
The answer is above
Subtract 6 from both sides, you get 1/3x=-14
Now multiply both sides by there, hence x=-42