Answer:
A) x ≤ -2 and 0 ≤ x ≤ 3
Step-by-step explanation:
g(x) is decreasing when g'(x) is negative.
Use second fundamental theorem of calculus to find g'(x).
g(x) = ∫₋₁ˣ (t³ − t² − 6t) / √(t² + 7) dt
g'(x) = (x³ − x² − 6x) / √(x² + 7) (1)
To find when g'(x) is negative, first find where it is 0.
0 = (x³ − x² − 6x) / √(x² + 7)
0 = x³ − x² − 6x
0 = x (x² − x − 6)
0 = x (x − 3) (x + 2)
x = -2, 0, or 3
Check the intervals before and after each zero.
x < -2, g'(x) < 0
-2 < x < 0, g'(x) > 0
0 < x < 3, g'(x) < 0
3 < x, g'(x) > 0
g(x) is decreasing on the intervals x ≤ -2 and 0 ≤ x ≤ 3.
The statement A is true since on forming the quadratic equations formed you will find the cost of a message being 15 cents and that of anytime as 10 cents.
Answer:y=Mx+b
Step-by-step explanation:
The rule is that :
B. The product of the means is equal to the product of the extremes
Answer:
Step-by-step explanation:
Given a curve defined by the function 2x²+3y²−4xy=36
The total differential of this function with respect to a variable x makes the function an implicit function because it contains two variables.
Differentiating both sides of the equation with respect to x we have:
4x+6ydy/dx-(4xd(y)/dx+{d(4x)/dx(y))} = 0
4x + 6ydy/dx -(4xdy/dx +4y) = 0
4x + 6ydy/dx - 4xdy/dx -4y = 0
Collecting like terms
4x-4y+6ydy/dx - 4xdy/dx = 0
4x-4y+(6y-4x)dy/dx = 0
4x-4y = -(6y-4x)dy/dx
4y-4x = (6y-4x)dy/dx
dy/dx = (4y-4x)/6y-4x
dy/dx = 2(2y-2x)/2(3y-2x)
dy/dx = 2y-2x/3y-2x proved!
