At point of intersection the two equations are equal,
hence, 6x³ =6x²
6x³-6x²=0
6x²(x-1)=0 , the values of x are 0 and 1
The points of intersection are therefore, (0,0) and (1,6)
To find the slopes of the tangents at the points of intersection we find dy/dx
for curve 1, dy/dx=12x, and the other curve dy/dx=18x²
At x=0, dy/dx=12x =0, dy/dx=18x² = 0, hence the angle between the tangents is 0, because the tangents to the two curves have the same slope which is 0 and pass the same point (0,0) origin.
At x=1, dy/dx =12x = 12, dy/dx= 18x² =18, Hence the angle between the two tangents will be given by arctan 18 -arctan 12
= 86.8202 - 85.2364 ≈ 1.5838, because the slope of the lines is equal to tan α where α is the angle of inclination of the line.
Answer:
b
Step-by-step explanation:
Answer:
f(g(x)) = 9x^2 + 15x - 6
Step-by-step explanation:
We are using function g(x) = 3x - 1 as the input to function f(x) = x^2 + 7x.
Starting with f(x) = x^2 + 7x, substitute g(x) for x on the left side and likewise substitute x^2 + 7x for each x on the right side. We obtain:
f(g(x)) = (3x - 1)^2 + 7(3x - 1).
If we multiply this out, we get:
f(g(x)) = 9x^2 - 6x + 1 + 21x - 7, or
f(g(x)) = 9x^2 + 15x - 6
|PR| = |PQ| + |QR|; |PQ| = |QR| conclusion |PR| = 2|PQ|
|PQ| = 3y; |PR| = 42; |QR|=?
subtitute
42 = 2(3y)
6y = 42 |divide both sides by 6
y = 7
|QR| = 3(7) = 21
-9.1
It all rounds up to the tenth place once you try to round to the hundredth's place
