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:
4x2–12x+33=0
Resta 4x
2
en los dos lados. Cualquier valor restado de cero da como resultado su valor negativo.
−12x+33=−4x
2
Resta 33 en los dos lados.
−12x=−4x
2
−33
Divide los dos lados por −12.
−12
−12x
=
−12
−4x
2
−33
Al dividir por −12, se deshace la multiplicación por −12.
x=
−12
−4x
2
−33
Divide −4x
2
−33 por −12.
x=
3
x
2
+
4
11
Resolver para x_2
x
2
=3x−
4
33
Step-by-step explanation:
Answer:
I believe it's A)
Step-by-step explanation:
5/2 or 2.5
Hope this helps!
