(6x^3 + 4x-2) + (4x^2 + 3x^3 - 6)=
6x^3 + 4x-2 + 4x^2 + 3x^3 - 6=
9x^3+4x^2+4x-8
Answer:
f^-1 (x) = (1 - x)/3.
Step-by-step explanation:
f(x) = -3x + 1
3x = 1 - f(x)
x = (1 - f(x)) / 3
So f^-1 x = (1 - x)/3.
Distance formula = sqrt[(x1 - x2)^2 + (y2-y1)^2]
D = sqrt[(1 + 2)^2 + (5 - 1)^2]
D = sqrt[3^2 + 4^2]
D = sqrt(9 + 16)
D = sqrt(25)
D = 5
Positive sine, negative tangent, means we have a negative cosine. We're talking about the second quadrant.
We know it's negative,
Answer: -(1/3)√5
Answer:
linear
Step-by-step explanation:
The easiest (if not the only) way to solve this, is to check whether the slope at each point stays constant, or not. And in general, see how the rate rate of change evolves. In other words:
- 6.5 / 5 = 1.3
- 13 / 10 = 1.3
- 19.5 / 15 = 1.3
- 26 / 20 = 1.3
- 32.5 / 25 = 1.3
- 39 / 30 = 1.3
As you can see " in every step to the right, the function goes up by 1.3 " --> constant
So, the function is linear
