Answer:
I think that is going to be x-3
Please note that your x^3/4 is ambiguous. Did you mean (x^3) divided by 4
or did you mean x to the power (3/4)? I will assume you meant the first, not the second. Please use the "^" symbol to denote exponentiation.
If we have a function f(x) and its derivative f'(x), and a particular x value (c) at which to begin, then the linearization of the function f(x) is
f(x) approx. equal to [f '(c)]x + f(c)].
Here a = c = 81.
Thus, the linearization of the given function at a = c = 81 is
f(x) (approx. equal to) 3(81^2)/4 + [81^3]/4
Note that f '(c) is the slope of the line and is equal to (3/4)(81^2), and f(c) is the function value at x=c, or (81^3)/4.
What is the linearization of f(x) = (x^3)/4, if c = a = 81?
It will be f(x) (approx. equal to)
Answer:
The answer is below
Step-by-step explanation:
a) If O(x, y) is the midpoint between two points X(x₁, y₁) and Y(x₂, y₂)m then the coordinates of O is:
x = (x₁ + x₂)/2; y = (y₁ + y₂) / 2
Since M(x, y) is the midpoint of AB, then the coordinates of M is:
x = (0 + 5) / 2 = 2.5; y = (1 + 0)/2 = 0.5
M = (2.5, 0.5)
b) The center of gravity is gotten by finding the average of the x coordinates and the y coordinates of the vertices of the triangle.
Let O(x, y) be the center of gravity of triangle ABC, hence;
x = (0 + 5 + 4) / 3 = 3
y = (1 + 0 + 2) / 3 = 1
O = (3, 1)
Answer:
i think it c since it is involving fractions
Step-by-step explanation:
