Answer:
(x, y) = (-0.6, 0.8) or (1, 4)
Step-by-step explanation:
Use the second equation to substitute for y in the first.
(x -1)² +((2x +2) -2)² = 4
x -2x +1 + 4x² = 4 . . . . . . . eliminate parentheses
5x² -2x -3 = 0 . . . . . . . . . . subtract 4, collect terms
Now we can rearrange the middle term to ease factoring by grouping.
(5x² -5x) +(3x -3) = 0
5x(x -1) +3(x -1) = 0
(5x +3)(x -1) = 0
The values of x that make these factors zero are ...
x = -3/5, x = 1
The corresponding values of y are ...
y = 2(-3/5)+2 = 4/5 . . . . for x = -3/5
y = 2(1) +2 = 4 . . . . . . . . for x = 1
The solutions are: (x, y) = (-3/5, 4/5) or (1, 4).
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A graphing calculator verifies these solutions.
She ran 5 miles last month
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)
I believe the diameter would be 12
d=2r
