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:
8 - 2n > 18
Step-by-step explanation:
8 less than 2 * a number so represent it with a variable 8 - 2*n
then more than so > 18
Step-by-step explanation:
47kg of potatoes is $517
That means 1kg of potatoes is 517/47 = $11.
$154 / $11 = 14.
Hence with $154, you can get 14 kilograms of potatoes.
Answer:
15.2
Step-by-step explanation:
Well so your finding the hypotnuse or “c”
You are already given A and B so add em up and follow the theroum
6^2+14^2=c^2
36+196=c^2
232=c^2
Find C by square rotting 232
√232= 15.2 rounded to the nearest tenths