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:
(-3/10,-16/5)
Step-by-step explanation:
One way to solve a system of equations is by elimination. We can take one equation and subtract the other equation from it to cancel out on of the variables making the equation into something we can solve. In this case the Y's will cancel out because we have 1 Y in each equation
-4x+y=-2
6x+y=-5 subtract
-4x-6x=-10x
y-y=0
-2-(-5)=3
so our equation is -10x=3 and we can solve for x giving us x=-3/10
now we can plug -3/10 in for x into either of the original equations and solve for Y
y+5=-6(-3/10)
y+5=9/5
y=-16/5
so our answer is (-3/10,-16/5)
I hope this helps and please don't hesitate to ask if there is anything still unclear!
Before winter, apples used to cost $1.21 per pound, now they cost $1.96 per pound.
Answer:
mass changes,volume remains same,density changes
