The correct structure of the question is as follows:
The function f(x) = x^3 describes a cube's volume, f(x) in cubic inches, whose length, width, and height each measures x inches. If x is changing, find the (instantaneous) rate of change of the volume with respect to x at the moment when x = 3 inches.
Answer:
Step-by-step explanation:
Given that:
f(x) = x^3
Then;
V = x^3
The rate whereby V is changing with respect to time is can be determined by taking the differentiation of V
dV/dx = 3x^2
Now, at the moment when x = 3;
dV/dx = 3(3)^2
dV/dx = 3(9)
dV/dx = 27 cubic inch per inch
Suppose it is at the moment when x = 9
Then;
dV/dx = 3(9)^2
dV/dx = 3(81)
dV/dx = 243 cubic inch per inch
Answer:
x = 2 ± (i)√5 (Answer b)
Step-by-step explanation:
x²-4x+9=0 can be solved in a variety of ways; the first two that come to mind that are also appropriate are (1) completing the square and (2) using the quadratic formula.
Completing the square is fast here:
Rewrite x²-4x+9=0 as x²-4x +9=0
Identify the coefficient of the x term: it is -4
Take half of that, obtaining -2
Square this result, obtaining 4
Add 4 to x²-4x +9=0, in the blank space in the middle, and then subtract 4: x²-4x +4 -4 +9=0
Rewrite x²-4x +4 as the square of a binomial:
(x - 2)² - 4 + 9 = 0 → (x - 2)² = -5
Take the square root of both sides: x - 2 = ±√(-5) = ± (i)√5
Then x = 2 ± (i)√5
X and y =4? is there more information about the question?
