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:
3
Step-by-step explanation:
To turn the word problem into an equation, when we read:
"I am thinking of a number" we write "x"
"when I double my number" we write "2x"
"and then subtract the result from 5" we write "5 - 2x"
"I get negative one" we write "-1 = 5 - 2x"
Now we solve for the number, which is x.
<em>Equation: </em>-1 = 5 - 2x
-1 = 5 - 2x
subtract 5 from both sides
-6 = -2x
divide both sides by -2
3 = x
<h3>
There we go! The number is </h3><h2>
3</h2>
Answer:
y = 3/4x + 19/2
Step-by-step explanation:
m = 3/4. point (-6,5)
Point slope form:
y-5=3/4(x- -6)
y -5 = 3/4x + 18/4. 18/4 = 9/2
Y = 3/4x + 9/2 + 5. 5 times 2 plus 9 over 2 = 19/2
y = 3/4x + 19/2
Using a table of values, the outputs of f(x) for whole numbers are 0, 1, 4, 9, 16, 25, 36, and so on. For the same input values, g(x) has outputs of 1, 2, 4, 8, 16, 32, and 64. Continuing to double the output each time results in larger outputs than those of f(x). The exponential function, g(x), has a constant multiplicative rate of change and will increase at a faster rate than the quadratic function.
(ed. just click all of them)