(Grad point )What is m
1 answer:
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X=750+125.75x
subtract to variable (x) on one side
x-125.75x=750+125.75x-125.75x
-124.75x=750
divide -124.75 to get variable (x) by itself
-124.75x/-124.75=750/-124.75
x=-3000/499
or
x=-6 6/499
or
x=-6.01202
subtract to variable (x) on one side
x-125.75x=750+125.75x-125.75x
-124.75x=750
divide -124.75 to get variable (x) by itself
-124.75x/-124.75=750/-124.75
x=-3000/499
or
x=-6 6/499
or
x=-6.01202
Answer:
When we have a function f(x), the average rate of change in the interval (a, b) is:
In this case, we have the function:
f(x) = (x + 3)^2 - 2
(but we do not have the interval, and I couldn't find the complete question online)
So if for example, we have the interval (2, 4)
The average rate of change will be:
If instead, we want the rate of change in a differential dx around the value x, we need to differentiate the function (this is way more complex, so I will define some rules first).
Such that the rate of change, in this case, will be:
f'(x) = df/dx
For a function like:
g(x) = x^n + c
g'(x) = n*x^(n - 1)
And for:
h(x) = k( g(x))
h'(x) = k'(g(x))*g'(x)
So here we can write our function as:
f(x) = k(g(x)) = (x + 3)^2 - 2
where:
g(x) = x + 3
k(x) = x^2 - 2
Then:
f'(x) = 2*(x + 3)*1 = 2*x + 6
That is the rate of change as a function of x (but is not an "average" rate of change)