I assume you are asking about using differentiation. In this case, where dy-by-dx=0, there is a stationary point on the graph i.e. where the gradient is equal to 0.
unsquaring a number is to chage it back
To find the average rate of change of the function from ( x = 0 ) to ( x = 4 ), you can use the formula: [ \text{Average Rate of Change} = \frac{f(b) - f(a)}{b - a} ] Here, ( f(0) = 4 ) and ( f(4) = 4 ). Thus, the average rate of change is: [ \frac{4 - 4}{4 - 0} = \frac{0}{4} = 0 ] Therefore, the average rate of change from ( x = 0 ) to ( x = 4 ) is 0.
with y=mx+b dy/dx=m d^2.y/dx^2=0 The rate of change is 0
The instantaneous rate change of the variable y with respect to x must be the slope of the line at the point represented by that instant. However, the rate of change of x, with respect to y will be different [it will be the x/y slope, not the y/x slope]. It will be the reciprocal of the slope of the line. Also, if you have a time-distance graph the slope is the rate of chage of distance, ie speed. But, there is also the rate of change of speed - the acceleration - which is not DIRECTLY related to the slope. It is the rate at which the slope changes! So the answer, in normal circumstances, is no: they are the same. But you can define situations where they can be different.
0. Differentiation of a constant gives f'(x)=0.
Rate of Change
i need chage the clutch of my car
You have to reach around and grab the chage, and twist.
"Chage and Aska" is a popular Japanese duo composed of Chage and Ryo Aska. They were formed in 1979 and are still active to this date, with over 31 million albums and singles sold in Japan.
you cant
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he was the leader of India
home page
Because of the gold rush.
Italy
yes
no