How do I find the average rate of change?
The average rate is:[math]\qquad\qquad\qquad\qquad\qquad\displaystyle\frac{\Delta y }{\Delta x} = \frac{24}{6} = 4[/math]In general:[math]\qquad\qquad\qquad\qquad\qquad\displaystyle\frac{\Delta y }{\Delta x} = \frac{f(b) - f(a)}{b - a}[/math]The formula you mentioned was correct, you probably made a simple arithmetic error.
Find rate of change of price!?
p(x)=1024/x+1000=1024x^-1+1000 p(x)=1024x^-1+1000. Differntiate p'(x) = -1024x^-2 Plug in x as 2 p'(x)=256
How can I find the relative rate of change?
Relative rate of change in given by[math] \frac{f’(x)}{f(x)} [/math]If f(x) =[math] x^2 [/math]Then f’(x) = [math]2x [/math]Relative rate of change = [math]\frac{f’(x)}{f(x)}[/math]= [math]\frac{2x}{x^2} [/math]= [math]\frac{2}{x}[/math]It can also be used in the form:[math]\frac{f’(x)}{f(x)} = [ln f(x)]’[/math]If[math] f(x) = x^2 [/math][math][ln f(x)]’[/math]=[math] [ln x^2]’[/math]=[math] [2lnx]’ [/math]= [math]\frac{2}{x} [/math]If [math]f(x) = e^{x}[/math][math]\frac{f’(x)}{f(x)} = [ln f(x)]’[/math]=[math] [ln e^{x}]’ [/math]= [math]x’[/math]=1
How do I find the rate of change?
Rate of change is how much per unit of x, y will change. In other words, it is your slope. Slope is defined as your change in y divided by your change in x.(y1 - y2) / (x1 - x2) Where (x1, y1) are coordinates of one point on your graph and (x2, y2) are the coordinates of another point on the graph. Finding the rate of change will then lead you to the correct answer.Here, you can use two points (1994, 28.1) and (2004, 22.6) as your coordinates for the formula. Once you plug it in you will find that the answer is -0.55.The rate of change here shows that every year the number of recipients dropped that 0.55 million. We know this because it is how much for per unit of x, y will change. Every year, which is a unit of x, y will drop 0.55. Thus, C is the right answer.
Find the rate of change of the amount A with respect to the rate r for the following values of r?
If 800 dollars is invested at an annual interest rate r compounded monthly, the amount in the account at the end of 4 years is given by A=800(1+ 1/12 * r) ^48. Find the rate of change of the amount A with respect to the rate r for the following values of r? r=3 percent: r=6.5 percent:
Find the rate of change of the area of a rectangle...?
Write a function for area with respect to time, and differentiate it with respect to "t". A = length * height A = (6t + 5)(t^1/2) I would use the product rule here... dA/dt = (6t + 5)(1/2)(t^-1/2) + (6)(t^1/2) dA/dt = (6t + 5)/[2*sqrt(t)] + 6*sqrt(t) dA/dt = (6t + 5)/[2*sqrt(t)] + 12t/[2*sqrt(t)] dA/dt = (18t + 5)/[2*sqrt(t)] EDIT: Simplified it into one fraction. EDIT2: If you're wondering what I did at the third dA/dt I multiplied the right term by 2*sqrt(t)/2*sqrt(t)
How do you find an average rate of change in a parabola?
You can start with the typical approach to finding slope for a line: start with points you know satisfy the equation (pick a value of x and solve for y, and do this for 3 points. Using two of these points x1, y1) and (x2, y2), the slope is (y2 - y1)/(x2 - x1). Then do the same calculation for (x2, y2) and (x3, y3). Slope is (y3 - y2)/(x3 - x2). Take these two slopes that you calculated and take the average of the two.
How would you find the rate of change of momentum, if we also assumed the mass was changing (like in a rocket)?
You would need to use calculus…differential equations. That is the branch of mathematics that deals with those kinds of problems were there is continuous and non-linear change.
Use the demand function to find the rate of change in the demand x for the given price p.?
x = 200 − p − 2p/(p + 1) dx/dp = -1 − 2/(p + 1)² For p = 3 dx/dp = -1 − 2/(3 + 1)² = --1.125
How would I find the rate of change with time of the voltage across a capacitor?
You could manually figure it Capacitor Charge and Time Constant CalculatorOr by using an oscilloscope and a square wave generator, How to measure a capacitor with an oscilloscope.