"integration"
By differentiating the answer and plugging in the x value along the curve, you are finding the exact slope of the curve at that point. In effect, this would be the slope of the tangent line, as a tangent line only intersects another at one point. To find the equation of a tangent line to a curve, use the point slope form (y-y1)=m(x-x1), m being the slope. Use the differential to find the slope and use the point on the curve to plug in for (x1, y1).
The answer is 32 you find the answer by simply multiplying 8*4. Hope this helps your problem!
The main utility of a cumulative frequency curve is to show the distribution of the data points and its skew. It can be used to find the median, the upper and lower quartiles, and the range of the data.
How do you find the area of an unequal quadrangle?
"integration"
Integration.
if its the mymaths one type integration into the box!
integral.
37.6
if this is for www.mymaths.co.uk, integration is all you need. best regards, lokilotus
In statistics you can find the area under a curve to establish what to expect between two input numbers. If there is a lot of area under the curve the graph is tall and there is a higher probability of things occurring there than when the graph is low.
well, it would help to clarify your question, but integrate the fuction and solve on the interval given
(3ab*pi)
To find the position from a velocity-vs-time graph, you need to calculate the area under the velocity curve. If the velocity is constant, the position can be found by multiplying the velocity by the time. If the velocity is changing, you need to calculate the area under the curve using calculus to determine the position.
The area under N(0,1) from -1 to 1 = 0.6826
To find the displacement from a negative velocity-time graph, you need to calculate the area under the curve for the portion representing displacement. If the velocity is negative, the displacement will be in the opposite direction. The magnitude of the displacement is equal to the absolute value of the area under the curve.