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1.Drawn a straight line AB=7 cm with the help of ruler.
2.With the help of compass drawn an arc from A and at the point where it cuts AB from that point made another arc drawn an arc cutting the previous arc.
3.From A drawn a straight line joining the arc and extend it to M.
4.With the help of ruler measured 5 cm and mark it as AC.
5.Joined BC and we get the required triangle.
6.From C drawn an arc and make it cut on AC and BC and from the point it cuts AC and BC drawn arc cutting each other and extend a line from point C extend a line to the point point of intersection of two arc.
7.Similarly we do for A and the point where the two line intersect denoted as O.
8.Made a perpendicular from O on AB this perpendicular will be radius and taking O as centre we draw a circle this is our incircle.
9.And AN is our locus of points equidistant from two lines AB and AC.
We need to construct a circle inscribed in triangle that is incircle it can be done by making angle bisector of two sides the point where it intersect will be incentre. The centre of required circle.
The angle bisector is the locus where points are equidistant from two sides.
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Answer:
see attached
Step-by-step explanation:
One way to approximate the derivative at a point is by finding the slope of the secant line between points on either side. That is what is done in the attached spreadsheet.
f'(0.1) ≈ (f(0.2) -f(0.0))/(0.2 -0.0) = -5 . . . for example
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Another way to approximate the derivative is to write a polynomial function that goes through the points (all, or some subset around the point of interest), and use the derivative of that polynomial function.
These points are reasonably approximated by a cubic polynomial. The derivative of that polynomial at the points of interest is given in the table in the second attachment. (f1 is a rounding of the derivative function f')
