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Step-by-step explanation:
<em>Given: Side lengths of a right triangle 3,4 and 5 units. </em>
<em> To draw: A right triangle with the given side length. </em>
<em> Solution: </em>
<em> We know, in a right angle triangle hypotenuse is the longest side and satisfying Pythagoras theorem. </em>
<em> From the given side length, </em>
<em> Hypotenuse = 5 unit </em>
<em> We can take any of the base and perpendicular. </em>
<em> Let, Base = 3 unit </em>
<em> Perpendicular = 4 unit. </em>
<em> It a right-angle triangle with a hypotenuse 5 unit. </em>
<em> Now we draw a right angle triangle taking in the first 3 base and 4 perpendicular and second 3 perpendicular and 4 bases.</em>
You can go through the effort of determining the zero of the function analytically and evaluating an analytic expression for the derivative at that point, or you can let a graphing calculator do that heavy lifting. Since the numbers have to be "nice" for your equation to have the desired form, it is easy to know what to round to in the event that is necessary (it isn't).
We find the positive zero-crossing at x=2, and the slope of the curve at that point to be 8. Thus the line will have slope -1/8 and can be written as
.. x +8y -2 = 0
We find the positive zero-crossing at x=2, and the slope of the curve at that point to be 8. Thus the line will have slope -1/8 and can be written as
.. x +8y -2 = 0