true
So that the arc is mid-way in perpendicular to the line segment
14.53 cm ======= the figure must be a square in order to have that line to a minimum value. then, if the area is 169, every side is 13 cm then, we use pytagoras: c2 = a2 + b2 a = 13 ; b = 13/2 solving for c c=14.53 Edit- I really do not have the time right now to work this problem out. However, I will say that the answer above is far from correct. First, the question does not ask from corner to corner, which would require the response above, rather it asks for the line to be drawn from one corner to the midpoint of one of the two more distant sides. Because it is the midpoint, you would have to create the area equation solve for a variable. Then create the Pythagorean theorem equation and find the derivative. Then plug in the value of the variable solved for in the area equation. Now you have to take this and find the critical points. After that plug them in to figure out which is the minimum value and you have your answer!
The line must be solid if the inequality is strict (less than or greater than). It must be a dashed line if otherwise (less than or equal to, greater than or equal to).
It could be. In addition to a Midpoint Rectangular Riemann Sum there is also a Left Rectangular Riemann Sum, a Right Rectangular Riemann Sum, and a Trapezoidal Riemann Sum. When you are asked to compute a Riemann Sum, you must choose from the above list depending on the specific question, your teacher's preferences, and/or your own preferences.
Any function of the form f(x) = ax + b, or any relation of the form Ax + By = C.This is the function that forms a line graphed. The slope of line can be taken out as C/A. * * * * * The above answer assumes that a line MUST be a straight line! Since the graph is a line, the domain must be an interval in the Real numbers. The interval may be finite, or infinite in one or both directions. In order that the graph does not have breaks in it the function must be continuous. Any such function will do.
It bisects the line segment at midpoint at 90 degrees and its slope is the reciprocal of the line segment's slope plus or minus.
No. Since a line is infinite, it has no mid-point. A bisector must go through a midpoint so nothing can bisect a line (not even a segment).
Yes
Equilateral triangles
With a straight-edge and a compass:Swing arcs from each end of the segment with the compass (without changing the settings)Connect the intersections of these arcs.The resultant is a perpendicular bisector of the segment.
That means that it is not a line segment.
A line segment can be defined as having two endpoints
A line of symmetry must go from one vertex to the midpoint of the opposite side.
Can't be sure what you're asking, but it would be two circles with equal radii longer that half the length of the segment, using the endpoints as the origins of the two circles.
The perpendicular bisector of the line XY will meet it at its midpoint at right angles.
A segment has two end points. If a line has one end point, then it must be called ray.
Any line that enters a circle (and is not a tangent) must cross its boundary twice; once to enter, once to exit. Since a secant is a line segment that joins two different points on a curve, such a line as above is a secant.