<span>A line segment LM is drawn. Two arcs equidistant from L on line LM are drawn at points P and T. A point Q is just above L. A compass is fixed at point Q and is shown making an arc at T.
</span>You draw arcs centered at P and T, such that they intersect above and below L. Note that the arcs must have radius greater than LT=LP.
<span>I get the feeling that Q is where the arcs intersect above L, so the intersection below L (at, say, S) means that QS is perpendicular to LM. </span>
<span>Incidentally, constructing LM does not really fit the bill to start with, since either P or T must be beyond the line segment. What it should say is that points L and M are marked on a line extending beyond LM.</span>
Step-by-step explanation:
I hope this is right, I used demos
Answer:
48y⁶w¹¹
Step-by-step explanation:
[-3y⁵][4y] → -12y⁶
×
[w³][-4w⁸] → -4w¹¹
______
48y⁶w¹¹
I am joyous to assist you anytime.
A because 180-130=50 u have to add 35 n 95 n subtract from 180!
Answer:
(5,0) is the only solution out of those points
Step-by-step explanation:
If you graph the 2 inequalities you'll notice the point (0,-2) falls on the dotted line, since the line is dotted this can not be a solution as a dotted line entails that anything on that line is not a solution, for the point (3,13) it doesn't fall in a solution for either inequality due to what I just stated, the point (5,0) on the other hand falls in the domain and range of both inequalities.
