Answer:
Mark point E where the circle intersects segment BC
Step-by-step explanation:
Apparently, Bill is using "technology" to perform the same steps that he would use with compass and straightedge. Those steps involve finding a point equidistant from the rays BD and BC. That is generally done by finding the intersection point(s) of circles centered at D and "E", where "E" is the intersection point of the circle B with segment BC.
Bill's next step is to mark point E, so he can use it as the center of one of the circles just described.
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<em>Comment on Bill's "technology"</em>
In the technology I would use for this purpose, the next step would be "select the angle bisector tool."
Answer:
Yes
Step-by-step explanation:
ΔMNL ≅ ΔQNL by ASA or AAS
by ASA
Proof:
∠ LNM = ∠LNQ =90
LN = LN {Common}
∠MLN = ∠QLN {LN bisects ∠ L}
By AAS
∠Q + ∠QLN + ∠LNQ = 180 {Angle sum property of triangle}
∠Q + 32 + 90 = 180
∠Q + 122 = 180
∠Q = 180 -122 =
∠Q = 58
∠Q = ∠M
∠MNL =∠QNL = 90
LN = LN {common side}
7 × __ = 94
Divide by 7 on both side.
__ = 94 ÷ 7
Answer = 13 3/7
Answer:
(3, 5)
Step-by-step explanation:
Point (3, -5) is 5 units below the x-axis.
After it is reflected, it will be 5 units above the x-axis.
Answer: (3, 5)
Answer:I'm pretty sure it would be C but that might be wrong
Step-by-step explanation:
