Answer:
16Km due east of school P
Step-by-step explanation:
Given
A school P is 16km due west of a school Q
Thus, we can say that distance PQ = 16 km.
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now we have to find the bearing of Q from P
As distance is same
distance PQ = distance QP
Thus,
Distance will remain same of 16 km.
For direction,
If Q is west of P, then P will be east of Q
as shown in figure P is west of Q,
now from point P , Q is west P.
Thus,
Bearing of School Q from P is 16Km due east of school P
Answer:
AB/DE=BC/EF=AC/DF
Step-by-step explanation:
Corresponding segments are designated by letters in corresponding positions in the triangle names. For example, of one segment is designated using the 1st and 2nd letters of one triangle name (such as AB), then the corresponding segment is designated using the 1st and 2nd letters of the other triangle name (such as DE).
Corresponding segments are proportional in length. Corresponding angles are congruent.
speed = 23000* 3.14= 72220 inches/ minute
* I don't have a phone, so I am not able to show my work