You can solve the median by putting all the numbers in order. Then cross off one at a time at the start, then cross off the one at the end, then keep going until you get to one.
You can solve the mode by looking for the most frequent number. I remembered that by looking at mode and seeing MO and remembering most often.
I hope this helps :-)
Answer:
M' (3 , 5)
Step-by-step explanation:
(3 , 7) ---> 2 unit down --> (3, 7-2) => (3, 5)
Fine the length BC using the cosine law. Google the law and sub it in
If points A, E and C are colinear, then they lie on the same line. The same statement you can say about points B, F and D.
1. Consider triangles AOC and BOD. In these triangles:
- AO≅OB (given);
- CO≅OD (given);
- ∠AOC≅∠BOD (as vertical angles).
Thus, ΔAOC≅ΔBOD by SAS Postulate (If any two corresponding sides and their included angle are the same in both triangles, then the triangles are congruent). Corresponding parts of congruent triangles are congruent, then
- AC≅BD;
- ∠ACO≅∠BDO;
- ∠CAO≅∠DBO.
Since angles ACO and BDO are alternate interior angles between lines AE and BF with transversal CD and these angles are congruent, then lines AE and BF are parallel.
This gives you that
2. Consider triangles ECO and FDO. In these triangles
- ∠CEO≅∠OFD (previous proof);
- CO≅OD (given);
- ∠ECO≅∠ODF (previous proof).
Therefore, ΔECO≅ΔFDO by AAS Postulate (if two angles and the non-included side one triangle are congruent to two angles and the non-included side of another triangle, then these two triangles are congruent). Then CE≅FD.
3. Note that
Since AC≅BD and CE≅DF, then AE=AC+CE=BD+DF=BF.
