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snow_tiger [21]
3 years ago
14

Could someone help me out please!

Mathematics
1 answer:
cluponka [151]3 years ago
7 0

Answer:

  1. BC = 6
  2. x = 5
  3. CE = 16
  4. Yes, BC║DE

Step-by-step explanation:

1. The parallel lines make the various triangles similar, so the corresponding sides are in proportion.

... BC/AB = FE/A.F

... BC/9 = 4/6

... BC = 9·4/6 = 6

2. As in problem 1, the triangles are similar, so ...

... x/(x+10) = 10/30

... 30x = 10x + 100 . . . . multiply by 30(x+10)

... 20x = 100 . . . . . . . . . subtract 10x

... x = 100/20 = 5

3. As in problems 1 and 2, the triangles are similar, so ...

... AD/DB = CE/EB

... 24/27 = CE/18

... 18·24/27 = CE = 16

4. If the lines of interest are parallel, the triangles will be similar and corresponding measures will be in proportion.

Compare AD/DB = 15/12 = 5/4 to AE/EC = 10/8 = 5/4. These are equal, so corresponding measures are proportional. Therefore we conclude the triangles are similar and BC║DE.

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In our case,

\begin{gathered} (\hat{i}+\hat{j})=1\hat{i}+1\hat{j}+0\hat{k} \\ \text{and} \\ (\hat{i}\times\hat{j})=(1,0,0)\times(0,1,0)=(0)\hat{i}+(0)\hat{j}+(1-0)\hat{k}=\hat{k} \\ \Rightarrow(\hat{i}\times\hat{j})=\hat{k} \end{gathered}

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\begin{gathered} (\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=(1,1,0)\times(0,0,1) \\ =(1\cdot1-0\cdot0)\hat{i}+(0\cdot0-1\cdot1)\hat{j}+(1\cdot0-0\cdot1)\hat{k} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=1\hat{i}-1\hat{j} \\ \Rightarrow(\hat{i}+\hat{j})\times(\hat{i}\times\hat{j})=\hat{i}-\hat{j} \end{gathered}

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