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FromTheMoon [43]
3 years ago
10

In Triangle XYZ below, x2+y2=z2.

Mathematics
1 answer:
choli [55]3 years ago
8 0

4, 2, 1, 3

we cannot use 2 before having another triangle, so 4 must be before 2

we can use 1, but it will be just a restatement of the given information is not done before 4 and 2

3 is the final goal, which is always the last step

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Bradley and Kelly are out flying kites at a park one afternoon. A model of Bradley and Kelly's kites are shown below on the coor
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The correct answer is:

C. They are similar because the corresponding sides of kites KELY and BRAD all have the relationship 2:1.

Using the distance formula,

d=\sqrt{(y_2-y_1)^2+(x_2-x_1)^2}

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\text{BR}=\sqrt{(7-6)^2+(3-4)^2}=\sqrt{1^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}\\\\\text{RA}=\sqrt{(6-3)^2+(4-3)^2}=\sqrt{3^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{AD}=\sqrt{(3-6)^2+(3-2)^2}=\sqrt{(-3)^2+1^2}=\sqrt{9+1}=\sqrt{10}\\\\\text{DB}=\sqrt{(6-7)^2+(2-3)^2}=\sqrt{(-1)^2+(-1)^2}=\sqrt{1+1}=\sqrt{2}

The lengths of the sides of KELY are:

\text{KE}=\sqrt{(2-0)^2+(11-9)^2}=\sqrt{2^2+2^2}=\sqrt{4+4}=\sqrt{8}=2\sqrt{2}\\\\\text{EL}=\sqrt{(0-2)^2+(9-3)^2}=\sqrt{(-2)^2+6^2}=\sqrt{40}=2\sqrt{10}\\\\\text{LY}=\sqrt{(2-4)^2+(3-9)^2}=\sqrt{(-2)^2+(-6)^2}=\sqrt{40}=2\sqrt{10}\\\\\text{YK}=\sqrt{(4-2)^2+(9-11)^2}=\sqrt{2^2+(-2)^2}=\sqrt{8}=2\sqrt{2}

Each side of KELY is twice the length of the corresponding side on BRAD.  This makes the ratio of the sides 2:1 and the figures are similar.

8 0
3 years ago
Read 2 more answers
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DedPeter [7]

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