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Otrada [13]
3 years ago
13

What is the slope of the line shown?

Mathematics
2 answers:
dem82 [27]3 years ago
4 0

Answer:1/2

Step-by-step explanation:

Oksanka [162]3 years ago
3 0

Answer: 0.5

Step-by-step explanation:

You subtract y2-y1 and x2-x1

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Answer:

\sf \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}=\dfrac{1}{7}

Step-by-step explanation:

If G is the midpoint of CD, and AC is parallel to DB, then AC = DH.

Therefore, G is the midpoint of AH and ΔACE is similar to ΔDBE.

As AC : DB = 1 : 3

⇒ Area of ΔACE : Area of ΔDBE = 1² : 3² = 1 : 9

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⇒ Area ΔACG = 2 × Area ΔACE

As AC = DH, and G is the midpoint of CD:

⇒ ΔACG ≅ ΔHDG

⇒ Area ΔHDG = 2 × Area ΔACE

Area of quadrilateral EGHB = Area of ΔDBE - Area ΔHDG

                                              = Area of ΔDBE - 2 × Area of ΔACE

Therefore:

\sf \implies \dfrac{Area\:of\: \triangle\: AEG}{Area \: of \: quadrilateral\:EGBH}

\sf \implies \dfrac{Area\:of\: \triangle\: ACE}{Area \: of \: \triangle\:DBE - 2 \times Area\:of\: \triangle ACE}

Using the ratio of Area ΔACE : Area ΔDBE =  1 : 9

\implies \sf \dfrac{1}{9-2}

\implies \sf \dfrac{1}{7}

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