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ikadub [295]
3 years ago
7

Given: AB parallel to DC prove: ABE ~ DCE

Mathematics
2 answers:
slava [35]3 years ago
8 0
The reasons would be:

alternate interior angles
vertical angles
AAA (all angles are congruent)

Since we know that the two line segments are parallel, this means that BC is a transversal.  Using that knowledge with the position of ABE and DCE, we see these are alternate interior angles.

AEB and DEC are vertical angles, since they share only a vertex.

This gives us 3 congruent angles, so the triangles must be similar.
Vesna [10]3 years ago
3 0
Answer: The 3 missing statements are listed below with explanations.

2) Alternate interior angles are congruent. Since the lines are parallel these angles are congruent.

3) Vertical angles are congruent. Intersecting angles across from each other are congruent.

4) AA Similarity Theorem. If 2 angles of a triangle are congruent, then the triangles are similar.
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BigorU [14]

Answer:

4

Step-by-step explanation:

The diagonals are equal length:

6x - 14 = 8x - 20\\-2x = -6\\x = 3

BD = 6x - 14\\BD = 6 * 3 - 14\\BD = 18 - 14\\BD = 4

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3 years ago
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Alla [95]

We\ know:\ i=\sqrt{-1}\to i^2=-1

\dfrac{2+4i}{3i}=\dfrac{2+4i}{3i}\cdot\dfrac{3i}{3i}=\dfrac{6i+12i^2}{9i^2}=\dfrac{6i+12(-1)}{9(-1)}\\\\=\dfrac{6i-12}{-9}=\dfrac{2i-4}{-3}=\dfrac{4}{3}-\dfrac{2}{3}i

\dfrac{3+2i}{4+i}=\dfrac{3+2i}{4+i}\cdot\dfrac{4-i}{4-i}=\dfrac{(3+2i)(4-i)}{4^2-i^2}\\\\=\dfrac{(3)(4)+(3)(-i)+(2i)(4)+(2i)(-i)}{16-(-1)}=\dfrac{12-3i+8i-2i^2}{16+1}\\\\=\dfrac{12+5i-2(-1)}{17}=\dfrac{12+5i+2}{17}=\dfrac{14+5i}{17}=\dfrac{14}{17}+\dfrac{5}{17}i

2i^{11}=2i^{10+1}=2i^{10}i^1=2i^{2\cdot5}i=2i(i^2)^5=2i(-1)^5=2i(-1)=-2i


Used:\ (a+b)(a-b)=a^2-b^2

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3 years ago
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See attachment

Step-by-step explanation:

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If the sales tax rate is 8%, what is the tax on a $5.75 cent purchase?
lora16 [44]

Answer:

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Step-by-step explanation:

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