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stiv31 [10]
4 years ago
12

Which postulate or theorem can be used to prove that ΔABC ΔEDC?

Mathematics
2 answers:
Reptile [31]4 years ago
8 0

Answer:

<h2>ASA</h2>

Step-by-step explanation:

According to the graph, givens are

BC \cong CD

\angle ABC \cong \angle EDC

Now, \angle ACB and \angle ECD are vertical angles, because they are in opposite sides of C, and they don't have any side in common. By congruence postulates, we know that \angle ACB \cong \angle ECD

So, at this point, we have three elements congruent:

BC \cong CD

\angle ABC \cong \angle EDC

\angle ACB \cong \angle ECD

This allow us to use ASA postulate, that is, Angle-Side-Angle postulate, because we have congruence between corresponding angles and the side in between.

In addition, the ASA postulate states that if two triangles have two corresponding angles congruent, and the sides between them are also congruent, then those triangles are congruent.

Therefore, the right answer is ASA.

ANEK [815]4 years ago
6 0
ASA because angle B and D are congruent, side BC and DC are congruent and angle C and angle C are congruent because they are vertical.

Hope this helps :)
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Write 2 eggs is to 3 cups of flour as 12 eggs is to 18 cups of flour as a proportion.
Yuliya22 [10]
The answer is <span>A. 2 eggs / 3 cups of flour = 12 eggs / 18 cups of flour because 2 x 18 = 3 x 12 = 36.</span>
6 0
4 years ago
Alice has two flying disks. One is 12 cm in diameter and the other is 24 cm in diameter. What is the difference between the area
Oksi-84 [34.3K]

Answer:

432π cm² or 1356.48 cm²

Step-by-step explanation:

The area of a circle is πr²

The area of the disks are 144π and 576π.

576π-144π = 432π

432π ≈ 1356.48 cm²

8 0
3 years ago
Kindly Answer This And Don't Waste My PTS! ☁️
GalinKa [24]

====================================

\large \sf \underline{Problems:}

  • PROBLEM NUMBER 1: Solve 5x+2=22
  • PROBLEM NUMBER 2: Solve 2x+5=12
  • PROBLEM NUMBER 3: Solve 2x-5=7

====================================

\large \sf \underline{Answers:}

\large \qquad \qquad \bold{ \#1. \: x \:  =  \: 4}

\large \qquad \qquad \bold{ \#2. \:  x  \: =  \: \frac{7}{2} }

\large \qquad \qquad \bold{ \#3. \: x \:  =  \: 6}

====================================

\large \sf \underline{Solutions:}

<h3>Problem Number 1:-</h3>

  • \large\rm{5x \:  +  \: 2 \:  =  \: 22}

  • \large\rm{5x \:  =  \: 22 \:  -   \: 2}

  • \large\rm{5x \:   =  \: 20}

  • \large\rm{x \:  =  \:  \frac{20}{5} }

  • \large\rm\green{x \:  =  \: 4}

\therefore The answer of you're problem is x = 4.

====================================

<h3>Problem Number 2:-</h3>

  • \large\rm{2x \:  +  \: 5 \:  =  \:   12}

  • \large\rm{2x \:  =  \: 12 \:  -  \: 5}

  • \large\rm{2x \:   =   \: 7}

  • \large\rm\green{x \:  =  \:  \frac{7}{2} }

\therefore The answer of you're problem is 7/2.

====================================

<h3>Problem Number 3:-</h3>

  • \large\rm{2x \:  -  \: 5 \:  =  \: 7}

  • \large\rm{2x \:  =  \: 7 \:  +  \: 5}

  • \large\rm{2x \:  =  \: 12}

  • \large\rm{x \:  =  \:  \frac{12}{2} }

  • \large\rm\green{x \:  =  \: 6}

\therefore The answer of you're problem is x = 6.

====================================

Hope this helps!

4 0
2 years ago
Help me plzzz it’s not that bad
Travka [436]
X=1498-202. 1 term was used. x=1296
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Eric starts with 10 milligrams of a radioactive substance. The amount of the substance
svet-max [94.6K]

Answer:

ok m

Step-by-step explanation:

ok

4 0
3 years ago
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