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Ganezh [65]
3 years ago
3

Read the proof. Given: AEEC; BDDC Prove: △AEC ~ △BDC Triangle A E C is shown. Line segment B D is drawn near point C to form tri

angle B D C. Statement Reason 1. AEEC;BDDC 1. given 2. ∠AEC is a rt. ∠; ∠BDC is a rt. ∠ 2. definition of perpendicular 3. ∠AEC ≅ ∠BDC 3. all right angles are congruent 4. ? 4. reflexive property 5. △AEC ~ △BDC 5. AA similarity theorem What is the missing statement in step 4? ∠ACE ≅ ∠BCD ∠EAB ≅ ∠DBC ∠EAC ≅ ∠EAC ∠CBD ≅ ∠DBC
Mathematics
2 answers:
Aleks04 [339]3 years ago
5 0

Answer:

Step-by-step explanation:∠ACE ≅ ∠BCD  

Naya [18.7K]3 years ago
3 0

Answer:

d

Step-by-step explanation:

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Natasha2012 [34]

x=\frac{12}{5} or x= twelve-fifths

Option A is correct.

Step-by-step explanation:

We need to find the solution of:

\sqrt{x^2+49}=x+5

Solving:

Taking square on both sides:

(\sqrt{x^2+49})^2=(x+5)^2\\x^2+49=x^2+2(x)(5)+25\\Simplifying:\\x^2+49-x^2-10x-25=0\\-10x+24=0\\-10x=-24\\x=\frac{-24}{-10}\\ x=\frac{24}{10}\\x=\frac{12}{5}

Verifying the solution by putting x = 12/5 in the given equation, we get true result.

So, x=\frac{12}{5} or x= twelve-fifths

Option A is correct.

Keywords: Solving Square root Equations

Learn more about Solving Square root Equations at:

  • brainly.com/question/4034547
  • brainly.com/question/1716201
  • brainly.com/question/10666510

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guapka [62]
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ValentinkaMS [17]

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