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netineya [11]
3 years ago
14

BA is congruent to BC and DA is congruent to EC. Prove BD is congruent to BE.

Mathematics
1 answer:
Readme [11.4K]3 years ago
5 0

Answer:

See below for answer.

Step-by-step explanation:

BA ≅ BC                         Given

DA  ≅ EC                       Given

BA + DA ≅ BC + EC      Congruent  parts are congruent

BD ≅ BE                        Addition

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If the blue graph shown is y=f(x), what is the red graph
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https://lmgtfy.com/?q=If+the+blue+graph+shown+is+y%3Df(x)%2C+what+is+the+red+graph%3F&iie=1

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4 years ago
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F(x)=x2+3x-5 and g(x)=x/4-6.g(x)=x/4-6
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3 years ago
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In ΔPQR,
irina1246 [14]

Step-by-step explanation:

Notice that, the angle QRS is external to the triangle and adjacent to the angle PRQ. According to the theorem of a external/adjacent angle, we have: m∠QRS = m∠PQR + m∠RPQ, where PQR and RPQ are internal angles.

From the hypothesis, we have:

m∠QRS =(10x−12)∘(10x−12)

m∠PQR = (3x+20)∘(3x+20)

m∠RPQ=(3x−8)∘(3x−8)

Using the first equation and replacing the hypothesis:

m∠QRS = m∠PQR + m∠RPQ

(10x−12)∘(10x−12) = (3x+20)∘(3x+20) + (3x−8)∘(3x−8)

Multiplying and applying the remarkable identity:

(10x - 12)^{2}  = (3x+20)^{2}  + (3x - 8)^{2} \\(10x)^{2} - 2(10x)(12) + (12)^{2} = (3x)^{2} + 2(3x)(20) + (20)^{2} + (3x)^{2} - 2(3x)(8) + (8)^{2} \\100x^{2} -240x+144=9x^{2} +120x+400 + 9x^{2} -48x+64\\100x^{2} -240x+144-18x^{2} -72x-464=0\\82x^{2} -312x-320=0\\

Then, we use a calculator to find the roots, which are:

x=4.7\\x=-0.8

In this case, we will see what root is the right one.

Now, we replace it into m∠QRS =(10x−12)∘(10x−12), because we need to find m∠QRS.

m∠QRS =(10x−12)∘(10x−12) = (10(4.7) - 12) (10(4.7) - 12) = (35) (35) = 1225

6 0
3 years ago
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