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ololo11 [35]
2 years ago
14

AB ≅ BC and AD ≅ CD What additional information would make it immediately possible to prove that triangles AXB and CXB are congr

uent using the HL theorem? What additional information would make it immediately possible to prove that triangles AXD and CXD are congruent using the SSS congruence theorem?
Mathematics
2 answers:
MrRissso [65]2 years ago
8 0

Answer:

1: C/AC and BD are perpendicular.

2: B/AX and CX are congruent.

Step-by-step explanation:

kobusy [5.1K]2 years ago
5 0

It is given that, B ≅ BC and AD ≅ CD  

We need BD perpendicular to AC, then only we can say  triangles AXB and CXB are congruent using the HL theorem.

If BD perpendicular to AC, means that AB and CB are the hypotenuse of triangles AXB and CXB respectively.

from the given information ABCD is a square

If BD and AC bisect each other then AX = CX

Then only we can  immediately possible to prove that triangles AXD and CXD are congruent by SSS congruence theorem

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Given:
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For a certain type of copper wire, it is knownthat, on the average, 1.5 flaws occur per millimeter.Assuming that the number of f
mario62 [17]

Answer:

The probability that no flaws occur in a certain portion of wire of length 5 millimeters =  1.1156 occur / millimeters

Step-by-step explanation:

<u>Step 1</u>:-

Given data A copper wire, it is known that, on the average, 1.5 flaws occur per millimeter.

by  Poisson random variable given that λ = 1.5 flaws/millimeter

Poisson distribution P(X= r) = \frac{e^{-\alpha } \alpha ^{r} }{r!}

<u>Step 2:</u>-

The probability that no flaws occur in a certain portion of wire

P(X= 0) = \frac{e^{-1.5 } \(1.5) ^{0} }{0!}

On simplification we get

P(x=0) = 0.223 flaws occur / millimeters

<u>Conclusion</u>:-

The probability that no flaws occur in a certain portion of wire of length 5 millimeters = 5 X P(X=0) = 5X 0.223 = 1.1156 occur / millimeters

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