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IgorLugansk [536]
3 years ago
6

Determine whether the two triangles can be proven congruent using the AAS congruence method. If they can, select the congruence

statement. answers: A) ΔABC ≅ ΔEDC B) ΔCBA ≅ ΔCED C) The triangles aren't congruent using AAS. D) ΔCAB ≅ ΔEDC

Mathematics
2 answers:
Ierofanga [76]3 years ago
5 0

Answer:

A) triangle ABC is congruent to triangle EDC

Step-by-step explanation:

The AAS method of proving congruence of triangles uses two angles and a non-included side of the triangle. If two angles and the non-included side of a triangle are congruent to the corresponding parts of another triangle, then the triangles are congruent.

Let's see what we have in this problem:

<ACB and <ECD are congruent since they are vertical angles.

<A and <E are congruent by given.

Sides AB and ED are non-included sides and are congruent.

Since we have two angles and a non-included side of a triangle and the corresponding parts of another triangle, the triangles are congruent by AAS.

Now we need the statement of congruence.

Angles ACB and ECD are corresponding angles, so the letter C must appear in both triangles in the same position.

Angles A and E are corresponding angles, so the letters A and E must appear in both triangles the same position.

We already have CA and CE. The last angles left are corresponding angles B and D, so we get triangle CAB and triangle CED. Since a triangle may be named using any order of the vertices, we can rename the triangles ABC and EDC and maintain the same corresponding vertices.

Answer: A) triangle ABC is congruent to triangle EDC

Dmitry_Shevchenko [17]3 years ago
3 0

Answer:

The A) ΔABC ≅ ΔEDC

Step-by-step explanation:

The AAS congruence method requires 2 angles and their un-included side to be congruent. ∠A ≅ ∠E due to the markings, ∠C ≅ ∠C because they are vertical angles, and AB ≅ ED due to the markings. 2 angles and their un-included side are congruent.

As for the congruence statement, A is the correct answer because ∠A ≅ ∠E, ∠B ≅ ∠D, and ∠C ≅ ∠C. The order of the naming of the triangles aligns to the angle's congruence.

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Step-by-step explanation:

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3 years ago
Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimeters,
tiny-mole [99]

Answer:

Probability that the sample mean would be greater than 141.4 millimetres is 0.3594.

Step-by-step explanation:

We are given that Thompson and Thompson is a steel bolts manufacturing company. Their current steel bolts have a mean diameter of 141 millimetres, and a standard deviation of 7.

A random sample of 39 steel bolts is selected.

Let \bar X = <u><em>sample mean diameter</em></u>

The z score probability distribution for sample mean is given by;

                            Z  =  \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } }  ~ N(0,1)

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Now, Percentage the sample mean would be greater than 141.4 millimetres is given by = P(\bar X > 141.4 millimetres)

      P(\bar X > 141.4) = P( \frac{ \bar X-\mu}{\frac{\sigma}{\sqrt{n} }  } } > \frac{141.4-141}{\frac{7}{\sqrt{39} }  } } ) = P(Z > 0.36) = 1 - P(Z \leq 0.36)

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The above probability is calculated by looking at the value of x = 0.36 in the z table which has an area of 0.6406.

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