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tatuchka [14]
3 years ago
14

Given: ∠BCD ≅ ∠EDC and ∠BDC ≅ ∠ECD Prove: Δ BCD ≅ Δ EDC

Mathematics
2 answers:
max2010maxim [7]3 years ago
8 0

Answer:  The proof is done below using ASA congruence postulate.

Step-by-step explanation:  We are given that in the figure shown :

∠BCD ≅ ∠EDC and ∠BDC ≅ ∠ECD.

We are to prove that :

Δ BCD ≅ Δ EDC.

From the figure, we can see that the side CD is the common side of both the triangles, ΔBCD and ΔECD.

In triangle BCD and ECD, we have

∠BCD ≅ ∠EDC,

∠BDC ≅ ∠ECD

and CD is the common side.

That is, two angles and the side between them of ΔBCD are congruent to the corresponding angles and the side between them of ΔECD.

Therefore, by ASA (Angle-side-Angle) postulate, we arrive at

Δ BCD ≅ Δ EDC.

Hence proved.

densk [106]3 years ago
6 0
Hopefully this is right!

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Answer:

9.8%

Step-by-step explanation:

Given the following information

\left|\begin{array}{c|cccccc}Ages &15-18&19-22&23-26&27-30&31-34&35-38\\$Number of students&6&4&5&7&10&9\end{array}\right|

Total Number of Students =6+4+5+7+10+9=41

Number of Students aged 19-22 =4

Therefore:

The relative frequency for the class with a lower class limit of 19

=\dfrac{4}{41}\times 100\\\\ =9.8\% $(to one decimal place)

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