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3 years ago

Given: AD = BC and BCD = ADC Prove: DE = CE

Mathematics

2 answers:

Romashka-Z-Leto [24]3 years ago

7 0

Answer:

1. AD ≅ BC [Given]

2. ∠ADC ≅ ∠BCD [Given]

3. DC ≅ DC [reflexive property]

4. ΔADC ≅ ΔBCD [SAS]

5. ∠DAE ≅ ∠CBE [CPCTC]

6. ∠DEA ≅ ∠CEB [def. of vertical angles]

7. ∠DEA ≅ ∠CEB [vertical angles theorem]

8. ΔAED ≅ ΔBEC [AAS]

9. DE ≅ CE [CPCTC]

Step-by-step explanation:

Just finished mine :D

erastova [34]3 years ago

3 0

Answer:

According to AD=BC So AC=BD

Step-by-step explanation:

AD = BC and angle ADC = angle BCD

we have to prove AC = BD

let's see ∆ADC and ∆BCD

AD = BC [ given ]

angle ADC = angle BCD [ given ]

CD = CD

from S - A - S

∆ADC congruence ∆BCD

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-13 + 24-46-( - 20 ) =

Naddik [55]

Answer

-15

Step-by-step explanation:

-13 +24=11

11-46=-35

-35+20=-15

3 0

2 years ago

Read 2 more answers

The lengths of nails produced in a factory are normally distributed with a mean of 4.84 centimeters and a standard deviation of

Helga [31]

Answer:

Top 3%: 4.934 cm

Bottom 3%: 4.746 cm

Step-by-step explanation:

Problems of normally distributed samples are solved using the z-score formula.

In a set with mean $\mu$ and standard deviation $\sigma$ , the zscore of a measure X is given by:

$Z = \frac{X - \mu}{\sigma}$

The Z-score measures how many standard deviations the measure is from the mean. After finding the Z-score, we look at the z-score table and find the p-value associated with this z-score. This p-value is the probability that the value of the measure is smaller than X, that is, the percentile of X. Subtracting 1 by the pvalue, we get the probability that the value of the measure is greater than X.

In this problem, we have that:

$\mu = 4.84, \sigma = 0.05$