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german
2 years ago
6

If ABC = ADC, which is NOT true by CPCTC ?

Mathematics
1 answer:
sdas [7]2 years ago
4 0
We cannot use CPCTC to show that BE is congruent to DE.

CPCTC refers to the angles and sides of two congruent triangles being congruent to their corresponding pieces.  The height of a triangle is not one of the given sides that counts in this.
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the answer is 13.5


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Triangle J K L is shown. Angle K J L is 58 degrees and angle J L K is 38 degrees. The length of J K is 2.3 and the length of J L
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What is the ratio of the afternoon to morning temperatures shown in the table? Temperatures Morning Afternoon 15 degrees. 30 deg
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The ratio of the afternoon to morning.

Solution:

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Deal with these relations on the set of real numbers: R₁ = {(a, b) ∈ R² | a > b}, the "greater than" relation, R₂ = {(a, b) ∈
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Answer:

a) R1ºR1 = R1

b) R1ºR2 = R1

c) R1ºR3 = \{ (a,b) \in R^2 \}

d) R1ºR4 = \{ (a,b) \in R^2 \}

e) R1ºR5 = R1

f) R1ºR6 = \{ (a,b) \in R^2 \}

g) R2ºR3 = \{ (a,b) \in R^2 \}

h) R3ºR3 = R3

Step-by-step explanation:

R1ºR1

(<em>a,c</em>) is in R1ºR1 if there exists <em>b</em> such that (<em>a,b</em>) is in R1 and (<em>b,c</em>) is in R1. This means that a > b, and b > c. That can only happen if a > c. Therefore R1ºR1 = R1

R1ºR2  

This case is similar to the previous one. (<em>a,c</em>) is in R1ºR2 if there exists <em>b</em> such that (<em>a,b</em>) is in R2 and (<em>b,c</em>) is in R1. This means that a ≥ b, and b > c. That can only happen if a > c. Hence R1ºR2 = R1

R1ºR3

(a,c) is in R1ºR3 if there exists b such that a < b and b > c. Independently of which values we use for a and c, there always exist a value of b big enough so that b is bigger than both a and c, fulfilling the conditions. We conclude that any pair of real numbers are related.

R1ºR4

This is similar to the previous one. Independently of the values (a,c) we choose, there is always going to be a value b big enough such that a ≤ b and b > c. As a result any pair of real numbers are related.

R1ºR5

If a and c are related, then there exists b such that (a,b) is in R5 and (b,c) is in R1. Because of how R5 is defined, b must be equal to a. Therefore, (a,c) is in R1. This proves that R1ºR5 = R1

R1ºR6

The relation R6 is less restrictive than the relation R3, if we find 2 numbers, one smaller than the other, in particular we find 2 different numbers. If we had 2 numbers a and c, we can find a number b big enough such that a<b and b >c. In particular, b is different from a, so (a,b) is in R6 and (b,c) is in R1, which implies that (a,c) is in R1ºR6. Since we took 2 arbitrary numbers, then any pair of real numbers are related.

R2ºR3

This is similar to the case R1ºR3, only with the difference that we can take b to be equal to a as long as it is bigger than c. We conclude that any pair of real numbers are related.

R3ºR3

If a and c are real numbers such that there exist b fulfilling the relations a < b and b < c, then necessarily a < c. If a < c, then we can use any number in between as our b. Therefore R3ºR3 = R3

I hope you find this answer useful!

5 0
2 years ago
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