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telo118 [61]
3 years ago
11

Given two triangles with three corresponding angles that are congruent, why can't the two triangles be proved congruent? A) Give

n three congruent angles, only the longest side can be congruent. B) Given three congruent angles, only the shortest side can be congruent. C) The triangles could have the same shape but not necessarily the same size. D) The triangles could have the same size but not necessarily the same shape.
Mathematics
2 answers:
Xelga [282]3 years ago
6 0
The answer is the triangles  could have the same shape( that is all the angles are equal) but the sides  may not be the same length . Thay are similar but not necessarily congruent.

C is the correct choice.
garri49 [273]3 years ago
6 0

Answer:

C) The triangles could have the same shape but not necessarily the same size.

Step-by-step explanation:

If the three angles of one triangle are congruent to the three corresponding angles of another triangle, then the two triangles are similar.

Similar triangles are triangles with the same shape but not necessarily the same size.

This means the correct answer is that the triangles could have the same shape but not necessarily the same size.

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Garwnicy nim
Vika [28.1K]

Answer:

What is the equation of a circle with a radius of 7 and centered at point (3, -4)?

Step-by-step explanation:

8 0
3 years ago
Find the number of 3-digit numbers formed using the digits 1 to 9, without repetition, such the numbers either have all digits l
Tasya [4]

Answer: 120

Step-by-step explanation:

The total number of digits from 1 to 9 = 10

The number of digits from less than 5 (0,1,2,3,4)=5

Since repetition is not allowed so we use Permutations , then the number of 3-digit different codes will be formed :-

^5P_3=\dfrac{5!}{(5-3)!}=\dfrac{5\times4\times3\times2!}{2!}=5\times4\times3=60

The number of digits from greater than 4 (5,6,7,8,9)=5

Similarly, Number of 3-digit different codes will be formed :-

^5P_3=60

Hence, the required number of 3-digit different codes = 60+60=120

3 0
3 years ago
Let n be a positive integer and define [n] to be the set of the first n positive integers. That is, [n] = {1, 2, 3, . . . , n}.
yaroslaw [1]

Answer: There are 2^{n-1} ways of doing this

Hi!

To solve this problem we can think in term of binary numbers. Let's start with an example:

n=5,  A = {1, 2 ,3},  B = {4,5}

We can think of A as 11100, number 1 meaning "this element is in A" and number 0 meaning "this element is not in A"

And we can think of B as 00011.

Thinking like this, the empty set is 00000, and [n] =11111 (this is the case A=empty set, B=[n])

This representation is a 5 digit binary number. There are 2^5 of these numbers. Each one of this is a possible selection of A and B. But there are repetitions: 11100 is the same selection as 00011. So we have to divide by two. The total number of ways of selecting A and B is the 2^{5-1} = 2^4.

This can be easily generalized to n bits.

6 0
4 years ago
Tch, hi ok help. thanks bye
Vikki [24]
Answer is C hope this helps
5 0
3 years ago
linda is building a rectangular playhouse. the width is x feet. the length is x + 3 the distance around the base of the playhous
Zanzabum
No. The value of x is 7.5 feet.
7 0
3 years ago
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