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ASHA 777 [7]
3 years ago
5

Each side of a triangle is a different length. One side is 6, one side is greater than 6, and one side is less than 6. The perim

eter of the triangle could not be
a) 13
b) 18
c) 22
d)24
Mathematics
2 answers:
riadik2000 [5.3K]3 years ago
6 0

Answer:

The perimeter of the triangle could not be 24 - which is our solution.

Step-by-step explanation:

The first approach to this problem that comes to my mind, is to find the most " restricting " triangles possible. By that I mean the largest length of the side greater than 6 and the side less than 6.

The length of the side greater than 6 has to be less than 12 - considering it has to stretch the sum 6 + 6 - by the Triangle Inequality Theorem. Therefore, the most restricting triangle possible to the 1st decimal place should be the following -

{ 6, 5.9, 11.8 } ... the catch here is that this triangle does not need to be composed of lengths being whole numbers. If that was true the most " restricting " triangle would have lengths { 6, 5, 10 } throwing you off track as it's perimeter is 21.

{ 6, 5.9, 11.8 } adds to a sum of 23.7, eliminating answer choice c. Therefore, we can conclude that the perimeter of the triangle can't be 24. In fact this is only to the first decimal place, you could be more specific and go to the 100th decimal place even. Either way, the perimeter of the triangle has to be less than 24.

GenaCL600 [577]3 years ago
5 0

Answer:

i think the correct answer is c sorry if im wrong

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3 years ago
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A number n, decreased by 10
Anettt [7]

Answer:

n - 10

Step-by-step explanation:

7 0
2 years ago
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Solve |12x + 1| = 10<br> (9/2, 11/2)<br> (-9/2, 9/2)<br> (-112.9/2)
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|12x + 1| = 10

Remove the absolute value term and make two equations:

12x +1 = 10

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Now solve for both x values:

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Divide both sides by 12:

x = 9/12

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Subtract 1 from both sides:

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Divide both sides by 12:

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