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ohaa [14]
3 years ago
13

Which set of numbers could be the length of the sides

Mathematics
1 answer:
Svetllana [295]3 years ago
3 0

Answer: Choice C)  {6, 9, 12}

=======================================================

Explanation:

Something like choice A can't form a triangle because the first two sides add to 6+9 = 15, which is <u>not</u> longer than the third side 15. We need to have x+y > z to be true. The same goes for choice B as well because 3+3 = 6 is too small compared to the third side 7. Choice D is similar to choice A.

In short, we can rule out choices A, B, and D.

The only thing left is choice C. Picking any two sides leads to having that sum be larger than the third side

  • 6+9 = 15 which is larger than 12
  • 9+15 = 24 which is larger than 6
  • 6+15 = 21 which is larger than 9

So the conditions of the triangle inequality theorem work out here, and we have a triangle.

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Step-by-step explanation:

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Translate the following words to algebra symbols, using "n" as the variable. The product of six and a number is 24.
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So n will be the unknown, and the expression is:
6n = 24
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In a shipment of 56 vials, only 13 do not have hairline cracks. If you randomly select 3 vials from the shipment, in how many wa
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Answer: 27434

Step-by-step explanation:

Given : Total number of vials = 56

Number of vials that do not have hairline cracks = 13

Then, Number of vials that have hairline cracks =56-13=43

Since , order of selection is not mattering here , so we combinations to find the number of ways.

The number of combinations of m thing r things at a time is given by :-

^nC_r=\dfrac{n!}{r!(n-r)!}

Now, the number of ways to select at least one out of 3 vials have a hairline crack will be :-

^{13}C_2\cdot ^{43}C_{1}+^{13}C_{1}\cdot ^{43}C_{2}+^{13}C_0\cdot ^{43}C_{3}\\\\=\dfrac{13!}{2!(13-2)!}\cdot\dfrac{43!}{1!(42)!}+\dfrac{13!}{1!(12)!}\cdot\dfrac{43!}{2!(41)!}+\dfrac{13!}{0!(13)!}\cdot\dfrac{43!}{3!(40)!}\\\\=\dfrac{13\times12\times11!}{2\times11!}\cdot (43)+(13)\cdot\dfrac{43\times42\times41!}{2\times41!}+(1)\dfrac{43\times42\times41\times40!}{6\times40!}\\\\=3354+11739+12341=27434

Hence, the required number of ways =27434

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(-5, 7) - Quadrant II

(8 1/2, -4) - Quadrant IV

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Step-by-step explanation:

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