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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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Can someone help me!?!?
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Check it carefully next time.!!

5 0
2 years ago
Let W be the subspace of R5 spanned by the vectors w1, w2, w3, w4, w5, where
Gwar [14]
To find W⊥, you can use the Gram-Schmidt process using the usual inner-product and the given 5 independent set of vectors. 

<span>Define projection of v on u as </span>
<span>p(u,v)=u*(u.v)/(u.u) </span>
<span>we need to proceed and determine u1...u5 as: </span>
<span>u1=w1 </span>
<span>u2=w2-p(u1,w2) </span>
<span>u3=w3-p(u1,w3)-p(u2,w3) </span>
<span>u4=w4-p(u1,w4)-p(u2,w4)-p(u3,w4) </span>
<span>u5=w5-p(u4,w5)-p(u2,w5)-p(u3,w5)-p(u4,w5) </span>

<span>so that u1...u5 will be the new basis of an orthogonal set of inner space. </span>

<span>However, the given set of vectors is not independent, since </span>
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7 0
3 years ago
A physics exam consists of 9 multiple-choice questions and 6 open-ended problems in which all work must be shown. If an examinee
katrin [286]

Answer: A) 1260

Step-by-step explanation:

We know that the number of combinations of n things taking r at a time is given by :-

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

Given : Total multiple-choice questions  = 9

Total open-ended problems=6

If an examine must answer 6 of the multiple-choice questions and 4 of the open-ended problems ,

No. of ways to answer 6 multiple-choice questions

= ^9C_6=\dfrac{9!}{6!(9-6)!}=\dfrac{9\times8\times7\times6!}{6!3!}=84

No. of ways to answer 4 open-ended problems

= ^6C_4=\dfrac{6!}{4!(6-4)!}=\dfrac{6\times5\times4!}{4!2!}=15

Then by using the Fundamental principal of counting the number of ways can the questions and problems be chosen = No. of ways to answer 6 multiple-choice questions x No. of ways to answer 4 open-ended problems

= 84\times15=1260

Hence, the correct answer is option A) 1260

5 0
3 years ago
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oee [108]

Answer:

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Step-by-step explanation:

6 0
3 years ago
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Ok i hope this helps and if it does pls give me brainliest!

So the longest line is 8 and makes a 90 degree angle with another smaller line which is labled 3.6. X is longer than 3.6 ( only by a little bit) But it looks like its half of the longest line (8). So i think the answer would be X equals 4. But i was thinking about it more i relized that you would prob just divide 8 by 3.6 which would be 2.22222 .... And that rounded tot he nearest 10th would be 2.2.

4 0
2 years ago
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