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NeTakaya
4 years ago
9

Can the measures 6cm 15cm 6cm make a triangle

Mathematics
2 answers:
trapecia [35]4 years ago
7 0

Answer:

Yes!

Step-by-step explanation:

Hi there! I'm glad I was able to help you answer this question!

You indeed can make a triangle with the given lengths, but it isn't going to be a perfect triangle, since one of the lengths is longer than the other two.

Also keep in mind that we are only given THREE lengths, and the prefix "tri" in triangle means three (sides). :)

I hope this helped you! If you have any other questions, leave a comment below!

blagie [28]4 years ago
5 0

Answer:tes

Step-by-step explanation:

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Every month, Jim makes $49 less than his best friend Keenan.
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Answer:

m+49=d ; d-49 = m

Step-by-step explanation:

Jim $ = m

Keenan $ = d

m+49=d ; d-49=m

btw please lmk if i get this wrong

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How to solve <br> 215x = 99.3
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Answer:

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Step-by-step explanation:

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3 years ago
You know how to play 21 songs on your guitar. (a) If you want to choose 8 of the songs to make a setlist to play for a small gat
UNO [17]

Answer:

<em>(a) 8,204,716,800</em>

<em>(b) 5,985</em>

Step-by-step explanation:

<u>Combinations and Permutations</u>

Combinatorics is the part of the discrete mathematics that studies the enumeration of groups or sorting of a determined number of elements.  The concept of combinations is tied to the different forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in a different order.

The concept of combinations is tied to the differents forms to group elements where the order of their arrangements is not important or does not differentiate from the very same set of elements picked in different order.

On the other hand, permutations or variations are sets selected in a specific order and another set with the same element but in different order is considered a different set.

If we have n elements available to pick from in sets of m elements each, there can be C(n,m) different combinations, and it's given by

\displaystyle C(n,m)=\frac{n!}{m!(n-m)!}

Similarly the number of permutations is given by

\displaystyle P(n,m)=\frac{n!}{(n-m)!}

(a) I have n=21 songs to pick from and I want to choose m=8 of them where the order matters, so it's a permutation:

\displaystyle P(21,8)=\frac{21!}{(13)!}=\frac{51,090,942,171,709,440,000}{6,227,020,800}=8,204,716,800

I can make more than 8 billion different setlists

(b) To choose m=4 songs from n=21 songs where the order does not matter, we compute the combination

\displaystyle C(21,4)=\frac{21!}{4!(17)!}=\frac{21\cdot 20\cdot 19\cdot 18\cdot 17!}{4\cdot 3\cdot 2\cdot 1(17)!}

\displaystyle C(21,4)=\frac{143,640}{24}=5,985

I can make almost 6,000 sets of 4 songs

5 0
3 years ago
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