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qaws [65]
3 years ago
14

There are 9 balls in a hat the balls are numbered 1-9 you need to choose 3 of the balls. How many possible combinations are ther

e
Mathematics
1 answer:
choli [55]3 years ago
8 0
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There are 57 variations that are above and then times that by 9 because the various will never repeat the same number like when you move onto 2's you'd replace all the 2 with 1's and so on and so forth

The answer is 513 variations 
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5 0
3 years ago
Please answer this correctly without making mistakes
Irina18 [472]

Answer:

theater

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3 0
3 years ago
AC DF and YZ are straight lines and all angles are in degrees.
sweet-ann [11.9K]

Answer:

Detail in steps

Step-by-step explanation:

∠ABY + ∠YBC = 180

(x + 25) + (2x + 50) = 180

3x + 75 = 180

3x = 105

x = 35

m∠YBC = 2 x 35 + 50 = 120

m∠BEF = 5 x 35 - 55 = 120

m∠YBC ≅ m∠BEF

∴ AC // DF

4 0
3 years ago
Simplify 7x/x-4(-2)​
igomit [66]

Answer:

15

Step-by-step explanation:

5 0
3 years ago
SOME ONE HELP ME I REALLY NEED A SOLUTION FOR THIS PROBLEM WITH A CLEAR EXPLANATION
Sergio [31]

I'm going to answer this question using logic.  Let x be Jenny's favorite number.  We are going to square root this number, \sqrt{x} and then multiply it by \sqrt{2}.

This product needs to be an integer.  How is that obtainable? To be an integer, we need to get rid of the nasty \sqrt{2} part.  The only way I can think of to get rid of it, is to multiply it by \sqrt{2}, because \sqrt{2} *\sqrt{2} = 2.  Thus we have some conditions we need to fulfill when choosing Jenny's favorite number.

When we take the square root of Jenny's favorite number, x, it must contain a perfect square and a 2 in its prime factorization.  For example, 8 works because 8 = 2 x 2 x 2, or 2² x 2.  

You notice 8 is made up of a perfect square multiplied by 2.  So when we take the square root of 8, we get:

\sqrt{8} = \sqrt{2^{2}*2 } = \sqrt{ 2^{2} }*\sqrt{2} = \sqrt{4} *\sqrt{2} = 2*\sqrt{2}

So 8 is the same thing as 2\sqrt{2}

So when we multiply this by \sqrt{2}, we will get an integer! So as long as Jenny's favorite number consists of a perfect square and two in its prime factorization, we will have an integer!

So possible choices are: 2,8,18.

Why does 18 work? Because \sqrt{18} = \sqrt{9}* \sqrt{2} = 3\sqrt{2}

When we multiply this by \sqrt{2}, we get 6, which is an integer.

b) Suppose instead of multiplying by \sqrt{2}, we divided by \sqrt{2}. Is the resulting quotient still an integer?

YES, because we can get rid of the \sqrt{2} part by dividing by \sqrt{2} as well.  This leaves only the "perfect square" part left in our square root, and obviously a perfect square is an integer when we square root it.

I hope that made sense! (⌐■_■)

3 0
3 years ago
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