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Free_Kalibri [48]
2 years ago
6

For f(x) = -4x+ 2. find f(x) when x = = -1.

Mathematics
1 answer:
ad-work [718]2 years ago
6 0
The answer is D :)
-4(-1)+2
4+2=
8
Hope this helpeddd
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Which of the following terms can be added to 8b^2c?
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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! (⌐■_■)

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