Y = 2x - 17, comparing to y = mx + c, slope m = 2.

If perpendicular, the new slope would be -1/2, that is the negative reciprocal of 2.

And passing through (-8 , 1).

using y = mx + c, and x = -8, y = 1, m = -1/2

1 = -1/2*-8 + c

1 = 4 + c

1 - 4 = c

c = -3

y = mx + c, substituting m = -1/2, and c = -3, y = -(1/2)x - 3.

Option C.

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