HELPMEEEEEEEEEEEEE

For this question, getting at most 1 head means either getting 0 or 1 heads. The probability of getting 0 heads is 1/2 * 1/2= 1/4 and the probability of getting 1 head is 1/2 * 1= 1/2. So the probability of getting at most 1 head is 1/4 + 1/2= 3/4. 

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Can you write this in laTex, it is hard to read otherwise.

Thanks!

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Edit:

Let three of the \(x_i\) be equal to 0, and let the remaining six be equal to \(\frac{2 \pi}{3}.\) Then

\(\cos x_1 + \cos x_2 + \dots + \cos x_9 = 3 \cos 0 + 6 \cos \frac{2 \pi}{3} = 3 + 6 \left( -\frac{1}{2} \right) = 0.\)
Also, 

\(\cos 3x_1 + \cos 3x_2 + \dots + \cos 3x_9 = 3 \cos 0 + 6 \cos 2 \pi = \boxed{9}.\)
Since \(\cos x \le 1\) for all \(x,\) this is clearly the maximum.

Answer from AOPS's Alcumus.

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"Please do not give answers only.

We do not want to overtly encourage cheating and how is that supposed to help someone learn?

It is better to give partial answers."

Melody

Thanks for the feedback!

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When written in vertex form, h= -1.5.

Hope this helps!

Edit: 
Step 1: Group the first 2 terms together, separating them from the constant term.

Step 2: Factor out leading coefficient, for completing the square to work, the coefficient of x^2 must be 1.

Step 3: Complete the square, Take half of x coefficient and square it. Notice to keep equation balanced you must add this number and subtract it making the net effect zero.

Step 4: Distribute and add constants.

So x-(-1.5) is the "h" you're looking for.

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Sounds good. Thanks!