All Answers 355923 Answers

Isn't it the (+/-) sign?

By the quadratic formula, the roots are (-5 +- sqrt(69))/2, so the answer is ((-5 + sqrt(69)/2 + 3)((-5 - sqrt(69)/2 + 3) = -13.

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

6 i believe

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.

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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1/8  /3 = 1/24 = 0.0416666666666667      this last digit is rounded up to 7

                            .04166      would be   .0417   rounded

                            .0416        would be   .042

                             .041         would be  .04

Sounds good. Thanks!

How did you come by those values Geno?

Ok,

It is a learning curve.

I actually did not think you were intentionally rude. 

I have seen how you have behaved here and I think you are a valuable addition to our forum.;

Next time you want to repost please do not add the full question again.

Just add a link to the original question,

say you still would still very much appreciate an answer

 and ask people to answer on the original thread.

Once the original post is answered you can put a note on both to say thanks to the answerer.

You can use this identity:  a³ - b³  =  (a - b)(a² + a·b + b²)

In this case:  a = cubr(3)  and  b = cubr(2)

So:  multiply the numerator and denominator by:  cubr(3²) + cubr(3·2) + cubr(2²}  =  cubr(9) + cubr(6) + cubr(4)

The numerator becomes:  cubr(9) + cubr(6) + cubr(4)

while the denominator becomes 3 - 2  =  1

A + B + C + D  =  cubr(9) + cubr(6) + cubr(4) + 1

A solution exists if and only if 8 is invertible modulo p. In other words, \(\gcd(8,p)=1\). Since \(8=2^3\) is a power of 2, 8 is invertible modulo q if and only if q is an odd integer. All primes except for 2 are odd, so the number we are looking for is 2.

Your answers are usually pretty good.

That phrase is very common among the younger crowd, so it’s not a unique trade mark, but it is always irritating, and it’s never enthusiastic. 

The enthusiasm should come from your answer not your valediction.     

Maybe you could use a trademark similar to what CPhill uses .  But don’t copy his. 

Heureka uses this 

Omi67 also uses 

Asinus uses this  !

Maybe you could use some combination of two or three emos that identifies you, but doesn’t distract from your answer. 

Why have you deleted your question Helpmeeeee?

I can see it has not been answered but even so this greatly discouraged as it is not polite.

Can you please put it back.

If you have the answer you could add that in as an answering post.

You'll need to combind conjugates with rationalizing a cube root.

2 sin (4x) = 1                      Divied by 2

sin (4x) = 1/2                      arcsin

4 x =  pi/6  and   5pi/6         divide by  4    

x = pi/24    5pi/24