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There are 240 ways to arrange the knights.

Use Vieta's formulas.

sum of roots = -b/a

product of roots = c/a

then do some manipulation, don't be lazy

Why are you posting AoPS questions? I think I know which class you are in. ;) And also dude don't post AoPS questions AoPS is for learning not for posting on web2.0calc.

If the arithmetic mean of a two numbers is 10 and the harmonic mean is 360. Find the geometric mean of the numbers.

SInce there are two numbers, and the arithmetic mean is 10, then the sum of the two numbers is 20.

The harmonic mean is \(2/(1/a+1/b)\) where a and b are the two numbers. Since it equals 360, then 1/a + 1/b must equal 1/180.

We can rewrite 1/a + 1/b to get \((a+b)/ab\)  and since we know a+b = 20, we get 20/ab = 1/180. Therefore ab = 3600.

The geometric mean is \(\sqrt{ab}\) so we get an answer of 60. 

I hope this helped.

Using a recursive approach, (e_1,e_2) = (4,6).

Let a and b be the solutions to \(5x^2-11x+4=0\). Find \(1/a + 1/b\) .

According to Vieta's Formula, the sum of the two roots are \(-b/a \) and the product of the roots is \(c/a\) for an equation in the form of\(ax^2+bx+c=0\) .

In this case, \(a=5, b= -11,\) and \(c=4\) .

We can factor \(1/a + 1/b\) to become \((a+b)/ab\). We can plug our values for a, b, and c in for the sum and product of the roots, giving us the answer of \(11/4\) . We don't need to solve for the roots.

Therefore our answer is 11/4. 

I hope this helped.

In a right triangle, the hypotenuse is 65 and the altitude to the hypotenuse is 300/13. Find the legs of the right triangle.

We first want to find the area of the triangle, which is \((65*300/13)/2 = 750\)

Note that we divide by two since this is a triangle. We could set up an algebraic equation to solve for the legs, but by some observation, we can find that this is a 25-60-65 triangle, with the area being 750 and the squares of the bases summing to to the hypotenuse's square.

Thus the legs of the right triangle are 25 and 60. 

I hope this helped.

The 3rd and 6th terms of an arithmetic progression are 94 and 85 respectively. Find the value of the 15th term.

From the 3rd term to the 6th term we have a 3 term skip, with a decrease of 9. Since this is an arithmetic progression, for every next term, we will subtract 3 from the pervious term since 9/3 =3.

15 - 6 = 9 more terms from the 6th term, so 9 x 3 = 27. Since the 6th term is 85, 85 - 27 = 58.

I hope this helped.

(x,y,z) = (2,-1,1)

Just use the formula for |z|^2.

a=0;b=0;c=0;p=0; cycle:d=a*100+b*10+c;if(d>1 and (d/3 + 15) > (d/2 +1), goto loop, goto next); loop:printd," ",;p=p+1; next:c++;if(c<10, goto cycle, 0);c=0;b++;if(b<10, goto cycle, 0);c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT:  2  3  4  5  6  7  8  9  10  11  12  13  14  15  16  17  18  19  20  21  22  23  24  25  26  27  28  29  30  31  32  33  34  35  36  37  38  39  40  41  42  43  44  45  46  47  48  49  50  51  52  53  54  55  56  57  58  59  60  61  62  63  64  65  66  67  68  69  70  71  72  73  74  75  76  77  78  79  80  81  82  83 > Total =  82 such numbers.

How many positive integers less than 2020 aren’t divisible by 2, 3, or 5?

a=0;b=0;c=0;d=0;p=0; cycle:n=a*1000+b*100+c*10+d;if(n%2==1 and n%3!=0 and n%5!=0 and n<=2020, goto loop, goto next); loop:printn,", ",;p=p+1; next:d++;if(d<10, goto cycle, 0);d=0;c++;if(c<10, goto cycle, 0);d=0;c=0;b++;if(b<10, goto cycle,0);b=0;c=0;d=0;a++;if(a<3, goto cycle,0);print"Total = ",p

OUTPUT = 538 - numbers that are NOT divisible by 2 or by 3 or by 5.

The fifth term of an arithmetic sequence is -18, and the sum of the first 32 terms is 1448. Find the ninth term.

- 18 =[F + 4D],  1448 =32/2 * [2*F + 31D], solve for F, D
F = - 40 - This is the first term
D =11/2 =5.5 - This is the common difference
- 40 + [8 x 5.5] = 4 - This is the 9th term.

Thank you

The sequence is: 2 + 2 +2 +2..............for 25 twos =2 x 25 =50

The smallest n =256. Because 2 x 5 x 6 =60

Can anyone find an integer smaller than 345?

a=0;b=0;c=0;p=0; cycle:d=a*100+b*10+c;if(d%2==0 and d%5==1 and d%7==0 and d<1000 and(a+b+c==23), goto loop, goto next); loop:printd," ",;p=p+1; next:c++;if(c<10, goto cycle, 0);c=0;b++;if(b<10, goto cycle, 0);c=0;b=0;a++;if(a<10, goto cycle,0);print"Total = ",p

OUTPUT = n =896

Since they lose 72% of their weight, they retain only 28% of their weight.

If x is the total weight of fresh mushrooms:  0.28 · x  =  12.25

     --->     x  =  12.25 / 0.28     --->     x  =  43.75  pounds of fresh mushrooms

You need 12.25/0.72 = 17.01 pounds of dried mushrooms.

Thank you for the response, but I don't believe that is correct. is there any way you can show me how you got there, so we can figure out, where it went wrong :)

The answer works out to (7*sqrt(2))^2 = 98.

The roots are (15 +/- sqrt(145))/10, and 1/a + 1/b = 7/5.

No... Can you help?