Melody

Your answer and your graphs are very impressive Solviet but I do not understand what you have done.

I do not understand the question either   

Write the equation of hyperbole which is tangent with the straights 5x-6y-16=0 and 13x-10y-48=0.

This appears to be translated from another language.

I think it might mean

Write the equation of hyperbola where theses straights lines    5x-6y-16=0 and 13x-10y-48=0.    are the asymptotes.

This is an interesting question ://

UNANSWERED  CHEMISTRY question here.

Thanks for the translation Solveit :)

I'd do it with Desmos Graphing calculator

Here is one I did as an example.  You have to add the open and closed circles seperataly I think.

Vf=682 km/h Vi=465 km/h T=15 secs A=?

I don't know what your i and f stand for but here are the motion formulas

I asssume you want the first one.   you need to commit these to memory!

5[x] (9[x^2]2 + 4)

The meaning is not very clear :/

Do you mean

\(5*x*(9x^2*2 + 4)\;\;\;?\)

ok i will assume that you do. 

\(5*x*(9x^2*2 + 4)\\ =5x(18x^2+ 4)\\ =90x^3+20x\)

The meaning of your question is not very clear.:/

How many sets of seven digits? How many sets, if the numbers in the set are different?

You have seven digits, you can choose 0,1,2,3,4,5,6,or all seven of them

Order does not matter because it is just a set of digiets.  So that woluld be

7C0+7C1+7C2+7C3+7C4+7C5+7C6+7C7

=7C0+7C7+7C1+7C6+7C2+7C5+7C3+7C4

=1+1+7+7+21+21+35+35

=2+14+70

= 86  ways

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Суть вашего вопроса не ясно.: /

Сколько наборов семь цифр? Сколько наборов, если разные числа внаборе?

У вас есть семь цифр, вы можете выбрать 0,1,2,3,4,5,6, или все семь из них

Порядок не имеет значения, потому что это просто набор digiets. Таким образом, чтобы быть woluld

7C 0 + 7 C 1 + 7 C 2 + 7 C 3 + 7 C 4 + 7 C 5 + 7 C 6 + 7 C 7

= 7C 0 + 7 C 7 + 7 C 1 + 7 C 6 + 7 C 2 + 7 C 5 + 7 C 3 + 7 C 4

= 1 + 1 + 7 + 7 + 21 + 21 + 35 + 35

= 2 + 14 + 70

= 86 пути

Here is one way:)

There are 24 Persons (p1,p2,p3,......) from which a team of 12 Persons has to be made. p1 and p2 can only come together in the team. p3 and p4 can never come together in the team. In how many ways can the team be formed?

If there were no restrictions at all then there would be  24C12 = 2,704,156 ways.

It has to be a lot less then that

This is what I am thinking.

Say we take those 4 out and deal with them seperately.  Then we will only be chosing from 20 people

P1 and P2 are  chosen,  P3 is chosen and P4 is NOT chosen           3 in aready             20C9

P1 and P2 are  chosen,  P3 is NOT chosen and P4 is chosen           3 in aready             20C9

P1 and P2 are  chosen,  P3 is NOT chosen and P4 is NOT chosen    2 in already           20C10

P1 and P2 are  NOT chosen,  P3 is chosen and P4 is NOT chosen        1 in already          20C11 = 20C9

P1 and P2 are  NOT chosen,  P3 is NOT chosen and P4 is chosen      1 in already          20C11 = 20C9

P1 and P2 are  NOT chosen,  P3 is NOT chosen and P4 is NOT chosen  0 in already      20C12 

20C9+20C9+10C10+10C9+20C9+20C12 = 4 * 20C9 +  20C10 + 20C12

These are the possibilities4*nCr(20,9)+nCr(20,10)+nCr(20,12) = 982566

So I think that there are  982,566 possible choices.    ://

Hi Solveit,

If you do not differentiate between AM and PM then obviously your answer of 18 hours is correct.

I am not sure about the formula either. :/

Is the principal root the one with the smallest argument   (angle) in the complex plane ?

I suppose it must be ://