All Answers 358090 Answers

There are one googol or 10¹⁰⁰ or 10000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000000  zeroes in a googolplex.

It's 21

The answer is 19.

Yes, you are exactly right. He will exceed his goal by $319.57. To find out how you get this amount, we will use this formula to do that: FV=P{[1 + R]^N - 1/ R}=FV OF $1 PER PERIOD.

FV=200 {[(1 + .06/12)^(4*12) - 1] / .005}

FV=200{[(1.005)^48 - 1] / .005}

FV=200{[1.2704891 -1 ] / .005}

FV=200{.2704891 / .005}

FV=200{54.0978322}

FV=$10,819.57

And that is the proper way of calculating this particular investment. You have to be careful about your brackets and which operation is done first, and the next step after that until you do the final calculation.

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.    ://

7P0+7P1+7P2+7P3+7P4+7P5+7P6+7P7

где р перестановки

Я использовал переводчик гугл.
Так жаль, если язык не ясно.

Cosine of a negative angle is same as that of positive angle.

So, cos^2(-120) is cos^2(120).

Cos(120)= Cos(180-60)=-Cos 60=-1/2

Square of which will be 1/4.

Depends on what level you are working at, for a-levels or university i'd recommend using the Casio fx-991ES PLUS. This allows for all basic functions plus functionality for matrices, intergrals and basic plotting functions.

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. :/

it is an analog clock 

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

I suppose it must be ://

cos(A)  = A     when    A ≈ 0.739085  [radians ]

Melody  turns......Molehills

Into

[ Hehehe  !!!!!!]

I have to let you win sometimes Chris.  

It would be rude if I didn't !     ://

Yep....you sure did......!!!!!

HAHAHA!!!!!

I beat Melody....!!!

I beat Melody .....!!!

LOL....!!!!!!

Hi Chris,

I did a good job of turning that mole hill into a little mountain didn't I     LOL    :D

Yes, I can see it is a very conlfusing question. :)

Ali, Bala, Clement and David shared some marbles. Ali took 28% of the marbles and Bala took 21% of the marbles. Clement took 1800 marbles more than Bala. David took the remaining 2700 marbles. How many marbles were there altogether?(help please, really hard question)

Let Ali's share be A

Bala share is B

Clements share is C

David's share is D

Total is T and      T=A+B+C+D

A=0.28T

B=0.21T

C=B+1800

D=2700

21/3*4=28   so   21*  4/3 = 28         B * 4/3 = A

so

A= 4/3 * B

B=B

C=B+1800

D=2700

T= A+B+C+D

\(T=A+B+C+D\\ T=\frac{4B}{3}+B+B+1800+2700\\ T=\frac{4B}{3}+2B+4500\\ T=\frac{4B+6B}{3}+4500\\ T=\frac{10B}{3}+4500\\~\\ but\\ B=0.21T\\ so\\ T=\frac{10*0.21T}{3}+4500\\ \frac{3T}{3}=\frac{2.1T}{3}+4500\\ \frac{0.9T}{3}=4500\\ 0.3T=4500\\ T=4500\div 0.3\\ T=15000 \)

There were 15000 marbles.

I have checked this answer and you should too :)

Let N  be the number of total marbles

And we have that....

.28N + .21N  + [.21N + 1800]  + 2700  = N   simplify

.7N + 4500  = N       subtract .7N from both sides

4500  = .3N     divide both sides by .3

4500  / .3   = N  = 15,000

Sun 18/1/16

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@@ What is Happening?  [Wrap4]   Sat 16/1/16   Sydney,  Australia Time 5:32 pm   ♪ ♫

Hi all,

Our great answers today have been supplied by CPhill, Heureka, Omi67, Geno3141, Hayley1, xvxvxv, Solveit, rarinstraw1195, BreeDaBoss and Einstein Jr.    Thank you.    

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