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8  =  4 + 3 + 1

8  =  5 + 2 +1

what is the smallest number that can be written as a sum of three natural numbers in two different ways?

How about 11.  (1 + 3 + 7) and (2 + 4 + 5)   I can't swear that 11 is the smallest but it's smaller than 251.  

_.

1 -  R ==10^12  x  1/2^(50/462)

      R==10^12   x   0.92772870876

      R  ==927,728,708,762 - atoms remain after 50 days.

2 - R ==10^12  x  1/2^(500/462)

      R ==10^12  x    0.47229131720

      R ==472,291,317,203 - atoms remain after 500 days.

3 - R ==10^12  x  1/2^(5000/462)

      R ==10^12  x    5.5220408368 x 10^-4

      R ==552,204,084 - atoms remain after 5000 days.

4x^3-8x^2+16x-34=71/x+2

Here you go  ........   you can finish it.  

I have responded on the original.  :)

Henry sells H, Liam sells L  and  Jack sells J

L=0.3(J+H)     (1)

H=0.2(J+L)     (2)

J=L+87           (3)

Substitute (3) inot the other 2 and then solve them simulaneously.

Can you take it from there?

the answer is 251

A club has 30 members: 15 boys and 15 girls. A 4-person committee is chosen at random. What is the probability that the committee has at least 1 boy and at least 1 girl?

1- P(all girls) - P(all boys)

=\(=1-2*\frac{15C4}{30C4}\\ =1-2*\frac{1365}{27405}\\ =1-\frac{26}{261}\\ =\frac{235}{261}\)

Is 51 a prime number ... People think it's prime  

Well, those people are wrong.  You can factor 51 to (3 • 17)  

-y-10=6x

5x+y=-10

solve by elimination 

Let's move some of those term around a bit, for clarity.  

–y – 10 = 6x         rearranges to             –6x – y = 10  

5x + y = –10         just leave it like it is      5x + y = –10  

                                                               ——————  

                          add the two together     –x        = 0   

so  x = 0  

substitute into an original equation     5(0) + y = –10   

so  y = –10  

_.

Answer: \(x=-7\), \(y=-2.8\)

Solution:

Adding both equations gives -x=7, so x=-7.

Plugging that back into the first equation gives -5y=14, so y=-2.8.

Did you proof-read your question??

\(120^\circ=\frac{2\pi}{3}\;radians\)

hint:

(this example is a bit harder than your ones but the method is the same.)

\(A=2.5e^{\frac{2\pi}{3}i}\\ A=2.5[(cos\frac{2\pi}{3})+isin(\frac{2\pi}{3})]\\ A=2.5[-(cos\frac{\pi}{3})+isin(\frac{\pi}{3})]\\ A=2.5[-(\frac{\sqrt3}{2})+i(\frac{1}{2})]\\ A=1.25[-(\sqrt3)+i)]\\ A=-1.25\sqrt3+1.25i\\\)

2573

Answer:

310+310+310+310+310+310+310+310+31+31+31 = 2573

31 x 83 = 2573

2573/83 =31

No, the factors of 51 are 1, 3, 17, 51

The two points are 3 units across from each other.

For the 2 points to be exactly 5 units apart from each other, we need a 3 - 4 - 5 right triangle, because one of the legs is a 3, and the hypotenuse is a 5. 

The only points that meet these conditions are \((2,7) \) and \((2, -1)\).

Thus, \(n\) can equal \(-1\) or \(7\), and the answer is \(-1 \times 7 = \)___.

\(\sqrt {160} = \sqrt{16} \times \sqrt{10} = {4 \sqrt10}\)

\(\sqrt{252} = \sqrt{36} \times \sqrt7 = 6\sqrt7\)

\(\sqrt{245} =\sqrt{49} \times \sqrt5 = 7\sqrt5\)

\(\sqrt{75} = \sqrt{25} \times \sqrt{3} = 5\sqrt3\)

\({4\sqrt{10} \over 6\sqrt7} \times {7\sqrt5 \over 5\sqrt3} = {28\sqrt{50} \over 30\sqrt{21}}\)

\({28\sqrt{50}\over30\sqrt{21}} = {14\sqrt{50} \over 15\sqrt{21}} = {14 \times \sqrt{25} \times \sqrt2 \over 15\sqrt{21}} = {14 \times 5 \times \sqrt2 \over 15\sqrt{21}} = {70\sqrt2 \over 15\sqrt{21}}={14\sqrt2 \over 3\sqrt{21}}\)

\({14\sqrt2 \over 3\sqrt{21}} \times {3\sqrt{21}\over3\sqrt{21}} = {42\sqrt{42}\over 189} = \color{brown}\boxed{2\sqrt{42}\over 9}\)