Melody

Depends on the accent but this pronounciation is good.

If we express x^2+4x+5 in the form a(x-h)^2+k then what is h. 

\(x^2+4x+5\\ =x^2+4x+4+1\\ =(x-(-2))^2+1\)

edited:

Thanks Chris, yes    h = -2

Yes the answers should be 2,4 and 6 as you said,

16C15 = 16   when you get more familiar with this the answer will become obvious to you.

It is like saying how many ways can 15 obects be selected from 16 (order doesn't count)

and this is the same as saying how many ways can one object be omitted from the original 16. There is 16 objects so there are obviously 16 ways that one can be left out.

16Cr = 16C(16-r)   it is always symmetrical like this. 

There are different ways to enter this on the web2 calc 

I would do it like this (if I had needed to)

nCr(16,15)=

nCr(16,15) = 16

"The mass of a dust particle is approximately 7.5x10^-10 kilograms

and the mass of an electron is 9.1x10^-31 kilograms.

Approximately how many electrons have the same mass as one dust particle?"

This is just a division problem. You want electrons are in each dust particle

So how many times 9.1x10^-31 kilograms. goes into 7.5x10^-10 kilograms

7.5x10^-10 kilograms      divided by     9.1x10^-31 kilograms 

(7.5x10^-10) / (9.1x10^-31)  

=(7.5/9.1) * (10^-10 / 10^-31) electrons per particle

(7.5/9.1) = 0.8241758241758242 = 8.241758241758242*10^-1

(10^-10 / 10^-31) = 10^21

So tha answer is   

\(\approx 8.24\times 10^{-1}\times 10^{21}\\ \approx 8.24\times 10^{20}\;\;\text {electons in one dust particle}\)

Come on guest, don't be mean :)

So long as you learned from your mistake that is all that is important :)

Can you prove that the LHS converges to 2.38423 ?

Yes I can now see I am wrong.

I took a and b to be pronumerals when they are cleaarly marked as directional vectors.

If a and b were ordinary pronumerals then the question would be nonsense but as directional vectors (as is stated in the question) it is my answer that is nonsense.  Sorry all.

Thanks Tiggsy and Heureka. 

Also CPhill found the answer as a video clip on a seperate site.

This is the site