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\(1)~\text{answered in other thread}\)

\(2)~\dfrac{2}{10}\dfrac 1 2 = \dfrac{1}{10}\)

\(3)~\dfrac{\dbinom{2}{1}\dbinom{3}{1}}{\dbinom{10}{2}} = \dfrac{6}{45}=\dfrac{2}{15}\)

\(4)~\dfrac 1 2 \dfrac 1 6 = \dfrac{1}{12}\)

\(5)~\dfrac{10}{18}\dfrac{8}{17} = \dfrac{40}{153}\)

101₂ = 1*2² + 0*2¹ +1*2⁰ = 4 + 0 + 1 = 5₁₀

"A ball thrown up from the top of a building is given  ... etc."

Note that Rom used engineering notation to depict the numerical values.   Engineering notation is a specialized subset of scientific notation, and its use is common (and preferred) in all branches of physical sciences. 

Reviewing this will help you understand the (different) notations presented by Rom and Melody

Yea maybe, that makes good sense but the question did not say that.   

There was no (3+x).       I interpreted what was written.

Perhaps the guest asker will learn to be more precise next time.

I answered the question that was asked. 

If the asker wants to they can come back and appologise for not including the brackets and ask us to answer the different question.

The question actually does not ask for a graph either. It does not ask any question at all. It is just a statement.

Melody, since the problem stipulated that x could not be –3, I think the equation was intended to be y = 7/(3+x).  If not, then why the stipulation? 

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THIS IS NOT A DEFINATIVE ANSWER.

It would have to be at least 4 weeks

If A hugs B and C   the first week

A hugs D and E the seconds week

A hugs Fand G the third week

A hugs H and one othe on the 4th week

I don't know for sure if itthe minimun is MORE  more than 4 weeks.

I just know that it will be 4 weeks or maybe more.

Here is the graph

Find the derivative of the trig function.

f(t)=t²sint

\(\begin{array}{|rcll|} \hline \mathbf{f(t)} & \mathbf{=} & \mathbf{t^2\sin(t)} \\\\ f'(t) &=& t^2 * \dfrac{d\ \sin(t)}{dt}+ \dfrac{d\ t^2}{dt}*\sin(t) \\\\ && \dfrac{d\ \sin(t)}{dt} = \cos(t) \qquad \dfrac{d\ t^2}{dt}=2t \\\\ f'(t) &=& t^2 *\cos(t) +2t*\sin(t) \\ \mathbf{f'(t)} & \mathbf{=} & \mathbf{ t\Big(2\sin(t) +t\cos(t) \Big) } \\ \hline \end{array}\)

I wasn't meaning to be mean Emerald. 

I am not good at wording things tactfully.

I really do like that you are active on the forum and I am sorry that we have got off on the wrong foot. :)

Maybe we can start over?

What is the remainder when \((x+1)^{2010}\) is divided by \(x^2+x+1\)?

I assume

Look at the low powers of x.

Eventually they will cycle.

\(\begin{array}{|r|c|c|} \hline n & & \text{remainder} \\ \hline 0 & \dfrac{(x+1)^0}{x^2+x+1} & \color{red}1 \\ \hline 1 & \dfrac{(x+1)^1}{x^2+x+1} & x+1 \\ \hline 2 & \dfrac{(x+1)^2}{x^2+x+1} & x \\ \hline 3 & \dfrac{(x+1)^3}{x^2+x+1} & -1 \\ \hline 4 & \dfrac{(x+1)^4}{x^2+x+1} & -x-1 \\ \hline 5 & \dfrac{(x+1)^5}{x^2+x+1} & -x \\ \hline 6 & \dfrac{(x+1)^6}{x^2+x+1} & \color{red}1 \\ \hline 7 & \dfrac{(x+1)^7}{x^2+x+1} & x+1 \\ \hline 8 & \dfrac{(x+1)^8}{x^2+x+1} & x \\ \hline \ldots &\ldots & \ldots \\ \hline 2010 & \dfrac{(x+1)^{2010}}{x^2+x+1} & \color{red}1 \\ \hline \end{array}\)

\((x+1)^n / (x^2+x+1) \) is cyclic of order 6
So we get \((x+1)^n \equiv (x+1)^{n\pmod{6}}\pmod{x^2+x+1}\)

\((x+1)^{2010} \equiv (x+1)^{6·335+0} \equiv(x+1)^{0} \equiv 1 \pmod{x^2+x+1}\)

The remainder when \((x+1)^{2010}\) is divided by \(x^2+x+1\) is 1

\(\lambda = \frac{3*10^8}{6*10^{14}}metres\\ \lambda = 0.5*10^{-6}metres\\ \text{there are 10^9 nm in a metre}\\ \lambda = 0.5*10^{-6}*10^9 nanometres\\ \lambda = 0.5*10^{3}nanometres\\ \lambda =500 \;nm\\ \)

Just like Rom has said     

1/2 x 10^-6m = 500 x 10^-9m = 500 nm is the correct answer

sorry, what is the right answer? I did the math myself. but now that you posted this i'm not sure if i'm right or wrong.

Hi Rom, I am not trying to hijack your question, I started this ages ago.

I will post it, even though you have already given a good answer.

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Hi SmartMathMan :)

I think you are expected to memorize the speed of light. 

You can google it of course, which is what i just did.

\(1)~ \dfrac{26}{52}\dfrac{13}{51} = \dfrac{13}{102}\)

\(2) ~\dfrac{3}{30}\dfrac{2}{29} = \dfrac{1}{145}\)

\(3)~\dfrac{26}{52}\dfrac{13}{52}=\dfrac 1 8\)

\(4)~\dfrac 2 6 \dfrac 1 6 = \dfrac {1}{18}\)

\(deg(p(x))\neq deg(q(x))\Rightarrow deg((p+q)(x)) = \max(deg(p(x)),deg(q(x)))\\ \text{In this case }11>7\Rightarrow deg((p+q)(x))=11\)

If the two polynomials have the same degree it's possible that terms can cancel making the resulting

degree of the sum polynomial dependent upon the individual polynomials being summed.

Thx for explaining it. I'm not trying to call you out, I was just confused.