heureka

What is the wavelength of a wave having a frequency of 3.76 multiplied by 10^14 s^-1?

(please go show ALL work to get to the answer)

\(\begin{array}{llll} \text{Here is the formula of the wavelength } (\lambda) \\ \text{depending on } v (\text{velocity of the wave }) \text{ and } f (\text{frequency}). \\ \end{array} \)

\( \begin{array}{rrr} && \mathbf{\lambda = \dfrac{v}{f} } \\ \end{array}\)

\(\begin{array}{llll} \mathbf{\lambda} & \text{ (Lambda) is the wavelength } \\ \mathbf{v} & \text{ is the velocity of the wave (default is velocity of light in vacuum: } 299792458 \frac{m}{s} ) \\ \mathbf{f} & \text{ is the frequency } \\ \end{array}\)

\(\begin{array}{|rcll|} \hline \lambda &=& \dfrac{v}{f} \quad & | \quad v=299792458 \frac{m}{s} \qquad f = 3.76\ \cdot 10^{14} \frac{1}{s} \\\\ &=& \dfrac{299792458 \frac{m}{s}} {3.76\ \cdot 10^{14} \frac{1}{s}} \\\\ &=& \dfrac{299792458 } {3.76\ \cdot 10^{14} } \cdot \frac{m}{s} \cdot \frac{s}{1} \\\\ &=& \dfrac{299792458 } {3.76\ \cdot 10^{14} } \ m \\\\ &=& \dfrac{2997.92458\cdot 10^{5} } {3.76\ \cdot 10^{14} } \ m \\\\ &=& \dfrac{2997.92458} {3.76}\cdot 10^{5-14} \ m \\\\ &=& \dfrac{2997.92458} {3.76}\cdot 10^{-9} \ m \quad & | \quad nm = 10^{-9} \ m \\\\ &=& \dfrac{2997.92458} {3.76}\ nm \\\\ &\mathbf{=}& \mathbf{797.320367021 \ nm} \\ \hline \end{array}\)

The wavelength of a wave is 797 nm

Is this number prime or composite? If not prime, can you find its prime factors? Thank you.

8555 551665 556208 555624 588899 297778 663111 342433 333481 111129

\(\begin{array}{|rcll|} \hline 8555 &=& 5 * 29 * 59 \\ 551665 &=& 5 * 19 * 5807 \\ 556208 &=& 2^4 * 34763 \\ 555624 &=& 2^3 * 3^2 * 7717 \\ 588899 &=& 127 * 4637 \\ 297778 &=& 2 * 13^2 * 881 \\ 663111 &=& 3^2 * 73679 \\ 342433 &=& 7 * 13 * 53 * 71 \\ 333481 &=& 37 * 9013 \\ 111129 &=& 3 * 17 * 2179 \\ \hline \end{array}\)

6 × 6! + 7 × 7! + 8 × 8! + 9 × 9! + 10 × 10! + 11 × 11! + 12 × 12! + 13 × 13! + 14 × 14! = a! - b!

a + b= ?

\(\begin{array}{|rcll|} \hline && n\times n! \\ &=& (n+1-1)\times n! \\ &=& [(n+1)-1]\times n! \\ &=& (n+1)\times n! -1\times n! \\ &=& n!\times (n+1) -n! \\ &=& (n+1)! -n! \\\\ &&\mathbf{ n\times n! = (n+1)! -n! } \\ \hline \end{array}\)

\(\begin{array}{|rcrcr|} \hline 6\times 6! &=& 7!- 6! \\ 7\times 7! &=& 8!- 7! \\ 8\times 8! &=& 9!- 8! \\ 9\times 9! &=& 10!- 9! \\ 10\times 10! &=& 11!-10! \\ 11\times 11! &=& 12!-11! \\ 12\times 12! &=& 13!-12! \\ 13\times 13! &=& 14!-13! \\ 14\times 14! &=& 15!-14! \\ \hline \text{sum} &=& 7! &\mathbf{-}& \mathbf{6!} \\ &+& 8! &-& 7! \\ &+& 9! &-& 8! \\ &+& 10! &-& 9! \\ &+& 11! &-& 10! \\ &+& 12! &-& 11! \\ &+& 13! &-& 12! \\ &+& 14! &-& 13! \\ &+& 15! &-& 14! \\\\ &=& \not{ {\color{green}7!}} && -6! \\ && {\color{red}{-}}\not{ {\color{red}7!}} && {\color{green}{+}} \not{ {\color{green}8!}} \\ && {\color{green}{+}}\not{ {\color{green}9!}} && {\color{red}{-}}\not{ {\color{red}8!}} \\ && {\color{red}{-}}\not{ {\color{red}9!}} && {\color{green}{+}} \not{ {\color{green}10!}} \\ && {\color{green}{+}} \not{ {\color{green}11!}} && {\color{red}{-}}\not{ {\color{red}10!}} \\ && {\color{red}{-}}\not{ {\color{red}11!}} && {\color{green}{+}} \not{ {\color{green}12!}} \\ && {\color{green}{+}} \not{ {\color{green}13!}} && {\color{red}{-}}\not{ {\color{red}12!}} \\ && {\color{red}{-}}\not{ {\color{red}13!}} && {\color{green}{+}} \not{ {\color{green}14!}} \\ && \mathbf{+15!} && {\color{red}{-}}\not{ {\color{red}14!}} \\\\ &\mathbf{=}& \mathbf{15! - 6!} \\ \hline \end{array} \)

so a = 15 and b = 6

a+b = 15+6 = 21

what is pi to the power of pi

\(\pi^{\pi} = 36.46215960720791177099082602269212366636550840222881873870...\)

108251299468315690105155229 how do i prime factorize this?

factor  \(108\ 251\ 299\ 468\ 315\ 690\ 105\ 155\ 229\)

Result: \(4466771201309 \times 24234798378881\) (2 distinct prime factors)

solve for x

8e^x=2297

\(\begin{array}{|rcll|} \hline 8e^x &=& 2297 \quad & | \quad : 8 \\ e^x &=& \frac{2297}{8} \\ e^x &=& 287.125 \quad & | \quad \ln() \text{ both sides}\\ ln( e^x ) &=& ln( 287.125) \\ ln( e^x ) &=& 5.65991766101 \quad & | \quad \ln( e^x ) = x\ln(e) \\ xln(e) &=& 5.65991766101 \quad & | \quad \ln(e) = 1 \\ x &=& 5.65991766101 \\ \hline \end{array}\)

2sin(2x)/1-sin^2(x)=5 can convert to tan(x)=1,25

a) Show that the equation \(\tfrac{2\sin(2x)} { 1-\sin^2(x) } = 5\) can convert  to tan(x)=1,25

\(\begin{array}{|rcll|} \hline \dfrac{2\sin(2x)} { 1-\sin^2(x) } &=& 5 \quad & | \quad : 2 \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=& \frac{5}{2} \\\\ \dfrac{\sin(2x)} { 1-\sin^2(x) } &=&2.5 \quad & | \quad \sin(2x) = 2\sin(x)\cos(x) \\\\ \dfrac{2\sin(x)\cos(x)} { 1-\sin^2(x) } &=& 2.5 \quad & | \quad 1-\sin^2(x) = \cos^2(x) \\\\ \dfrac{2\sin(x)\cos(x)} { \cos^2(x) } &=& 2.5 \quad & | \quad : \cos(x) \\\\ \dfrac{2\sin(x)} { \cos(x) } &=& 2.5 \quad & | \quad : 2 \\\\ \dfrac{\sin(x)} { \cos(x) } &=& 1.25 \quad & | \quad \dfrac{\sin(x)} { \cos(x) } = \tan(x) \\\\ \tan(x) &=& 1.25 \\ \hline \end{array}\)

b) Solv the ecvation tan(x)=1,25 fully

\(\begin{array}{|rcll|} \hline \tan(x) &=& 1.25 \quad & | \quad \arctan() \text{ both sides }\\ x &=& \arctan(1.25) + n\cdot \pi \quad & \quad n \in Z \\ x &=& 51.3401917459^{\circ} + n\cdot \pi \quad & \quad n \in Z \\ \hline \end{array}\)

Going To The Picnic

Let p = people
Let w = wagon

\(\begin{array}{|lrcll|} \hline (1) & p &=& (w-10)(\frac{p}{w}+1) \\ & p &=& p + w - 10\frac{p}{w}-10 \\ & 0 &=& w - 10\frac{p}{w}-10 \\ & 10\frac{p}{w} &=& w -10 \\ &\mathbf{ p } & \mathbf{=} & \mathbf{\frac{w}{10} (w -10) } \\\\ (2) & p &=& (w-25)(\frac{p}{w}+3) \\ & p &=& p+3w - 25\frac{p}{w}-75 \\ & 0 &=& 3w - 25\frac{p}{w}-75 \\ & 25\frac{p}{w} &=& 3w-75 \\ &\mathbf{ p } & \mathbf{=} & \mathbf{\frac{w}{25} (3w-75) } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline p= \mathbf{\frac{w}{10} (w -10) } &\mathbf{=}& \mathbf{\frac{w}{25} (3w-75) } \\ \frac{w}{10} (w -10) &=& \frac{w}{25} (3w-75) \\ 25 (w -10) &=& 10 (3w-75) \\ 5 (w -10) &=& 2 (3w-75) \\ 5w-50 &=& 6w-150 \\ \mathbf{ w } &\mathbf{ =}& \mathbf{ 100 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline \mathbf{ p } & \mathbf{=} & \mathbf{\frac{w}{10} (w -10) } \\ &=& \frac{100}{10} (100 -10) \\ &=& 10\times 90 \\ \mathbf{ p } &\mathbf{ =}& \mathbf{ 900 } \\ \hline \end{array}\)

Thank you Melody and CPhill

Algebra

Compute the sum \(\mathbf{\frac{2}{1 \cdot 2 \cdot 3} + \frac{2}{2 \cdot 3 \cdot 4} + \frac{2}{3 \cdot 4 \cdot 5} + \cdots}\)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ ? } } \\\\ \begin{array}{|lcll|} \hline s_n = \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\ \hline \end{array} \\ \end{array}\\\)

Formula:

\(\begin{array}{|lcll|} \hline \text{in general}:\ \frac{1}{n(n+d)} = \frac{1}{d}\left(\frac{1}{n}- \frac{1}{n+d} \right) \\ \hline \\ \begin{array}{lrcll} \text{we need}: & \dfrac{1}{(n+1)(n+2)} &=& \dfrac{1}{n+1}-\dfrac{1}{n+2} \\ & \dfrac{1}{n(n+1)} &=& \dfrac{1}{n}-\dfrac{1}{n+1} \\ & \dfrac{1}{n(n+2)} &=& \dfrac{1}{2} \left( \dfrac{1}{n}-\dfrac{1}{n+2} \right) \\ \end{array} \\ \hline \end{array}\)

we rearrange:

\(\begin{array}{|rcll|} \hline \dfrac{2}{n \cdot (n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{(n+1) \cdot (n+2)} \\\\ &=& \dfrac{2}{n}\times \left( \dfrac{1}{n+1}-\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n}\times \dfrac{1}{n+1} - \dfrac{2}{n}\times \dfrac{1}{n+2} \\\\ &=& 2\times \left(\dfrac{1}{n}-\dfrac{1}{n+1} \right)- 2\times \dfrac{1}{2} \times \left(\dfrac{1}{n} -\dfrac{1}{n+2} \right) \\\\ &=& \dfrac{2}{n} - \dfrac{2}{n+1} -\dfrac{1}{n} + \dfrac{1}{n+2} \\\\ \mathbf{\dfrac{2}{n \cdot (n+1) \cdot (n+2)} } & \mathbf{=} & \mathbf{ \dfrac{1}{n} - \dfrac{2}{n+1} + \dfrac{1}{n+2} } \\ \hline \end{array}\)

telescoping series

\(\begin{array}{|rcll|} \hline s_n &=& \mathbf{\dfrac{1}{1}} &\mathbf{-}& \mathbf{\dfrac{2}{2}} &\color{red}+& \color{red}\dfrac{1}{3} \\\\ &\mathbf{+}& \mathbf{\dfrac{1}{2}} &\color{red}-& \color{red}\dfrac{2}{3} &\color{blue}+& \color{blue}\dfrac{1}{4} \\\\ &\color{red}+& \color{red}\dfrac{1}{3} &\color{blue}-& \color{blue}\dfrac{2}{4} &\color{red}+& \color{red}\dfrac{1}{5} \\\\ &\color{blue}+& \color{blue}\dfrac{1}{4} &\color{red}-& \color{red}\dfrac{2}{5} &\color{green}+& \color{green}\dfrac{1}{6} \\\\ && \ldots \\\\ &+\color{red}& \color{red}\dfrac{1}{n-2} &\color{green}-& \color{green}\dfrac{2}{n-1} &\color{red}+& \color{red}\dfrac{1}{n} \\\\ &\color{green}+& \color{green}\dfrac{1}{n-1} &\color{red}-& \color{red}\dfrac{2}{n} &\mathbf{+}& \mathbf{\dfrac{1}{n+1}} \\\\ &\color{red}+& \color{red}\dfrac{1}{n} &\mathbf{-}& \mathbf{\dfrac{2}{n+1}} &\mathbf{+}& \mathbf{\dfrac{1}{n+2}} \\ \hline \end{array}\)

The part of each term cancelling with part of the next two diagonal terms:

Example:

\(\begin{array}{|lcll|} \hline \frac{1}{3}-\frac{2}{3}+\frac{1}{3} = 0 \\ \frac{1}{4}-\frac{2}{4}+\frac{1}{4} = 0 \\ \frac{1}{5}-\frac{2}{5}+\frac{1}{5} = 0 \\ \ldots \\ \frac{1}{n}-\frac{2}{n} + \frac{1}{n} = 0 \\ \hline \end{array}\)

So \(s_n\) is, we have all black terms left :

\(\begin{array}{|rcll|} \hline s_n &=& \dfrac{1}{1}-\dfrac{2}{2}+\dfrac{1}{2} + \dfrac{1}{n+1} - \dfrac{2}{n+1} + \dfrac{1}{n+2} \\\\ \mathbf{s_n} &\mathbf{=}& \mathbf{\dfrac{1}{2} - \dfrac{1}{n+1} + \dfrac{1}{n+2}} \\ \hline \end{array} \)

 \(\lim \limits_{n\to \infty} { \dfrac{1}{n+1}} = 0 \quad \text{ and } \quad \lim \limits_{n\to \infty} { \dfrac{1}{n+2} } = 0 \)

\( \begin{array}{|rcll|} \hline \lim \limits_{n\to \infty} s_n &=& \dfrac{1}{2} - 0 + 0 \\ &=& \dfrac{1}{2} \\ \hline \end{array} \)

\(\begin{array}{lcll} \mathbf{ \dfrac{2}{1 \cdot 2 \cdot 3} + \dfrac{2}{2 \cdot 3 \cdot 4} + \dfrac{2}{3 \cdot 4 \cdot 5} + \dfrac{2}{4 \cdot 5 \cdot 6} + \cdots \ + \dfrac{2}{n \cdot (n+1) \cdot (n+2)} + \cdots =\ \mathbf{ \dfrac{1}{2} } } \\ \end{array}\\\)