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Am I the first one to discover this solution?

(Edited on 5 May 2022: Probably not, this is basic calculus.)

yes its right

ABCDE = 42857.

This is a really common question in Hong Kong's Maths competition and I have to remember this special number so that once I see the question I know it is related to 142857.

UM THIS IS NOT A MATH QUESTION

This is called asymptotic notation.

(((5000*1.08^8))(*1.10^10)*1.07^6)

(((5000*1.08^8))(1.10^10)*1.07^6) = 36023.8035764331287797

− 6+( − 3 − 3)2÷ 3

There should be a times sign between the ) and the 2.

= − 6+( − 3 − 3)*2÷ 3

= − 6+( -6)*2÷ 3

= − 6+-12÷ 3

= − 6+-4

= -6-4

= -10

a d**k

An MaxWong,

thanks for the hint.
I did not want to have a term with "lg" in the result.

In addition, I was "persecuted" and had to get ready quickly.

Greeting asinus :- ) !

Find the reciprocal of

\(\frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \)    

( The letter "i" refers to the complex number)

Formula:

\(\begin{array}{|rcll|} \hline z &=& x+i\cdot y \\ \frac{1}{z} &=& \frac{x}{x^2+y^2} - i\cdot \frac{y}{x^2+y^2} \\ \hline \end{array}\)

We have:

\(x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2}\)

so the reciprocal is:

\(\begin{array}{|rcll|} \hline z &=& x+i\cdot y \quad & | \quad x =\frac{\sqrt{2}} {2} \quad \text{and} \quad y = -\frac{\sqrt{2}} {2} \\ z &=& \frac{\sqrt{2}} {2} - i\cdot \frac{\sqrt{2}} {2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ (\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{(\frac{\sqrt{2}} {2})^2+(-\frac{\sqrt{2}} {2})^2} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ \frac24+\frac24} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{\frac24+\frac24} \\ \frac{1}{z} &=& \frac{\frac{\sqrt{2}} {2}}{ 1} - i\cdot \frac{-\frac{\sqrt{2}} {2}}{1} \\ \mathbf{\dfrac{1}{z}} &\mathbf{=}& \mathbf{\dfrac{\sqrt{2}} {2} + i\cdot \dfrac{\sqrt{2}} {2} } \\ \hline \end{array}\)

7log(3x)+3log(5x)=15 what is x

\(7(lg3+lgx)+3(lg5+lgx)=15\)

\(\large x=9.04324132413\)

\(\large 10.03412+4.96588=15\)

  !

\(\dfrac{\left(3x^2y^3\right)\left(2xy^2\right)^2}{-2x^2y}\\ =\dfrac{\left(3\cdot 2\right)\cdot\left(x^2\cdot x^2\right)\cdot\left(y^3\cdot y^4\right)}{-2x^2y}\\ =\dfrac{6x^4y^7}{-2x^2y}\\ =\left(\dfrac{6}{-2}\right)\left(\dfrac{x^4}{x^2}\right)\left(\dfrac{y^7}{y}\right)\\ =-3x^2y^6\)

~The smartest cookie in the world.

To asinus:

The 6th step:

\(\lg x^{10} = 15 - \lg 273375\\ 10 \lg x = 15 - \lg 273375\\ x = 10^{\frac{15-\lg 277375}{10}}\)

Because the right hand side is already a numerical value and the left hand side is x, so this can be the solution. 

No need to do all that..... LOL 

~The smartest cookie in the world

\(\quad h\\ = \dfrac{18\tan 9 \tan 3.5}{\tan 9}-\tan 3.5\\ =18\tan 3.5 - \tan 3.5\\ = 17\tan 3.5\)

7log(3x)+3log(5x)=15 what is x

i assume lg is the 10 - logarithm

\(7(lg3+lgx)+3(lg5+lgx)=15\)

\(7lg3+7lgx+3lg5+3lgx=15\)

\(7lgx+3lgx=15-(7lg3+3lg5)\)

\(lgx^7+lgx^3=15-lg \ 273375\)

\(lg(x^7\times x^3)=15-lg \ 273375\)

\(lgx^{10}=15-lg \ 273375\)

\(lgx^{10}+lg273375=15\)

\(lg \ (273375 \ x^{10})=15\)

\(273375 x^{10}=10^{15}\)

\(\large x^{10}=\frac{10^{15}}{273375}\)

\(\Large x=\sqrt[10]{\frac{10^{15}}{273375}} \)

\(\large x=9.04324132413\)

  !

7log(3x)+3log(5x)=15 what is x

i assume log is the natural logarithm

\(\begin{array}{|rcll|} \hline 7\cdot log(3x)+3 \cdot log(5x) &=& 15 \\ 7\cdot (~log(3)+log(x) ~)+3 \cdot (~log(5)+log(x) ~) &=& 15 \\ 7\cdot log(3) + 7\cdot log(x) +3\cdot log(5) + 3\cdot log(x)&=& 15 \\ 10\cdot log(x) + 7\cdot log(3) +3\cdot log(5) &=& 15 \\ 10\cdot log(x) + log(3^7) + log(5^3) &=& 15 \\ 10\cdot log(x) + log(3^7\cdot 5^3) &=& 15 \quad & | \quad -log(3^7\cdot 5^3) \\ 10\cdot log(x) &=& 15 -log(3^7\cdot 5^3) \quad & | \quad : 10 \\ log(x) &=& \frac{ 15 -log(3^7\cdot 5^3) } {10 } \\ log(x) &=& \frac{ 15}{10} - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \frac{ \log(3^7\cdot 5^3) } {10 } \\ log(x) &=& 1.5 - \log(~(3^7\cdot 5^3)^{\frac{1}{10}}~) \\ log(x) &=& 1.5 - \log(3^\frac{7}{10}\cdot 5^\frac{3}{10}) \\ log(x) &=& 1.5 - \log(3^{0.7}\cdot 5^{0.3}) \quad & | \quad e^{()} \\ x &=& e^{1.5 - \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot e^ {- \log(3^{0.7}\cdot 5^{0.3}) } \\ x &=& e^{1.5} \cdot \frac{ 1 } { e^{\log(3^{0.7}\cdot 5^{0.3}) } } \\ x &=& e^{1.5} \cdot \frac{ 1 } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ e^{1.5} } { 3^{0.7}\cdot 5^{0.3} } \\ x &=& \frac{ 4.48168907034 } {2.15766927997\cdot 1.62065659669 } \\ x &=& \frac{ 4.48168907034 } {3.49684095207 } \\ \mathbf{x} & \mathbf{=} & \mathbf{1.28163938016178...} \\ \hline \end{array}\)

(3x2y3)(2xy2)2

____________

       -2x2y

\(\Large \frac{(3x^2y^3)(2xy^2)^2}{-2x^2y}=\frac{(3x^2y^3)(4x^2y^4)}{-2x^2y}=-6x^2y^6\)

  !

Lovepreet is selling tickets for the school play. She sold 10 more adult tickets than child tickets, and she sold twice as many senior tickets as children tickets. adult tickets cost $5, Senior tickets cost $3, and child tickets cost $2. If Lovepreet made $700, determine how many for each ticket she sold.

child tickets = x

\(\$5(x+10)+\$2\times x+\$3\times2x=\$700\) [divided by $;

\(5x+50+2x+6x=700\)     [ alternate sides and signs                                                                                                              add 

\(13x=650\)                                               [ divided by 13

\(x=50\)

\(Lovepreet \ sold \ 50 \ tickets \ for \ children, \)

\( 60 \ tickets \ for \ adults \ and \ 100 \ tickets \ for \ seniors.\)

  !

this is from that big red book

h= [18(tan9)*tan(3.5)]/tan(9)-tan(3.5)

\(h=\frac{18(tan 9)(tan3.5)}{tan 9}-tan3.5\)

deg    \(h=\frac{18(tan 9)(tan3.5)}{tan 9}-tan3.5=1.03976454256\)

rad     \(h=\frac{18(tan 9)(tan3.5)}{tan 9}-tan3.5=6.36795588271\) 

  !