член: heureka

$\begin{array}{|rcll|} \hline A &=& x\cdot y \\ &=& x\cdot (4-3x^2) \\ &=& 4x-3x^3 \\\\ A' &=& 4-9x^2 \quad | \quad A' = 0 \\ 4-9x^2 &=& 0 \\ 9x^2 &=& 4 \quad | \quad \text{square root both sides} \\ 3x &=& 2 \\\\ \mathbf{x} &=& \mathbf{\dfrac{2}{3}} \quad | \quad A''=-18x =-18\cdot \dfrac{2}{3}=-12\ (A''<0 \text{ maximum}) \\ \hline \end{array}$

heureka28 авг. 2019 г.

+26398

Suppose the width of your index finger is 0.6 inches and the length of your arm is 31.9 inches.

Based on these measurements, what will be the angular width (in degrees) of your index finder held at arm’s length?

$\begin{array}{|rcll|} \hline \mathbf{\tan{\dfrac{\alpha}{2}} } &=& \mathbf{\dfrac{\dfrac{0.6}{2}}{31.9} } \\ \\ \tan{\dfrac{\alpha}{2}} &=& \dfrac{0.3}{31.9} \\ \\ \tan{\dfrac{\alpha}{2}} &=& 0.00940438871 \\ \\ \dfrac{\alpha}{2} &=& \arctan{ 0.00940438871 } \\\\ \dfrac{\alpha}{2} &=& 0.53881589788 \\ \\ \alpha &=& 2\cdot 0.53881589788 \\ \mathbf{\alpha} &=& \mathbf{1.07763179577^\circ} \\ \hline \end{array}$

heureka27 авг. 2019 г.

+26398

Let $\mathbf{ f(x) = (x^2 + 6x + 9)^{50} - 4x + 3 }$,
and let $\mathbf{ r_{1}, r_{2}, \ldots , r_{100} }$ be the roots of $\mathbf{f(x)}$.
Compute $\mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} }$.

$\begin{array}{|rcll|} \hline \mathbf{ f(x) } &=& \mathbf{(x^2 + 6x + 9)^{50} - 4x + 3} \quad | \quad x^2 + 6x + 9 = \left(x+3\right)^2 \\\\ f(x) &=& \Big(\left(x+3\right)^2\Big)^{50} - 4x + 3 \\ f(x) &=& (x+3)^{2\cdot 50} - 4x + 3 \\ f(x) &=& (x+3)^{100} - 4x + 3 \\ \hline \end{array} $

$\begin{array}{|lrcll|} \hline f(r_1) = 0: & \left(r_1+3\right)^{100} -4r_1 + 3 &=& 0 \\ & \left(r_1+3\right)^{100} &=& \mathbf{4r_1 - 3} \\\\ f(r_2) = 0: & \left(r_2+3\right)^{100} -4r_2 + 3 &=& 0 \\ & \left(r_2+3\right)^{100} &=& \mathbf{4r_2 - 3} \\ \\ f(r_3) = 0: & \left(r_3+3\right)^{100} -4r_3 + 3 &=& 0 \\ & \left(r_3+3\right)^{100} &=& \mathbf{4r_3 - 3} \\ \ldots & \ldots \\ f(r_{100}) = 0: & \left(r_{100}+3\right)^{100} -4r_{100} + 3 &=& 0 \\ & \left(r_{100}+3\right)^{100} &=& \mathbf{4r_{100} - 3} \\ \\ & (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\ \hline \end{array}$

Vieta theorem:

$\begin{array}{|lrcll|} \hline & \mathbf{0} &=& \mathbf{x^n+a_{n-1}x^{n-1}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & a_{n-1} &=& -\sum \limits_{k=1}^{n}r_k \\ & a_{n-1} &=& -(r_1+r_2+\ldots + r_n ) \\ \hline n = 100: & \mathbf{0} &=& \mathbf{x^{100}+a_{99}x^{99}+\ldots+a_2x^2+a_1x^1+a_0} \\\\ & \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \\ \hline \end{array} $

$\begin{array}{|rcll|} \hline (x+3)^{100} - 4x + 3 &=& 0 \\ \Big( \binom{100}{0}x^{100} + \binom{100}{1}x^{99}\cdot 3^1 + \ldots + \binom{100}{100} 3^{100} \Big) - 4x + 3 &=& 0 \\ \Big( x^{100} + \underbrace{300}_{=a_{99}}x^{99} + \ldots + 3^{100} \Big) - 4x + 3 &=& 0 \\\\ \mathbf{a_{99}} &=& \mathbf{300} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{a_{99}} &=& \mathbf{-(r_1+r_2+\ldots + r_{100} )} \quad | \quad a_{99} = 300 \\\\ 300 &=& -(r_1+r_2+\ldots + r_{100} )\\ \mathbf{r_1+r_2+\ldots + r_{100}} &=& \mathbf{-300} \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{4\cdot ( r_1+r_2+\ldots +r_{100} ) -3 \cdot 100} \\\\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& 4\cdot ( -300 ) -3 \cdot 100 \\ (r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} &=& -1200 - 300 \\ \mathbf{(r_{1} + 3)^{100} + (r_{2} + 3)^{100} + \ldots + (r_{100} + 3)^{100} } &=& \mathbf{-1500 } \\ \hline \end{array}$

Edited, thank you Rom !

heureka27 авг. 2019 г.

+26398

In a chess variant, a "lord" can move one space at a time, either upward, or to the right, or diagonally up and to the right.

How many ways are there for a lord to move from the bottom left to top right corner of the 8 by 8 chessboard?

$\begin{array}{|c|c|c|c|c|c|c|c|} \hline 1 & 15 & 113 & 575 & 2241 & 7183 & 19825 & \mathbf{48639} \\ \hline 1 & 13 & 85 & 377 & 1289 & 3653 & 8989 & 19825 \\ \hline 1 & 11 & 61 & 231 & 681 & 1683 & 3653 & 7183 \\ \hline 1 & 9 & 41 & 129 & 321 & 681 & 1289 & 2241 \\ \hline 1 & 7 & 25 & 63 & 129 & 231 & 377 & 575 \\ \hline 1 & 5 & 13 & 25 & 41 & 61 & 85 & 113 \\ \hline 1 & 3 & 5 & 7 & 9 & 11 & 13 & 15 \\ \hline \mathbf{1} & 1 & 1 & 1 & 1 & 1 & 1 & 1 \\ \hline \end{array} $

heureka26 авг. 2019 г.

+26398

Thank you, Melody !

heureka25 авг. 2019 г.

+26398

Find the equation of the asymptote of the graph of $r = \cos ( 2\theta) \sec( \theta)$.

I assume the function is a polar function:
https://www.wolframalpha.com/input/?i=polar+plot+r%3Dcos%282x%29sec%28x%29

The asymptote of the graph is $\mathbf{x = -1}$.

source: https://www.youtube.com/watch?v=7Ae3VHPyMMQ

\(\begin{array}{|rcll|} \hline \mathbf{r} &=& \mathbf{\cos (2\theta) \sec( \theta)} \\\\ r &=& \dfrac{\cos (2\theta)} {\cos( \theta)} \quad | \quad \theta=\dfrac{\pi}{2} \Rightarrow r\to \infty \\\\ && \text{Set $\alpha = \dfrac{\pi}{2}$ } \\\\ \dfrac{dr}{d\theta } &=& \dfrac{-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } {\cos^2(\theta)} \\\\ \dfrac{d\theta }{dr} &=& \dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} r^2\dfrac{d\theta}{dr}\\ \\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {\cos^2( \theta)} \left(\dfrac{\cos^2(\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) }\right) \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\theta)} {-2\sin(2\theta)\cos(\theta)+\cos(2\theta)\sin(\theta) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(2\dfrac{\pi}{2})} {-2\sin(2\dfrac{\pi}{2})\cos(\dfrac{\pi}{2})+\cos(2\dfrac{\pi}{2})\sin(\dfrac{\pi}{2}) } \\\\ p &=& \lim \limits_{\theta\to \dfrac{\pi}{2}} \dfrac{\cos^2(\pi)} {-2\sin(\pi)\cos(\dfrac{\pi}{2})+\cos(\pi)\sin(\dfrac{\pi}{2}) } \\\\ p &=& \dfrac{(-1)^2} {-2\cdot 0\cdot 0+(-1)(1)} \\\\ p &=& \dfrac{1} {-1} \\\\ \mathbf{p} &=& \mathbf{-1} \\ \hline \end{array}\)

$\begin{array}{|rcll|} \hline \mathbf{p} &=& \mathbf{r \sin(\alpha-\theta)} \quad | \quad r=-1,\ \alpha = \dfrac{\pi}{2} \\\\ -1 &=& r \sin(\dfrac{\pi}{2} -\theta) \\\\ -1 &=& r \cos( \theta) \\\\ \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} \quad | \quad \text{the asymptote of the graph }} \\ \hline \end{array}$

polar plot asymptote see: https://www.wolframalpha.com/input/?i=polar+plot+r%3D+-1%2Fcos%28x%29

Asymptote in Cartesian Coordinates:
To convert from Polar Coordinates (r,$\theta$) to Cartesian Coordinates (x,y) :
$x = r \times \cos(\theta) \\ y = r \times \sin(\theta)$

$\begin{array}{|rcll|} \hline \mathbf{r} &=&\mathbf{ -\dfrac{1}{\cos( \theta)} } \quad | \quad \cos{\theta}=\dfrac{x}{r} \\\\ r &=& -\dfrac{1}{ \dfrac{x}{r} } \\ r &=& -\dfrac{r}{ x } \\ xr &=& -r \quad | \quad :r \\\\ \mathbf{x} &=& \mathbf{-1 } \\ && \mathbf{\text{the asymptote of the graph in Cartesian Coordinates}} \\ \hline \end{array}$

hint: The function $r = \cos ( 2\theta) \sec( \theta)$ in cartesian coordinates: $(x^2+y^2) (1+x) = 2x^2 \quad | \quad r^2 = x^2+y^2,\ x = r \times \cos(\theta) ,\ y = r \times \sin(\theta) $

heureka25 авг. 2019 г.

+26398

Find the equation of the asymptote of the graph of $r = \cos 2 \theta \sec \theta$.

I assume the function is a polar function:

polar plot see: https://www.wolframalpha.com/input/?i=polar+plot+r%3Dcos%282x%29sec%28x%29

The asymptote of the graph is $\mathbf{x = -1}$.

Source: https://www.youtube.com/watch?v=7Ae3VHPyMMQ

polar plot asymtote see: https://www.wolframalpha.com/input/?i=polar+plot+r%3D+-1%2Fcos%28x%29

Asymptote in Cartesian Coordinates:

To convert from Polar Coordinates (r,θ) to Cartesian Coordinates (x,y) :
$x = r \times \cos( \theta ) \\ y = r \times \sin( \theta )$

heureka25 авг. 2019 г.

+26398

Thank you, CPhill !

heureka23 авг. 2019 г.

+26398

A six digit number begins with one.

But if the one put at the other end, the resulting six digit number is 3 times greater than the original number.

Find the number.

$\text{The six digit number begins with one: $\mathbf{1rstuv}$ } \\ \text{The resulting six digit number ends with one: $\mathbf{rstuv1}$ } $

$\begin{array}{|lcll|} \hline \mathbf{rstuv1} = 3\cdot \mathbf{1rstuv} \\ \hline \end{array}$

$\small{ \begin{array}{|rcll|} \hline r\cdot 10^5+s\cdot 10^4+t\cdot 10^3+u\cdot 10^2+v\cdot 10+1 &=& 3\cdot (1\cdot 10^5+r\cdot 10^4+s\cdot 10^3+t\cdot 10^2+u\cdot 10+v ) \\\\ r\cdot 10^5+s\cdot 10^4+t\cdot 10^3+u\cdot 10^2+v\cdot 10+1 &=& 3\cdot 10^5+ r\cdot3\cdot 10^4+ s\cdot 3\cdot 10^3+t\cdot 3\cdot 10^2+ u\cdot 3\cdot 10+3\cdot v \\\\ r(\cdot 10^5-\cdot3\cdot 10^4) +s(\cdot 10^4-\cdot 3\cdot 10^3) +t(\cdot 10^3-\cdot 3\cdot 10^2 ) +u(\cdot 10^2-\cdot 3\cdot 10 ) +v(\cdot 10+1-3) &=& 3\cdot 10^5-1 \\\\ 70000r+7000s+700t+70u+7v &=& 299999 \\ 7\cdot r\cdot 10^4 +7\cdot s\cdot 10^3 +7\cdot t\cdot 10^2 +7\cdot u\cdot 10 +7v &=& 299999 \\ 7\cdot \left(r\cdot 10^4+s\cdot 10^3 +t\cdot 10^2 +u\cdot 10 +v \right) &=& 299999 \quad | \quad : 7 \\ r\cdot 10^4+s\cdot 10^3 +t\cdot 10^2 +u\cdot 10 +v &=& 42857 \\ \mathbf{ r s t u v } &=& \mathbf{ 42857 } \\ \hline \end{array} }$

The number is $\mathbf{142857}$

Check: $3\cdot 142857 = 428571$

heureka23 авг. 2019 г.

+26398

If two of the roots of $2x^3 + 8x^2 - 120x + k = 0$ are equal,

find the value of k given that k is positive.

An alternative solution:

so $y=0$ and $y'=0$

$\begin{array}{|rcll|} \hline y = 0 &=& 2x^3 + 8x^2 - 120x + k \\ y'= 0 &=& 6x^2 + 16x - 120 \\ \hline 6x^2 + 16x - 120 &=& 0 \quad & | \quad : 2 \\ 3x^2 + 8x - 60 &=& 0 \\\\ x&=& \dfrac{ -8\pm \sqrt{64-4\cdot 3\cdot(-60)} } {2\cdot 3} \\ x&=& \dfrac{ -8\pm \sqrt{64+720} } {6} \\ x&=& \dfrac{ -8\pm \sqrt{784} } {6} \\ x&=& \dfrac{ -8\pm 28 } {6} \\\\ x_1 &=& \dfrac{ -8+ 28 } {6} \\ x_1 &=& \dfrac{20} {6} \\ \mathbf{x_1} &=& \mathbf{\dfrac{10} {3}} \\\\ x_2 &=& \dfrac{ -8- 28 } {6} \\ x_2 &=& \dfrac{-36} {6} \\ \mathbf{x_2} &=& \mathbf{-6} \\ \hline \end{array} $

$\begin{array}{|l|rcll|} \hline x_1 = \dfrac{10} {3}: & 2\cdot \left(\dfrac{10} {3}\right)^3 + 8\cdot \left(\dfrac{10} {3}\right)^2 - 120\cdot \left(\dfrac{10} {3}\right) + k &=& 0 \\ & -237.\overline{037} + k &=& 0 \\ & \mathbf{k} &=& \mathbf{237.\overline{037}} \ \checkmark\quad & | \quad k>0! \\ \hline \end{array} $

$\begin{array}{|l|rcll|} \hline x_2 =-6: & 2 \left(-6\right)^3 + 8 \left(-6\right)^2 - 120 \left(-6\right) + k &=& 0 \\ & -432+288+720+ k &=& 0 \\ & 576+ k &=& 0 \\ & \mathbf{k} &=& \mathbf{-576} \quad & | \quad k<0\ \text{ no solution!}\\ \hline \end{array} $

heureka23 авг. 2019 г.

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