heureka

The length of a rectangular garden is double it's width.

The area of the garden is 162 m squared.

If the garden wishes to construct a fence diagonally across the garden,

how much fencing(in meters) will be required?

round your answer to the nearest meter.

Let d = diagonal

Let A = 162 m²

Let L = length

Let w = width

\(\begin{array}{|rcll|} \hline A &=& L\cdot w \quad & | \quad L = 2w \\ A &=& 2w\cdot w \\ A &=& 2w^2 \\ A &=& 2w^2 \quad & | \quad :2 \\ \frac{A}{2} &=& w^2 \\ \mathbf{w^2} & \mathbf{=} & \mathbf{\frac{A}{2}} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline d &=& \sqrt{L^2+w^2} \quad & | \quad L = 2w \\ d &=& \sqrt{(2w)^2+w^2} \\ d &=& \sqrt{4w^2+w^2} \\ d &=& \sqrt{5w^2} \quad & | \quad w^2=\frac{A}{2} \\ d &=& \sqrt{5\cdot \frac{A}{2}} \quad & | \quad A = 162\ m^2 \\ d &=& \sqrt{5\cdot \frac{162}{2}} \\ d &=& \sqrt{5\cdot 81} \quad & | \quad 81 = 9^2\\ d &=& \sqrt{5\cdot 9^2} \\ \mathbf{d} &\mathbf{=}& \mathbf{9\cdot \sqrt{5}\ m} \\ d &=& 20.1246117975\dots\ m \\ d &\approx & 20\ m \\ \hline \end{array} \)

how much fencing(in meters) will be required?  \(20\ m\)

\(\begin{array}{|rcll|} \hline d &=& \sqrt{5w^2} \\ d^2 &=& 5w^2 \\ 5w^2 &=& d^2 \quad & | \quad : 5\\ w^2 &=& \frac{d^2}{5} \\ w &=& \frac{d}{\sqrt{5}} \quad & | \quad 9\cdot \sqrt{5} \\ w &=& \frac{9\cdot \sqrt{5}}{\sqrt{5}} \\ \mathbf{w} &\mathbf{=}& \mathbf{9\ m} \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline L &=& 2w \quad & | \quad w = 9 \\ L &=& 2\cdot 9\\ \mathbf{L} &\mathbf{=}& \mathbf{18\ m} \\ \hline \end{array}\)

You mean that there is a maximum of 8 steps if the element exists in the array,

but 9 steps if the element does not exist?

Hello Melody!

YES!

\(\begin{array}{|lrcll|} \hline & \text{We search number 1:} & & \text{set } \{1,\dots,193 \}\\\\ (1) & 1 > \lfloor \frac{1+193}{2} \rfloor = 97 & no & \text{new set } \{1,\dots,97 \}\\ (2) & 1 > \lfloor \frac{1+97}{2} \rfloor = 49 & no & \text{new set } \{1,\dots,49 \}\\ (3) & 1 > \lfloor \frac{1+49}{2} \rfloor = 25 & no & \text{new set } \{1,\dots,25 \}\\ (4) & 1 > \lfloor \frac{1+25}{2} \rfloor = 13 & no & \text{new set } \{1,\dots,13 \}\\ (5) & 1 > \lfloor \frac{1+13}{2} \rfloor = 7 & no & \text{new set } \{1,\dots,7 \}\\ (6) & 1 > \lfloor \frac{1+7}{2} \rfloor = 4 & no & \text{new set } \{1,\dots,4 \}\\ (7) & 1 > \lfloor \frac{1+4}{2} \rfloor = 2 & no & \text{new set } \{1,2 \}\\ (8) & 1 > \lfloor \frac{1+2}{2} \rfloor = 1 & no & \text{we have found number 1 in the last set}\\ \hline \end{array} \\ \begin{array}{|lrcll|} \hline & \text{We search number 1.5:} & & \text{set } \{1,\dots,193 \}\\\\ (1) & 1.5 > \lfloor \frac{1+193}{2} \rfloor = 97 & no & \text{new set } \{1,\dots,97 \}\\ (2) & 1.5 > \lfloor \frac{1+97}{2} \rfloor = 49 & no & \text{new set } \{1,\dots,49 \}\\ (3) & 1.5 > \lfloor \frac{1+49}{2} \rfloor = 25 & no & \text{new set } \{1,\dots,25 \}\\ (4) & 1.5 > \lfloor \frac{1+25}{2} \rfloor = 13 & no & \text{new set } \{1,\dots,13 \}\\ (5) & 1.5 > \lfloor \frac{1+13}{2} \rfloor = 7 & no & \text{new set } \{1,\dots,7 \}\\ (6) & 1.5 > \lfloor \frac{1+7}{2} \rfloor = 4 & no & \text{new set } \{1,\dots,4 \}\\ (7) & 1.5 > \lfloor \frac{1+4}{2} \rfloor = 2 & no & \text{new set } \{1,2 \}\\ (8) & 1.5 > \lfloor \frac{1+2}{2} \rfloor = 1 & yes & \text{new set } \{ 2 \}\\ (9) & 1.5 = 2 & no & \text{we have not found number 1.5 in the last set}\\ \hline \hline \end{array}\)

Thanks Heureka but can you explain to me why you have to add the 1 ?

Hello Melody!

The last compare on equality, if not equal, the name does not exist in the array.

Compute how many steps binary search would take to find an item in arrays of various sizes.

Let n = size of the array

Let c = comparisons in the worse case

\(\begin{array}{|rcll|} \hline c &=& 1 + \log_2(n) \qquad & | \qquad \log_2(n) = \dfrac{ \ln(n) }{\ln(2)} \\ c &=& 1 + \dfrac{ \ln(n) }{\ln(2)} \\\\ n &=& 193\\ c &=& 1 + \dfrac{ \ln(193) }{\ln(2)} \\\\ c &=& 1 + \dfrac{ 5.26269018890 }{0.69314718056} \\\\ c &=& 1 + 7.59245703727 \\ c &=& 8.59245703727 \\ c &\approx& 9 \\ \hline \end{array} \)

No more than 9.

A rock is thrown off of the edge of a 9.8 meter high cliff with an initial horizontal velocity of 8.9 m/s.

It will land ____ m from the base of the vertical cliff.

Let d_x = horizotal distance

Let d_y = vertical distance

\(\begin{array}{|rcll|} \hline d_y &=& \frac12 \cdot g\cdot t^2 \qquad & | \qquad d_y = 9.8\ m \qquad g = 9.8\ \frac{m}{s^2}\\ 9.8\ m &=& \frac12 \cdot 9.8\ \frac{m}{s^2}\cdot t^2 \qquad & | \qquad : 9.8\\ 1\ m &=& \frac12 \cdot\frac{m}{s^2}\cdot t^2 \\ 1 &=& \frac12 \cdot\frac{1}{s^2}\cdot t^2 \qquad & | \qquad *2\\ 2s^2 &=& t^2 \\ \sqrt{2}\ s &=& t \\ \mathbf{t} & \mathbf{=} & \mathbf{\sqrt{2}\ s} \\ \hline \end{array} \)

The time the rock will land is \(t = \sqrt{2}\ s \)

\(\begin{array}{|rcll|} \hline d_x &=& v_0\cdot t \qquad & | \qquad v_0 = 8.9\ \frac{m}{s} \qquad t = \sqrt{2}\ s \\ d_x &=& 8.9\ \frac{m}{s}\cdot \sqrt{2}\ s \\ d_x &=& 8.9 \cdot \sqrt{2}\ m \\ d_x &=& 8.9 \cdot 1.41421356237\ m \\ d_x &=& 12.5865007051\dots \ m \\ \hline \end{array} \)

It will land 12.59 m from the base of the vertical cliff.

Wie kann ich den Term

\(\dbinom{n}{k} + \dbinom{n}{k+1}\)

vereinfachen?

\(\begin{array}{|rcll|} \hline \dbinom{n}{k} + \dbinom{n}{k+1} &=& \dbinom{n+1}{k+1}\\\\ \text{ oder }\\\\ \dbinom{n}{k} + \dbinom{n}{k+1} &=& \dfrac{n+1}{k+1} \cdot \dbinom{n}{k} \\ \hline \end{array}\)

7 über 3?

\(\begin{array}{|rcll|} \hline \binom{7}{ 3} &=&\frac{7}{3}\cdot \frac{6}{2}\cdot \frac{5}{1} \\ &=&\frac{7}{3}\cdot 3\cdot \frac{5}{1} \\ &=&7 \cdot \frac{5}{1} \\ &=&7 \cdot 5 \\ &=& 35\\ \hline \end{array}\)

solve for x

In a triangle the sum of the angles are \(180^{\circ}\)

So we have: 

\(\begin{array}{|rcll|} \hline 180^{\circ} &=& 90^{\circ} + (2x-10)^{\circ}+ ( x+25)^{\circ} \quad &|\quad -90^{\circ}\\ 180^{\circ} -90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90^{\circ} &=& (2x-10)^{\circ}+ ( x+25)^{\circ} \\ 90 &=& (2x-10) + ( x+25) \\ 90 &=& 2x-10 + x+25 \\ 90 &=& 2x+x -10 +25 \\ 90 &=& 3x +15 \quad &|\quad :3 \\ 30 &=& x +5 \quad &|\quad -5 \\ 30-5 &=& x \\ 25 &=& x \\ \mathbf{x} & \mathbf{=} & \mathbf{25} \\\\ \text{First angle: } (2x-10)^{\circ} &=& (2\cdot 25 -10)^{\circ} \\ &=& (50-10)^{\circ} \\ &=& 40^{\circ} \\\\ \text{Second angle: } (x+25)^{\circ} &=& (25 +25)^{\circ} \\ &=& 50^{\circ} \\ \hline \end{array}\)

1. ( 65pt-65t+13p-13 ) / ( 65p^2 *t-65pt+13p^2 -13p )  = 1/p^2   ?

2. ( 9p^2-9y^2 ) / ( 18y^2-36py+18p^2 ) = 1/2py      ?

3. (12*(11*w-g)*s^3 *y^2 *m^5 ) / ( -6*s^5 *y^3 *m^4 (g-11*w)  ) = -2m/s^2 *y  ?

1.

\(\begin{array}{|rcll|} \hline && \dfrac{ 65pt-65t + 13p-13 } { 65\cdot p^2\cdot t-65pt + 13p^2 -13p } \\\\ &=& \dfrac{ 65pt-6t + 13p-13 } { p\cdot(65pt-65t + 13p -13) } \\\\ &=& \dfrac{1}{p} \cdot \frac{ 65pt-65t + 13p-13 } { 65pt-65t + 13p -13 } \\\\ &=& \dfrac{1}{p} \cdot 1 \\\\ &=& \dfrac{1}{p} \\ \hline \end{array}\)

2.

\(\begin{array}{|rcll|} \hline && \dfrac{ 9p^2-9y^2 }{ 18y^2-36py+18p^2 } \\\\ &=& \dfrac{ 9\cdot (p^2-y^2) }{ 18\cdot ( y^2-2py+p^2 ) } \\\\ &=& \dfrac{ p^2-y^2 }{ 2\cdot ( y^2-2py+p^2 ) } \quad &| \quad (p^2-y^2) = (p-y)(p+y)\\\\ &=& \dfrac{ (p-y)(p+y) }{ 2\cdot ( y^2-2py+p^2 ) } \quad &| \quad y^2-2py+p^2 = (y-p)^2 \\\\ &=& \dfrac{ (p-y)(p+y) }{ 2\cdot ( y-p )^2 } \\\\ &=& \dfrac{ (p-y)(p+y) }{ 2\cdot ( y-p )(y-p) } \\\\ &=& \dfrac{ (p+y) }{ 2\cdot (y-p) } \\ \hline \end{array}\)

3.

\(\begin{array}{|rcll|} \hline && \dfrac{ 12\cdot(11w-g)\cdot s^3 \cdot y^2 \cdot m^5 }{ -6\cdot s^5 \cdot y^3 \cdot m^4 \cdot (g-11w) } \quad &| \quad -(g-11w) = (11w-g)\\\\ &=& \dfrac{ 12\cdot(11w-g)\cdot s^3 \cdot y^2 \cdot m^5 }{ 6\cdot s^5 \cdot y^3 \cdot m^4 \cdot (11w-g) } \\\\ &=& \dfrac{ 12\cdot(11w-g)\cdot s^3 \cdot y^2 \cdot m^5 }{ 6\cdot (11w-g) \cdot s^5 \cdot y^3 \cdot m^4 } \\\\ &=& \dfrac{ 12 \cdot s^3 \cdot y^2 \cdot m^5 }{ 6 \cdot s^5 \cdot y^3 \cdot m^4 } \\\\ &=& \dfrac{ 2 \cdot s^3 \cdot y^2 \cdot m^5 }{ s^5 \cdot y^3 \cdot m^4 } \\\\ &=& \dfrac{ 2 \cdot m^{5-4} }{ s^{5-3} \cdot y^{3-2} } \\\\ &=& \dfrac{ 2 \cdot m }{ s^2 \cdot y } \\\\ \hline \end{array}\)

\sqrt{15/4-2i}

Ohne Fehler, Sorry!