heureka

The base diameter of a right circular cone is \(18~\text{cm}\).
If the lateral area is \(516.4~\text{cm}^2\), find its volume in cubic centimeters

\(\text{Let the radius of the circle $ r = 9~\text{cm} $ } \\ \text{Let the lateral area $ F_\text{lateral} = 516.4~\text{cm}^2$ } \\ \text{Let the circle area $ F_\text{circle} = \pi r^2 =81\pi= 254.469\ldots~\text{cm}^2$ }\)

volume:

\(\begin{array}{|rcll|} \hline \mathbf{V_{\text{circular cone}}} &=& \mathbf{\dfrac{r}{3}*\sqrt{F_\text{lateral}^2 - F_\text{circle}^2}} \\\\ &=& \dfrac{9~\text{cm}}{3}*\sqrt{\left(516.4~\text{cm}^2\right)^2 - \left(254.469\ldots~\text{cm}^2\right)^2} \\\\ &=& 3~\text{cm}*\sqrt{266668.96~\text{cm}^4 - 64,754.4744755~\text{cm}^4} \\ &=& 3~\text{cm}*\sqrt{201,914.485524~\text{cm}^4} \\ &=& 3~\text{cm}*449.348957409~\text{cm}^2 \\ &=& 1348.04687223~\text{cm}^3 \\ \mathbf{ \mathbf{V_{\text{circular cone}}} } &=& \mathbf{1348.0~\text{cm}^3} \\ \hline \end{array} \)

Find the number of 10-digit numbers where the sum of the digits is divisible by 5.

The first digit cannot be zero, so 9 ways, the next 8 digits can be anything, so there are 10 ways for each of them.
We can then only choose two digits as last digit to satisfy the condition.

sum = \(9\times 10^8 \times 2 = 1~ 800~ 000~ 000\)

\(\begin{array}{|c|c|} \hline \text{the sum of the digits 1 until 9} & \text{last digit to satisfy the condition }\\ \hline \ldots 0 & 0~ \text{ or } ~5 \\ \ldots 1 & 4~ \text{ or } ~9 \\ \ldots 2 & 3~ \text{ or } ~8 \\ \ldots 3 & 2~ \text{ or } ~7 \\ \ldots 4 & 1~ \text{ or } ~6 \\ \ldots 5 & 0~ \text{ or } ~5 \\ \ldots 6 & 4~ \text{ or } ~9 \\ \ldots 7 & 3~ \text{ or } ~8 \\ \ldots 8 & 2~ \text{ or } ~7 \\ \ldots 9 & 1~ \text{ or } ~6 \\ \hline \end{array} \)

Example: 1 000 000 08(1) or 1 000 000 08(6)

                4 567 123 98(0) or 4 567 123 98(5)

Aufgabe 2.
In einem Basketballspiel der NBA hat Dirk Nowitzki in einem Spiel 43 Punkte erzielt.
Neun davon waren Freiwürfe, die je einen Punkt bringen.
Körbe aus dem Spiel heraus zählen zwei oder drei Punkte, je nachdem aus wlcher Entfernung sie erzielt werden.
Wie viele Dreier hatte Nowitzki in diesem Spiel, wenn er aus dem Spiel heraus 15 Körbe erzielt hat?

9 Freiwürfe ergeben 9 Punkte.

Wir setzen x = Anzahl der Körbe mit je 2 Punkten.
Wir setzen y = Anzahl der Körbe mit je 3 Punkten.

Die Anzahl der Körbe insgesamt ist 15.
Also \(x+y = 15\).

Die gesamte Punktanzahl 43  ergibt sich zu:

\(\begin{array}{|rcll|} \hline 9+ 2x+3y &=& 43 \quad & | \quad x+y=15~ \text{ oder }~ x = 15-y \\ 9+ 2(15-y)+3y &=& 43 \\ 9+ 30-2y+3y &=& 43 \\ 9+ 30+ y &=& 43 \\ y &=& 43-9-30 \\ \mathbf{y} &=& 4 \\ \hline \end{array}\)

Nowitzki hatte in diesem Spiel 4 Dreier.

Aufgabe 1.
Die SV der Realschule veranstaltet ein Konzert in der Turnhalle.
Die DJ macht zwei Angebote:
300€ pauschal und zuzüglich 1€ je Konzertbesucher oder
600€ pauschal und je 0,40€ je Konzertbesucher.
Ab wie vielen Konzertbesuchern lohnt sich das zweite Angebot?

Die Anzahl der Konertbesucher = x

Angebot 1: \(300€ + 1€ * x\)
Angebot 2: \(600€ + 0,40€ * x\)

Wann sind die Angebote gleich ?

\(\begin{array}{|rcll|} \hline 300€ + 1€ * x &=& 600€ + 0,40€ * x \\ 0.6€ * x &=& 300€ \\\\ x &=& \dfrac{300€}{0.6€} \\\\ \mathbf{x} &=& \mathbf{500} \\ \hline \end{array}\)

Die Angebote sind gleich bei 500 Konzertbesuchern.

Wir setzten \(x = 501\) Konzertbesucher.

Angebot 1: \(300€ + 1€ * 501 = 801€\)
Angebot 2: \(600€ + 0,40€ * 501 = 800.40€\)

Das zweite Angebot lohnt sich ab 501 Konzertbesuchern.

What is the sum of the angle measures of a 38-gon?​

You can divide the polygon into 38-2 triangles. Each triangle has 180 degrees.

So the sum is \((38-2)*180^\circ = \mathbf{6480^\circ}\)

Melissa read 126 pages in a book.
This represents 30% of the total numbers of pages.
How many pages are in her book?

\(\begin{array}{|rcll|} \hline x*30\ \% &=& 126 \\ x*0.3 &=& 126 \\ x &=& \dfrac{126}{0.3} \\ \mathbf{x} &=& \mathbf{420} \\ \hline \end{array}\)

In her book are 420 pages.

If \(x\) and \( y\) are positive integers for which \(3x + 2y + xy = 115\), then what is \(x + y\)?

\(\begin{array}{|lrcll|} \hline & 3x + 2y + xy &=& 115 \quad | \quad + 3*2 \\ & 3x + 2y + xy + 3*2 &=& 115 + 3*2 \\ \text{First solution}& \mathbf{(x+2)(y+3)} &=& \mathbf{121} \\ \text{Second solution}& \mathbf{[-(x+2)][-(y+3)]} &=& \mathbf{121} \\ \hline \end{array}\)

\(\text{The divisors of $\mathbf{121}: 1 |\ 11 |\ 121$ }\\ \begin{array}{|rclcl|} \hline 1&*&121 &=& 121 \\ 11&*&11 &=& 121 \\ 121&*&1 &=& 121 \\ \hline \end{array} \)

\(\text{First solution}\quad \mathbf{(x+2)(y+3)} = \mathbf{121} \)

\(\begin{array}{|r|r|r|r|r|} \hline x+2 & y+3 & x & y & x+y \\ \hline 1 & 121 & -1 & 118 & 117 \\ 11 & 11 & \color{red}9 & \color{red}8 & \color{red}17 & \checkmark \text{ $x$ and $y$ are positive integers} \\ 121 & 1 & 119 & -2 & 117 \\ \hline \end{array} \)

\(\text{Second solution}\quad \mathbf{[-(x+2)][-(y+3)]} = \mathbf{121}\)

\(\begin{array}{|r|r|r|r|r|} \hline -(x+2) & -(y+3) & x & y & x+y \\ \hline 1 & 121 & -3 & -124 & -127 \\ 11 & 11 & -13 & -14 & -27 \\ 121 & 1 & -123 & -4 & -127 \\ \hline \end{array}\)

The directrix of a parabola is x = 4. Its focus is (2,6)
What is the standard form?
Answer in the form x-h=a(y-k)^2

\(\text{Let a general point on the parabola $(x,y)$ }\)

1. The distance between \((x,y)\) and focus \((2,6) \) is \(\sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2}\)

2. The distance between \((x,y)\) and the directrix \((x=4,y)\) is \(\sqrt{\left(x-4 \right)^2+ \left(y-y \right)^2} =x-4\)

On the parabola, these distances are equal:

\(\begin{array}{rcll} x-4 &=& \sqrt{\left(x-2 \right)^2+ \left(y-6 \right)^2} \\ \left(x-4 \right)^2 &=& \left(x-2 \right)^2+ \left(y-6 \right)^2 \\ x^2-8x+16 &=& x^2-4x+4 + \left(y-6 \right)^2 \\ -8x+16 &=& -4x+4 + \left(y-6 \right)^2 \\ -4x+12 &=& \left(y-6 \right)^2 \quad | \quad :(-4) \\ x+ \dfrac{12}{-4} &=& \dfrac{ \left(y-6 \right)^2}{-4} \\ \mathbf{ x -3 } &=& \mathbf{ \left(-\dfrac{1}{4}\right) \left(y-6 \right)^2 } \\ \end{array}\)

Find constants A and B such that
\(\dfrac{x + 7}{x^2 - x - 2} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}\)

for all \(x\) such that \(x\neq -1\) and \(x\neq 2\).
Give your answer as the ordered pair \((A,B)\).

\(\text{Let $x^2 - x - 2=(x-2)(x+1)$ }\)

\(\begin{array}{|lrcll|} \hline &\mathbf{ \dfrac{x + 7}{x^2 - x - 2} } &=& \mathbf{ \dfrac{A}{x - 2} + \dfrac{B}{x + 1} } \quad | \quad x^2 - x - 2=(x-2)(x+1) \\\\ &\dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad \times (x-2)(x+1) \\\\ &\dfrac{(x + 7)(x-2)(x+1)}{(x-2)(x+1)} &=& \dfrac{A(x-2)(x+1)}{(x-2)} + \dfrac{B(x-2)(x+1)}{(x+1)} \\\\ & x + 7 &=& A(x+1) + B(x-2) \\\\ \mathbf{x=-1:}& -1 + 7 &=& A(-1+1) + B(-1-2) \\ & 6 &=& A\times 0 -3B \\ & 3B &=& -6 \\ & B &=& -\dfrac{6}{3} \\ & \mathbf{B} &=& -2 \\\\ \mathbf{x=2:}& 2 + 7 &=& A(2+1) + B(2-2) \\ & 9 &=& 3A +B\times 0\\ & 3A &=& 9 \\ & A &=& \dfrac{9}{3} \\ & \mathbf{A} &=& 3 \\ \hline \end{array}\)

\(\mathbf{\boxed{(A,\ B) = (3,\ -2)}}\)

check:

\(\begin{array}{|rcll|} \hline \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{A}{x - 2} + \dfrac{B}{x + 1} \quad | \quad A=3,\ B=-2 \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3}{x - 2} - \dfrac{2}{x + 1} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3(x+1)-2(x-2)}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{3x+3-2x+4}{(x-2)(x+1)} \\\\ \dfrac{x + 7}{(x-2)(x+1)} &=& \dfrac{x+7}{(x-2)(x+1)} \ \checkmark \\ \hline \end{array}\)

In hexagon ABCDEF, AB = DE = 2, BC = EF = 3, and CD = AF = 8, and all the interior angles are equal.

Find the area of hexagon ABCDEF.

\(\begin{array}{|rcll|} \hline \mathbf{A} &=& \mathbf{\dbinom{0}{0}} \\\\ \mathbf{B} &=& 2\dbinom{\cos(120^\circ)}{\sin(120^\circ)}= \mathbf{\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)}} \\\\ \mathbf{C} &=& 3\dbinom{\cos(120^\circ-60^\circ)}{\sin(120^\circ-60^\circ)} + \mathbf{\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)}}\\ &=& \dbinom{3\cos(60^\circ)}{3\sin(60^\circ)}+\dbinom{-2\cos(60^\circ)}{2\sin(60^\circ)} = \mathbf{\dbinom{\cos(60^\circ)}{5\sin(60^\circ)}} \\\\ \mathbf{D} &=& 8\dbinom{\cos(120^\circ-120^\circ)}{\sin(120^\circ-120^\circ)} + \mathbf{\dbinom{\cos(60^\circ)}{5\sin(60^\circ)}} \\ &=& \dbinom{8\cos(0^\circ)}{8\sin(0^\circ)} + \dbinom{\cos(60^\circ)}{5\sin(60^\circ)} = \mathbf{ \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ) } } \\\\ \mathbf{E} &=& 2\dbinom{\cos(120^\circ-180^\circ)}{\sin(120^\circ-180^\circ)} + \mathbf{ \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ) } } \\ &=& \dbinom{2\cos(60^\circ)}{-2\sin(60^\circ)} + \dbinom{8+\cos(60^\circ)}{5\sin(60^\circ)} = \mathbf{ \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } } \\\\ \mathbf{F} &=& 3\dbinom{\cos(120^\circ-240^\circ)}{\sin(120^\circ-240^\circ)} + \mathbf{ \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } } \\ &=& \dbinom{-3\cos(60^\circ)}{-3\sin(60^\circ)} + \dbinom{8+3\cos(60^\circ)}{3\sin(60^\circ) } = \mathbf{\dbinom{8}{0}} \\ \hline \end{array}\)

The area of hexagon ABCDEF

\(\begin{array}{|crr|} \hline \text{Point} & x & y \\ \hline A & 0 & 0 \\ B & -2\cos(60^\circ) & 2\sin(60^\circ) \\ C & \cos(60^\circ) & 5\sin(60^\circ) \\ D & 8+\cos(60^\circ) & 5\sin(60^\circ) \\ E & 8+3\cos(60^\circ) & 3\sin(60^\circ) \\ F & 8 & 0 \\ A & 0 & 0 \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline 2F &=& -2\cos(60^\circ)5\sin(60^\circ)-\cos(60^\circ)2\sin(60^\circ) \\ &+& \cos(60^\circ)5\sin(60^\circ)-(8+\cos(60^\circ))5\sin(60^\circ) \\ &+& (8+\cos(60^\circ))3\sin(60^\circ)-(8+3\cos(60^\circ))5\sin(60^\circ) \\ &+& (8+3\cos(60^\circ))*0-8*3\sin(60^\circ) \\\\ 2F &=& -10\sin(60^\circ)\cos(60^\circ)-2\sin(60^\circ)\cos(60^\circ) \\ &+& 5\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-5\sin(60^\circ))\cos(60^\circ) \\ &+& 24\sin(60^\circ)+3\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-15\sin(60^\circ)\cos(60^\circ) \\ &-& 24\sin(60^\circ) \\\\ 2F &=& -10\sin(60^\circ)\cos(60^\circ)-2\sin(60^\circ)\cos(60^\circ) -40\sin(60^\circ) \\ &+& 3\sin(60^\circ)\cos(60^\circ)-40\sin(60^\circ)-15\sin(60^\circ)\cos(60^\circ) \\\\ 2F &=& -24\sin(60^\circ)\cos(60^\circ) - 80\sin(60^\circ) \quad | \quad \cos(60^\circ) = \dfrac{1}{2} \\ 2F &=& -24\sin(60^\circ)\dfrac{1}{2} - 80\sin(60^\circ) \\ 2F &=& -12\sin(60^\circ)- 80\sin(60^\circ) \\ 2F &=& -92\sin(60^\circ) \quad | \quad : 2 \\ F &=& -46\sin(60^\circ) \quad | \quad \sin(60^\circ) = \dfrac{\sqrt{3}}{2} \\ F &=& -46\dfrac{\sqrt{3}}{2} \\ F &=& |-23 \sqrt{3}| \\ \mathbf{F} &=& \mathbf{23 \sqrt{3}} \\ \hline \end{array}\)