heureka

an einem tag essen  drei hunde 2 dosen hunde futter.

eine dose futter kostet 80 cent.

berechne den preis in monat. 

$$3~\rm{Hunde} \cdot \dfrac{2~\rm{Dosen}}{\rm{Hund}\cdot \rm{Tag}}
\cdot \dfrac{0,80~ \rm{Euro} } {1~\rm{Dose} } \cdot \dfrac{x ~ \rm{} Tage}{1~\rm{Monat}} \\\\\\
\text{Preis im Monat }=3 \cdot 2 \cdot 0,80 \cdot x \cdot
\dfrac{ \rm{Euro} } { 1~\rm{Monat} } \\$$

x = Anzahl der Tage im Monat

$$\small{\text{
$
\begin{array}{c|c}
x = $ Anzahl Tage im Monat $ & $ Preis im Monat \\
\hline
\\
28 & 4,8 \cdot 28 ~$ Euro $ = 134,40~ $Euro$ \\
29 & 4,8 \cdot 29 ~$ Euro $ = 139,20~ $Euro$ \\
30 & 4,8 \cdot 30 ~$ Euro $ = 144,00~ $Euro$ \\
31 & 4,8 \cdot 31 ~$ Euro $ = 148,80~ $Euro$ \\
\\
\hline
\end{array}
$}}$$

if f(x)=3x find (i)f(1) (ii) f(3) (iii) f(4) (iv) f(6)

$$\mathbf{~ f(x) = 3\cdot x ~}$$

(i) f(1) 

Set x = 1

$$\small{\text{$
f(1) = 3\cdot 1 = 3
$}}$$

(ii) f(3)

Set x = 3

$$\small{\text{$
f(3) = 3\cdot 3 = 9
$}}$$

(iii) f(4)

Set x = 4

$$\small{\text{$
f(4) = 3\cdot 4 = 12
$}}$$

(iv) f(6)

Set x = 6

$$\small{\text{$
f(6) = 3\cdot 6 = 18
$}}$$

If then the numerical value of a/b + b/a is equal to x. Find the value of x.

$$\small{\text{$
\begin{array}{rcll}
(a^2 +b^2)^3 &=& (a^3+b^3)^2 \\
(a^2)^3+3\cdot(a^2)^2\cdot (b^2)^1 +3\cdot (a^2)^1\cdot (b^2)^2 + (b^2)^3 &=& (a^3)^2 + 2\cdot(a^3)^1\cdot (b^3)^1 + (b^3)^2\\
a^6+3\cdot a^4\cdot b^2 +3\cdot a^2\cdot b^4 + b^6 &=& a^6 + 2\cdot a^3\cdot b^3 + b^6\\
\not{a^6}+3\cdot a^4\cdot b^2 +3\cdot a^2\cdot b^4 + \not{b^6} &=& \not{a^6} + 2\cdot a^3\cdot b^3 + \not{b^6}\\
3\cdot a^4\cdot b^2 +3\cdot a^2\cdot b^4 &=& 2\cdot a^3\cdot b^3 \\
3\cdot a^4\cdot b^2 +3\cdot a^2\cdot b^4 &=& 2\cdot a^3\cdot b^3 & | \quad : (a^3\cdot b^3) \\
3\cdot\frac{ a }{ b }+3\cdot \frac{ b } { a } &=& 2 \\
3\cdot\frac{ a }{ b }+3\cdot \frac{ b } { a } &=& 2 & | \quad : 3 \\
\frac{ a }{ b }+ \frac{ b } { a } &=& \frac{2}{3}
\end{array}
$}}\\
\mathbf{x=\dfrac{ a }{ b }+ \dfrac{ b } { a } &=& \dfrac{2}{3} }$$

Daniel kicks a soccer ball up into the air with an initial upward velocity of 63 feet per second. The ball is 2 feet above the ground when it is kicked. Write an equation to model the situation.

I. Write an equation to model the situation.

$$\small{\text{$~\boxed{~h(t) = \dfrac{g}{2}\cdot t^2 + v\cdot t + h_0 \qquad g = -32 ~\frac{ft.}{sec^2} \qquad v = 63~\frac{ft.}{sec} \qquad h_0 = 2~ft. ~}
~$}}$$

The height of the soccer ball at any given time, t, can be calculated according to the function,

$$\small\text{$~h(t) = -16\cdot t^2 + 63\cdot t + 2 ~$}}$$ , where time is measured in seconds and the height is measured in feet.

This is a parabola $$\small{\text{$~y = a\cdot x^2+b\cdot x+c~$}}$$.

II. How long will it take the ball to reach its maximum height ?

The vertex of the parabola is $$\small{\text{$~x=t=\frac{-b}{2a} = \frac{-63}{2\cdot(-16)}=1.96875 ~\rm{sec.} ~$}}$$

III. What is the maximum height of the ball ?

$$\small\text{$~
\begin{array}{rcl}
h(1.96875~\rm{sec.}) &=& -16\cdot {1.96875}^2 + 63\cdot 1.96875 + 2\\
&=& -62.015625~\rm{ft.} +124.03125~\rm{ft.} +2 ~\rm{ft.}\\
h(1.96875~\rm{sec.}) &=& 64.015625~\rm{ft.}
\end{array}
~$}}$$

An athletics track is to have a perimeter of 400m and straights of length 100m. Find the radius of the bend ?

$$\small{\text{$
\begin{array}{rcll}
2\cdot (\pi \cdot r + 100~ \rm{m} ) &=& 400 ~ \rm{m} \quad & | \quad :2 \\
\pi \cdot r + 100~ \rm{m} &=& 200~ \rm{m} \\
\pi \cdot r &=& 100~ \rm{m} \\
r &=& \dfrac { 100~ \rm{m} } { \pi } \\
r &=& 31.8309886184 ~ \rm{m} \\
r &\approx& 31.83 ~ \rm{m}
\end{array}
$}}$$

divide and write the result using scientific notation (4.0 x 10^7) / (5.0 x 10^9)

$$\small{\text{
$
\begin{array}{rcl}
\dfrac{4.0 \cdot 10^7} {5.0 \cdot 10^9} & \\
&=& \dfrac{4.0}{5.0} \cdot 10^7\cdot 10^{-9} \\
&=& 0.8 \cdot 10^{7-9} \\
&=& 0.8 \cdot 10^{-2} \\
&=& \dfrac{8}{10} \cdot 10^{-2} \\
&=& 8 \cdot \dfrac{ 10^{-2}}{10} \\
&=& 8 \cdot 10^{-2}\cdot 10^{-1} \\
&=& 8 \cdot 10^{-2-1} \\
&=& 8 \cdot 10^{-3 }
\end{array}
$}}$$

P.S.

$$\small{\text{Newton's Law of Gravitation: }} F=\boxed{\,G*\frac{m_1*m_2}{r^2}=m_2*g\,} \Rightarrow \boxed{\,
g=\frac{m \cdot G}{r^2} \,}$$

Man kann nicht durch einen Vektor dividieren!

Die Division eines Vektors ist nicht definiert !

I.

$$2x^2+4x-5=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{-4 \pm \sqrt{16-4\cdot 2\cdot (-5) } } {2\cdot2 }
= \dfrac{-4 \pm \sqrt{16+40 } } {4}
= \dfrac{-4 \pm \sqrt{56} } {4}
= \dfrac{-4 \pm \sqrt{16\cdot 3.5} } {4}
=-1\pm\sqrt{3.5}
$}}\\
\small{\text{
$
x_1 =-1+\sqrt{3.5} \qquad x_2 =-1-\sqrt{3.5}
$}}$$

II.

$$-4x^2-8x+5=3\\
-4x^2-8x+2=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ 8 \pm \sqrt{64-4\cdot (-4)\cdot 2 } } {2\cdot (-4)}
= \dfrac{ 8 \pm \sqrt{64+32 } } {-8}
= \dfrac{ 8 \pm \sqrt{96} } {-8}
= \dfrac{ 8 \pm \sqrt{64\cdot 1.5} } {-8}
=-1\mp\sqrt{1.5}
$}}\\
\small{\text{
$
x_1 =-1+\sqrt{1.5} \qquad x_2 =-1-\sqrt{1.5}
$}}$$

III.

$$4x^2-4x+1=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ 4 \pm \sqrt{16-4\cdot 4 \cdot 1 } } {2\cdot 4 }
= \dfrac{ 4 \pm \sqrt{16-16} } {8}
= \dfrac{ 4 \pm 0 } {8}
= \dfrac{ 4 } {8}
= \dfrac{ 1 } {2}
= 0.5
$}}\\
\small{\text{
$
x_1 = x_2 = 0.5
$}}$$

IV.

$$3x^2+2x+4=0$$

$$\small{\text{
$
x_{1,2}=\dfrac{ -2 \pm \sqrt{4-4\cdot 3 \cdot 4 } } {2\cdot 3 }
= \dfrac{ -2 \pm \sqrt{4-48} } {6}
= \dfrac{ -2 \pm \sqrt{-44} } {6}
$}}\\
\small{\text{
$
\sqrt{-44} $ not real, no solution$
$}}$$

how would I do this? A competitive high school swimmer takes 52 seconds to swim 100 yards. What is his rate in km/hr? (1 m = 1.0936 yards)

$$\small{\text{
$
\begin{array}{l}
\dfrac{100~\rm{yard}}{52~\rm{s} } \cdot
\dfrac{ 1~\rm{m} } { 1.0936~\rm{yard} } \cdot
\dfrac{ 1~\rm{km} } { 1000~\rm{m} } \cdot
\dfrac{ 60~\rm{s} } { 1~\rm{min} } \cdot
\dfrac{ 60~\rm{min} } { 1~\rm{h} } \\\\
=\dfrac{100\cdot 60\cdot 60}{52\cdot 1.0936\cdot 1000}\cdot \dfrac{ \rm{km}}{ \rm{h} } \\\\
=6.33053851781\cdot \dfrac{ \rm{km}}{ \rm{h} } \\\\
\approx 6.33 \cdot \dfrac{ \rm{km}}{ \rm{h} }
\end{array}
$}}$$