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kgV(210;140;315)=1260

Primzahl Faktoren:
$210 = 2*3*5*7 \\ 140 = 2^2*5*7\\ 315 = 3^2 *5*7$

$kgV(210;140;315)=2^2*3^2*5*7=1260$

What is the center and radius of 2x^2+2y^2=32?

$2x^2+2y^2=32 \quad | \quad : 2\\ x^2+y^2=16\\ (x-0)^2+(y-0)^2 = 4^2$

center = (0,0)

radius = 4

compute least-squares regression line for predicting y from x given the following summary statistics.
$\begin{array}{|lcl|} \hline \bar{x} &=& 6 \\ s_x &=& 3 \\ \bar{y} &=& 1350 \\ s_y &=& 101 \\ r &=& 0.7 \\ \hline \end{array}$

$\begin{array}{|rcl|rcl|rcl|} \hline y &=&\alpha+\beta x & \alpha &=& \bar{y}-\beta\bar{x} & \beta&=& r\cdot \frac{s_y}{s_x} \\ && & && & \beta&=& 0.7\cdot \frac{101}{3} \\ && & && & \beta&=& 23.5\bar{6} \\ && & \alpha &=& 1350-23.5\bar{6} \cdot 6 & \\ && & \alpha &=& 1208.6 \\ y &=&1208.6+23.5\bar{6} x \\ \hline \end{array}$

C(9,6)

$\begin{array}{|rcll|} \hline && C(9,6) \\ &=& C(9,9-6) \\ &=& C(9,3) \\ &=& \frac{9}{3}\cdot \frac{8}{2}\cdot \frac{7}{1} \\ &=& 3\cdot 4\cdot 7 \\ &=& 84 \\ \hline \end{array}$

When a certain medical drug is administered to a patient,

the number of milligrams remaining in the patient's bloodstream after t hours is modeled by

D(t) = 70e−0.5t.

How many milligrams of the drug remain in the patient's bloodstream after 3 hours?

(Round your answer to one decimal place.)

$\begin{array}{|rcll|} \hline D(t) &=& 70e^{-0.5t} \quad & | \quad t = 3\ hours \\ D(3) &=& 70e^{-0.5\cdot 3} \\ D(3) &=& 70e^{-1.5} \\ D(3) &=& 70\cdot 0.22313016015 \\ D(3) &=& 15.6191112104\ mg \\ \hline \end{array}$

There are 15.6 mg of the drug remain in the patient' bloodstream after 3 hours.

wo wird die Mantellänge bei einem kegel gemessen?

gegeben habe ich :ein kegelstumpf von 120,65mm mantellänge hat die durchmesser 60 und 85 mm.

Berechne die kegelhöhe.

Die Mantellänge s:

$\begin{array}{|lclcl|} \hline s &=&\text{ Mantellänge } &=& 120,65\ mm \\ h&=&\text{ Höhe } &=& ?\ mm \\ r_u&=&\text{ Radius unten } &=& \frac{85}{2} \ mm \\ r_o&=&\text{ Radius oben } &=& \frac{60}{2} \ mm \\ \hline \end{array}$

Ansatz:

$\begin{array}{|rcll|} \hline (r_u-r_o)^2 + h^2 &=& s^2 \\ \left(\frac{85}{2}-\frac{60}{2}\right)^2 + h^2 &=& 120,65^2 \\ \left(\frac{85-60}{2}\right)^2 + h^2 &=& 120,65^2 \\ \left(\frac{25}{2}\right)^2 + h^2 &=& 120,65^2 \\ \frac{25^2}{2^2} + h^2 &=& 120,65^2 \\ \frac{625}{4} + h^2 &=& 120,65^2 \\ 156,25 + h^2 &=& 120,65^2 \quad & | \quad - 156,25 \\ h^2 &=& 120,65^2 - 156,25 \\ h^2 &=& 14556,4225- 156,25 \\ h^2 &=& 14400,1725 \\ h &=& \sqrt{14400,1725} \\ \mathbf{ h } & \mathbf{=} & \mathbf{ 120,000718748\ mm } \\ \hline \end{array}$

Die Kegelhöhe beträgt 120,00 mm

Welche zwei Zahlen, die beide keine Null enthalten dürfen,

ergeben miteinander multipliziert die Zahl 1 000 000 000?

$\begin{array}{|rcll|} \hline 1\ 000\ 000\ 000 &=& 2^9 \times 5^9 \\ &=& 512 \times 1\ 953\ 125 \\ \hline \end{array}$

Find the x-coordinates of the points where the line y=5x-1 meets the curve y= 2x^3 + x^2 +1.

I keep solving but I can't seem to get anything right,

especially when I set these two next to each other.

Intersections:

$\begin{array}{|rcll|} \hline y_{\text{line}} &=& y_{\text{curve}} \\ 5x-1 &=& 2x^3 + x^2 +1 \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline 2x^3 + x^2 +1 &=& 5x-1 \quad & | \quad -5x+1 \\ 2x^3 + x^2 +1 -5x+1 &=& 0 \\ 2x^3 + x^2 -5x +2 &=& 0 \quad & | \quad :2 \\ x^3 + \frac12 x^2 -\frac52x +\color{red}{1} &=& 0 \\ \hline \end{array}$

We get the rational solutions, when we test all dividers from the absolut term 1:

We must test +1 or -1.

For +1 we get the first root $x_1$: $1^3 + \frac12 \cdot 1^2 -\frac52 \cdot 1 + 1 = 1+ \frac12 -\frac52+1 = 0$

$\begin{array}{|rcll|} \hline x_1 = 1 \\ \hline \end{array}$

We know that $(x-x_1)$ is a divider of  $x^3 + \frac12 x^2 -\frac52x +1$

$( x^3 + \frac12 x^2 -\frac52x +1) : (x-1) = (x^2+1.5x-1 )$

Because $( x^3 + \frac12 x^2 -\frac52x +1) = (x-1)\cdot (x^2+1.5x-1 ) = 0$
we find the roots $x_2$ and $x_3$, if we set  $x^2+1.5x-1 = 0$

$\begin{array}{|rcll|} \hline x^2+1.5x-1 &=& 0 \\ x &=& \frac{-1.5\pm \sqrt{1.5^2-4\cdot 1 \cdot (-1) } }{2} \\ x &=& \frac{-1.5\pm \sqrt{2.25+4} }{2} \\ x &=& \frac{-1.5\pm \sqrt{6.25} }{2} \\ x &=& \frac{-1.5\pm 2.5 }{2} \\\\ x_2 &=& \frac{-1.5+ 2.5 }{2} \\ x_2 &=& \frac{1 }{2} \\ \mathbf{x_2} &\mathbf{=}& \mathbf{0.5} \\\\ x_3 &=& \frac{-1.5 - 2.5 }{2} \\ x_3 &=& \frac{-4 }{2} \\ \mathbf{x_3} &\mathbf{=}& \mathbf{-2}\\ \hline \end{array}$

The x-coordinates of the intersection-points are: $x_1 = 1, x_2 = 0.5, ~ \text{and}~ x_ 3 = -2$

 Simplify c**p with radicals

$\begin{array}{|rcll|} \hline && 2\cdot[~\frac14 (\sqrt{5}-1)~]\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac24\cdot (\sqrt{5}-1)\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac12\cdot (\sqrt{5}-1)\cdot \sqrt{\frac18\cdot (5+\sqrt{5}) } \\ &=& \frac12\cdot (\sqrt{5}-1)\cdot \frac{\sqrt{ 5+\sqrt{5} } } { \sqrt{8} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot (\sqrt{5}-1)\cdot \sqrt{ 5+\sqrt{5} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (\sqrt{5}-1)^2\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (5-2\sqrt{5}+1)\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ (6-2\sqrt{5})\cdot (5+\sqrt{5}) } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 30+6\sqrt{5}-10\sqrt{5}-2\cdot 5 } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 30+6\sqrt{5}-10\sqrt{5}-10 } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 20-4\sqrt{5} } \\ &=& \frac{1}{2\cdot \sqrt{8} } \cdot \sqrt{ 4(5-\sqrt{5}) } \\ &=& \frac{\sqrt{4}}{2\cdot \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{2}{2\cdot \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{8} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{4\cdot 2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ \sqrt{4}\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{1}{ 2\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{\sqrt{2}} {\sqrt{2}}\cdot \frac{1}{ 2\cdot \sqrt{2} } \cdot \sqrt{ 5-\sqrt{5} } \\ &=& \frac{\sqrt{2}}{ 2\cdot 2 } \cdot \sqrt{ 5-\sqrt{5} } \\\\ &\mathbf{=}& \mathbf{ \frac{\sqrt{2}}{ 4 } \cdot \sqrt{ 5-\sqrt{5} } } \\ &=& 0.58778525229 \\ \hline \end{array}$

2) In a meeting, everyone shakes hands with another person.

If the total number of shaking hands is 2016 times,

how many people was in the meeting?

Let n = people in the meeting

$\begin{array}{|rcll|} \hline \frac{ n\cdot (n-1) } {2} &=& 2016\ \text{handshakes }\\ n\cdot (n-1) &=& 4032 \\ n^2-n &=& 4032 \\ n^2-n -4032 &=& 0 \\\\ n &=& \frac{1+\sqrt{1-4\cdot(-4032) } }{2} \\ n &=& \frac{1+\sqrt{ 16129 } }{2} \\ n &=& \frac{1+127 } {2} \\ n &=& \frac{ 128 } {2} \\ n &=& 64 \\ \hline \end{array}$