heureka

Hallo Omi67,

du hast vollkommen recht!

Alle Lösungen lauten: $$x = 239.032604916\ensurement{^{\circ}} \pm k\cdot 360 \ensurement{^{\circ}}$$  und $$x = 41.3562528996\ensurement{^{\circ}} \pm k\cdot 360 \ensurement{^{\circ}}$$

k=0,1,2,...

Viele Grüße

The perimeter of a rectangular traffic sign is 106 inches. Also its length is 15 inches longer than its width. What is the length and the width of this sign ?

Let L = length and W = width

perimeter :  106 = 2L + 2W     (1)

L = W + 15

L-W = 15  | *2

2L - 2W = 30      (2)

$$\\(1) \qquad 2L + 2W = 106 \\ (2) \qquad 2L - 2W = 30$$

$$\begin{array}{lrcl} \hline (1)+(2):& 4L &=& 136\\\\ & L &=& \dfrac{136}{4} \\\\ & \textcolor[rgb]{1,0,0}{L} & \textcolor[rgb]{1,0,0}{=} & \textcolor[rgb]{1,0,0}{34} } \\\\ \hline & W &=& L - 15 \\\\ & W &=& 34 - 15 \\\\ & \textcolor[rgb]{1,0,0}{W} & \textcolor[rgb]{1,0,0}{=} & \textcolor[rgb]{1,0,0}{19} } \end{array}$$

Differentiate y with respect to x:

y=6x^2+5x-10

$$y = 6x^{\textcolor[rgb]{1,0,0}{2}} +5x^{\textcolor[rgb]{1,0,0}{1}} -10x^{\textcolor[rgb]{1,0,0}{0}}}\\\\ y' = 6\cdot \textcolor[rgb]{1,0,0}{2} \cdot x^{ \textcolor[rgb]{1,0,0}{2} - 1 } + 5 \cdot \textcolor[rgb]{1,0,0}{1} \cdot x^{ \textcolor[rgb]{1,0,0}{1} - 1 } - 10 \cdot \textcolor[rgb]{1,0,0}{0} \cdot x^{ \textcolor[rgb]{1,0,0}{0} - 1 }\\\\ y'= 12x^1+5x^0+0\\\\ y' = 12x+5$$

How much is a year ?

Year Definitions

The most common definition in the western world of the year is based on the revolution of the Earth around the Sun and is therefore called a `Solar Year'. However, there are several possibilites to define beginning and end of one revolution and thus also several kinds of solar years:

• A tropical year is the interval between two successive passages of the Mean Sun through the mean vernal equinox and lasts 365.242199 days UT. The name refers to the changes of seasons (greek `tropai', the turning points) which are fixed in this kind of year. It is for this reason that the tropical year is of great importance in the construction of calendars.
• A siderial year is the interval between two successive passages of the Mean Sun at the same (fixed) star. It lasts 365.256366 days UT.
• An anomalistic year is the interval between two successive passages of the Earth through the perihelion (the point closest to the Sun) of its orbit and it lasts 365.259636 days UT.

The years so defined differ in length because of the precession of Earth's rotation and the tumbling of the Earth orbit.
Julian year (365.25 days UT) and Gregorian year (365.2425 days UT) as defined in the calendars of the respective name are solar years as well.

suppose a capacitor has XC =-8800 Ohms at f = 830 kHz. What is C ?

$$\boxed{X_C = \dfrac{ 1 }{ 2 \cdot \pi \cdot f \cdot C } \qquad C = \dfrac{ 1 }{ 2 \cdot \pi \cdot f \cdot X_C } } \\$$

Capacitive reactance, where: X_C = reactance in ohms (ohm), f = frequency in hertz (Hz), C = capacitance in farads (F)

We have:

reactance:  $$\small{\text{$X_C = 8800\ \Omega$}}$$

frequency:  $$\small{\text{$f = 830000\ \rm{Hz} $}}$$

capacitance:   $$\small{\text{$C=\dfrac{1}{2\cdot\pi\cdot f\cdot X_C} $}} \small{\text{$=\dfrac{1}{2\cdot\pi\cdot 830000\cdot (8800)} $}} \small{\text{$=0.00000000002179011\ \rm{F} = 21.79011\ \rm{pF}$}}$$

(2+3i) + (-4+5i)

$$\small{\text{$(2+3i) = \left( \begin{array}{c} 2 \\ 3 \end{array} \right)$}}\\ \small{\text{$(-4+5i) = \left( \begin{array}{c} -4 \\ 5 \end{array} \right)$}}\\ \small{\text{$(2+3i) + (-4+5i) = \left( \begin{array}{c} 2 \\ 3 \end{array} \right)+\left( \begin{array}{c} -4 \\ 5 \end{array} \right) $}}\\ \small{\text{$=\left( \begin{array}{c} 2-4 \\ 3+5 \end{array} \right) $}}\\ \small{\text{$=\left( \begin{array}{c} -2 \\ 8 \end{array} \right) = (-2+8i)$}}\\$$

Hallo Omi67,

Deine 2. Lösung sin(x2) = -0.85746025108 mit dem Winkel x2 = -59.0326049162

erfüllt nicht die Ausgangsgleichung:

0,9*cos(x)-0,75*sin(x)=0,18

0,9*cos(-59,0326049162) - 0,75 * sin(-59,0326049162)

= 0,9* 0,51455020923 -0.75*(-0.85746025108)

= 0.46309518831 + 0.64309518831 = 1.10619037662 $$\ne$$  0.18

Siehe Link zur Lösung mit WolframAlpha:

Die horizontale Sichtweite ( horizon distance ) aus 400 km  Höhe ( ISS ) auf die Erde soll nach einer Berechnung  2294 km  betragen

Wie kann man das berechnen ?

Mit Hilfe des Tangentensatzes!

The diameter of the bicycle is 22 inches. About how far will the tire go in one rotation ?

$$\small{\text{One Rotation $ = \pi \cdot d = 3.14\cdot 22\ \rm{inches} = 69 $}}$$

Choice D is okay!

The endpoint of line CD are given.find the coordinates of the Midpoint M.

1: C(3,-5) and D(7,9)

$$M(x,y) = \left(\frac{3+7}{2},\frac{-5+9}{2}\right) = (5,2)$$

2: C(-8,-6) and D(-4,10)

$$M(x,y) = \left(\frac{-8-4}{2},\frac{-6+10}{2}\right) = (-6,2)$$