asinus

Aufgabe 2

Hallo Toli7777!

Wassertank beheizt mit 2000 W über 450 Betriebsstunden.

a) \(W= P\cdot t=2000W\cdot 450h\cdot\frac{kW}{1000W}=\color{blue}900kWh\)

         \(900kWh\cdot \frac{kJ}{kW\cdot s}\cdot \frac{3600s}{h}=\color{blue}3.240.000kJ\)

b)  Preis bei 0,3€/kWh = 900kWh * 0,3€/kWh = 270€

c) Senkung der Energiekosten:

1. Wärmeisolierung des Wassertanks.

2. Angebote der Energielieferanten prüfen.

3. Eigene Sonnenkollektoren auf den Dächern anbringen.

  !

Akkordlohn, Zeitgrad, Lohnkosten berechnen?

Hallo Tolli7777!

Zweischichtbetrieb. Gearbeitet wird an 20 Tagen im Monat von.6.00 bis 23.00 Uhr.

Normalleistung 4000 Fla./h. Vollauslastung 5000 Fla./h.

5 Mitarbeiter pro 8-Stunden-Schicht.

Akkordrichtsatz 8,00 €/h.

Zeitgrad = Istleistung / Normalleistung.

Nachtzulage ab 22.00 Uhr 30% auf den Akkordrichtsatz.

Pausen (bezahlte) 45 min / Schicht

a) Zeitgrad = 5000 Fl./h / 4000Fl./h = 1,25

b) Akkordlohn 6.00h bis 22.00h = 8h * 8,00€/h * 1,25 = 80 €. (ganze Schicht)

c) Akkordlohn ab 22.00h = 8,00 €/h * 1,25 * 130% = 13 €/h.

d) Lohnkosten / Monat

                      = 20 Tage * 160% * 5MA * (8h * 8€/h + 7h * 8€/h + 1h * 13€/h)

                                       Nebenko.             Schicht 1       Schicht 2

                      = 21280 €

e) Lohnkosten / Betriebsstunde = 21280 € / (20 Tage * 17h) = 62,59 €/h

  !

A circle is inscribed in a quarter-circle sector. It is tangent to the arc of the sector and the two perpendicular radii of the sector. What is the ratio of the area of the inscribed circle to the area of the sector?

Hello Guest!

\(a=r+r\sqrt{2}\\ a=r(1+\sqrt{2})\)          Radius of the sector

\(A_{sec}=\pi r^2(1+\sqrt{2})^2/4\)

\(A_{insc}=\pi r^2\)

\(\frac{A_{insc}}{A_{sec}}=\frac{4}{(1+\sqrt{2})^2}\)

\(\frac{A_{insc}}{A_{sec}}=\frac{2}{1+\sqrt{2}}=1:2,414\)

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Two 4 x 4 squares intersect at right angles, bisecting their intersecting sides, as shown. The circle's diameter is the segment between the two points of intersection. What is the area of the shaded region created by removing the circle from the square?

Hello Guest!

\(A=2\cdot (4\cdot 4)-2^2=28\)

\(d=\sqrt{2^2+2^2}=\sqrt{8} \\d=2\cdot \sqrt{2}\\ r=\sqrt{2}\)

\(A_{shaded}=A-\pi r^2=28-\pi \cdot 2=28-6.28\)

\(A_{shaded}=21.72\)

  !

I prove it

\(x+y+z=2\\ xy-z=-3\\ x+y+z^2=4 \)

\(x=-1\\ y=1\\ z=2\)

\(-1+1+2=2\\ -1\cdot 1-2=-3\\ -1+1+2^2=4 \)

q.e.d.

Mistake!
The question asks for a unique solution. The equations are solved by both x = -1, y = 1, z = 2 and x = 1, y = -1, z = 2.
The correction follows.

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write a system of 3 equations and 3 variables that has the unique solution of (-1,1,2).

Hello Guest!

\(x+y+z=2\)

\(x y-z=-3\)

\(x+y+z^2=4\)

  !

Find the surface area.

Hello Guest!

\(A=2\times(\frac{12+8}{2}\times 3+8\times 10)+5\times 10+3\times 10\)

\(A=300\)

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Find the minimum value of y = 5x^2 - 12x + 15.

Hello Guest!

\( y = 5x^2 - 12x + 15\\ \frac{dy}{dx}=10x-12=0\\ x_{min}=\frac{12}{10}\)

\(x_{Pmin}=\frac{6}{5}\)

\(y_{min}=5\cdot \frac{36}{25}-12\cdot \frac{6}{5}+15\)

\(y_{min}=\frac{5\cdot 36-12\cdot6\cdot 5+15\cdot 25}{25}=\frac{195}{25}\)

\(y_{min}=\frac{39}{5}\)

\(P_{min}(1.2,7.8)\)

  !

what is x=4 in standard form. ax^2 +bx +c

Hello Guest!

\((x-4)(a-4)=(x-4)^2=x^2-8x+16\)

\(ax^2 +bx +c\)

\(a=1\\ b=-8\\ c=16\\ \)

\(x^2-8x+16\)

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One number x is twenty-five less than four times another number y. If each number were multiplied by six, their difference would be 246. What are the numbers?

Eine Zahl x ist fünfundzwanzig weniger als das vierfache einer anderen Zahl y. Wenn jede Zahl mit sechs multipliziert würde, wäre ihre Differenz 246. Was sind die Zahlen?

Hello Guest!

\(4y-25=x\\ \underline{6x-6y=246}\\ 6(4y-25)-6y=246\\ 24y-150-6y=246\)

\(18y=396\)

\(y=22\)

\(x=4y-25\\ x=4\cdot 22-25\)

\(x=63\)

\(4\cdot 22-25=63\\ 6\cdot 63-6\cdot 22=246\\ 4y-25=x\\ 6x-6y=246\)       q.e.d.

The numbers are x=63 and y=22.

  !