asinus

On the number line above, points J, K, L, and M are integers, and JK:KL:KM = 2:1:3. What is the value of KM?

Hllo Guest!

 --            24                --   

J          K     L               M

     2        1          3               | \(\times\frac{24}{6}\)

     8        4         12

                    16

The value of KM is 16.  (G)

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Hello Guest!

\(sin (8°)=\frac{8}{\overline{UX}}\\ \overline{UX}= \frac{8}{sin(8°)}=\frac{8}{0.13917}=\color{blue}76.53 (C)\)

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Subtract: \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)       [ \(2x^2-x-3=0\)   Assume: x = -1]

\((2x^2-x-3):(x+1)=2x-3\)

 \(\underline{2x^2+2x}\)

        \(-3x-3\)

        \(\underline{-3x-3}\)

                     \(0\)

 \(\frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\) \(\frac{2x-3}{2x-3}\cdot \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}\)\(=\frac{2x^2-4x-3x+6-(3-12x)}{2x^2-x-3}\)     [expand with  (2x-3)]

\(=\frac{2x^2+5x+3}{2x^2-x-3}\)

  \((2x^2+5x+3):(x+1)=2x+3\)

   \(\underline{2x^2+2x}\)

               \(3x+3\)

               \(\underline{3x+3}\)

                         \(0\)

  \(\frac{2x^2+5x+3}{2x^2-x-3}\cdot\frac{x+1}{x+1}=\frac{2x+3}{2x-3}\)    [ shorten with (x+1)]

  \(\large \frac{x-2}{x+1}-\frac{3-12x}{2x^2-x-3}=\frac{2x+3}{2x-3}\)     \(x\in \mathbb R\ |\ \ x\notin \{-1.5\ ;-1\}\)

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Wenn gilt: 50 kb/sec sind 50 KB/sec dann:

W = 30 GB

N = 50 KB/sek

1 GB = 10^6 KB

\({\color{blue}t=\frac{W}{N}}=\frac{30GB\cdot sec}{50KB}\cdot \frac{10^6KB}{GB}\cdot \frac{h}{3600cec}\)

\(t=166h\ 40 min\)

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How long will it take to download a 30 GB file if I download 50 kb per second? I used this tool to download Rechner but I think it is incorrect.

Wie lange dauert das Herunterladen einer 30-GB-Datei, wenn ich 50 kb pro Sekunde herunterlade?

Hallo Gast!

Es gilt: 50 kb/sec sind 50 kbit/sec.

\(W=N\cdot t\\ \color{blue}t=\frac{W}{N}\)

W = 30 GB

N = 50 kbit/sek

1 kbit = 125 Byte

1GB = 10^9 Byte

\(t=\frac{W}{N}=\frac{30\ GB\cdot sec}{50\ kbit}\cdot\frac{10^9Byte}{GB}\cdot \frac{kbit}{125Byte}\cdot \frac{h}{3600sec}\)

\(t=1333h\ 20min\)

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For each number from 1 to 99, Andrew wrote down the sum of its digits. What is the mean of the numbers Andrew wrote down?

Hello Guest!

\(\sum_{k=1}^n\ k=\frac{n^2+n}{2}\)

\(\sum_{k=1}^9\ k=\frac{9^2+9}{2}=45\)

The mean of the numbers from 1 to 99 is

\(\frac{10\cdot 45+9\cdot 45}{90+99}=\frac{855}{189}=\color{blue}4.\overline{523809}\)

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Hallo Probolobo,

nein, Pauls Ball hatte keinen Raketenantrieb installiert. Ich habe nur ein kleines Minus in der Parabelgleichung vergessen einzutragen. Die Richtige Gleichung heißt:

h(x) = -  0,02x^2 + 0,8x + 1,9

Ich habe das in der Fragestellung korrigiert.

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Show all work to factor x^4 − 17x^2 + 16 completely.

Hello Guest!

\(x^4-17x^2 +16=0\)

We determine the zeros.

Exchange:  \(x^2=u\)

\(u^2-17u+16=0\\ u^2-m\cdot u+n =0\)

\(u=-\frac{m}{2}\pm \sqrt{(\frac{m}{2})^2-n}\\ u=8.5\pm \sqrt{8.5^2-16}\\ u=8.5\pm7.5\)

\(u_1=16\\ u_2=1\)

Reinstate:  \(x^2=u\\ x=\sqrt{u}\)

\(x_1=+\sqrt{16}\\ x_2=-\sqrt{16}\\ x_3=+\sqrt{1}\\ x_4=-\sqrt{1}\)

\(x_1=4\\ x_2=-4\\ x_3=1\\ x_4=-1\)

\(x^4-17x^2 +16=(x-4)(x+4)(x-1)(x+1)\)

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Danke Omi67!

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Welche flächenbezogene Masse besitzt Karton, wenn eine Probe im Format 5cm x 5cm  0,875 g wiegt.

Hallo Gast!

Die flächenbezogene Masse der Kartonprobe ist

\(m_{/Fl}=\frac{0,875g}{5cm\cdot 5cm}\cdot\frac{10^4cm^2}{m^2}=\color{blue}350g/m^2\)

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