heureka

what is the solution of -2(8x-4)<2x+5

$$-2(8x-4)<2x+5 \\ -16x+8 < 2x+5 \\ -16x+8 < 2x+5 \quad | \quad +16x\\ 8 < 2x+16x+5 \\ 8 < 18x+5 \\ 8 < 18x+5 \quad | \quad -5\\ 8 -5< 18x \\ 3< 18x \\ 3< 18x \quad | \quad :18\\ \frac{3}{18}< x \\ \frac{1}{6}< x \\ x > \frac{1}{6}$$

Wie löst man y'=(1+y)/x

$$y'=\frac{1+y}{x} \quad | \quad y' = \frac{dy}{dx} \\\\ \frac{dy}{dx} = \frac{1+y}{x} \qquad \small{\text{ Trennung der Variablen }} \\\\ \left( \frac{1}{1+y} \right)\ dy = \left( \frac{1}{x} \right)\ dx \\\\ \left( \frac{1}{1+y} \right)\ dy = \left( \frac{1}{x} \right)\ dx \quad | \quad \int\\\\ \int\left( \frac{1}{1+y} \right)\ dy = \int\left( \frac{1}{x} \right)\ dx \qquad \small{\text{ Berechnung der Integrale }} \\\\$$

I.

$$\int\left( \frac{1}{x} \right)\ dx \quad \small{\text{ Wir substituieren: }} x=e^u \qquad dx=e^udu \\\\ \int\left( \frac{1}{x} \right)\ dx = \int\frac{e^u}{e^u}du = \int du=u \quad | \quad u=\ln{(x)}\\\\ \boxed{\int\left( \frac{1}{x} \right)\ dx = \ln{(x)}+ c_1 }$$

II.

$$\int\left( \frac{1}{1+y} \right)\ dy \quad \small{\text{ Wir substituieren: }} y=u-1 \qquad dy=du \\\\ \int\left( \frac{1}{1+y} \right)\ dy = \int\frac{1}{u}du = \ln{(u)} \quad | \quad u=y+1\\\\ \boxed{\int\left( \frac{1}{1+y} \right)\ dy = \ln{(y+1)}+ c_2 }$$

III.

$$\int\left( \frac{1}{1+y} \right)\ dy = \int\left( \frac{1}{x} \right)\ dx \\\\ \ln{(y+1)}+ c_2 = \ln{(x)}+ c_1 \\\\ \ln{(y+1)} = \ln{(x)}+ c_1 - c_2\\\\ \ln{(y+1)} = \ln{(x)}+ c_3 \\\\ \ln{(y+1)} = \ln{(x)}+ c_3 \quad | \quad e^{()}\\\\ y+1 = e^{ \ln{(x)}+ c_3 } \\\\ y+1 = e^{ \ln{(x)}}* e^{c_3} \\\\ y+1 = x*e^{c_3} \\\\ y+1 = x*c \\\\ \boxed{ \boxed{ y = x*c-1 } }$$

How to find surface area of a cylinder ?

$$\\ \small \begin{array}{ll} \text{ Area of the top is } & \pi r^2 \\ \text{ Area of the bottom is } & \pi r^2 \\ \text{ Area of the side is } & 2\pi rh \\ \boxed{A = 2\pi r^2 + 2\pi rh } \end{arry} }$$

X^2+7x=18

$$x^2+7x-18=0 \qquad (x-2)(x+9)=0\qquad x_1=2 \quad x_2=-9$$

In a factory, we know that 20% of produced goods will be replaced. How many products will have to be produced to the warehouse, so that 10 000 orders and the replacements can be sent be with probability of 0.9 ? The answer is 12 051, but I dont know how to calculate it, or even understand the question properly

$$\small{\text{binomcdf$\ (10000,0.2,\textcolor[rgb]{1,0,0}{2051}) \\\\ = \sum \limits_{i=0}^{\textcolor[rgb]{1,0,0}{2051}} \binom{10000}{ i}*0.2^i*0.8^{(10000-i)} =0.9008 \\ $ }}$$

find the slope of the line that passes through the points (4,5) and (11,8)

slope = $$\small{\text{ $ \frac{8-5}{11-4}=\frac{3}{7}=0.42857142857 $ }}$$

Given a triangle with C=108°, a=10, b=6.5, find side c

$$c^2=a^2+b^2-2ab*\cos(108\ensurement{^{\circ}}) \\ c^2 = 10^2+6.5^2-2*10*6.5*\cos(108\ensurement{^{\circ}}) \\ c^2 = 100 +42.25 - 130 * (-0.30901699437) \\ c^2 = 142.25 + 40.1722092687\\ c^2 = 182.422209269 \quad | \qaud \sqrt{} \\ c= 13.5$$

Wenn man 37 Zimmer hat und insgesamt müssen in den Zimmern 72 Betten untergebracht werden. Wie lautet da die Antwort ?

a = Anzahl der Zimmer mit 2 Betten

b = Anzahl der Zimmer mit einem Bett

(1)   a * 2 + b * 1 = 72 Betten

(2)   a      +  b      = 37 Zimmer

------------------------------------

(1) - (2):   2a-a + b-b = 72 - 37

                a                = 35

(2)     35 + b = 37

                b = 37 - 35

               b = 2

Wir haben 35 Zimmer mit 2 Betten und 2 Zimmer mit einem Bett.

Hi CPhill,

ditto!

a geometric progression with 28 as its first term has the sum to infinity of 70 .

find its common ratio

$$\text{sum: } \begin{array}{rlccl} s &=& 28 & + & 28r+ 28r^2 + 28r^3+28r^4+28r^5 +\dots+ \\ r*s &=& && 28r +28r^2+28r^3+28r^4+28r^5+\dots+ \\ \hline s-r*s &=& 28 &\\ &&&\\ \hline &&&\\ \end{array}$$

$$r*s = s - 28 \\ \\ r = \dfrac{s - 28}{s} \quad | \quad s = 70\\\\ r = \dfrac{70 - 28}{70} \\\\ r = \frac{42}{70} \\\\ r = \frac{21}{35} \\\\ r = \frac{3}{5} \\\\ r = 0.6$$