heureka

$$\textstyle{ \begin{array}{rcl} -\dfrac{1}{5}y-\dfrac{1}{2}&=&-\dfrac{3}{7} \quad | \quad \times(-1)\\\\ \dfrac{1}{5}y+\dfrac{1}{2} &=&\dfrac{3}{7} \\\\ \dfrac{2y+5}{10} &=&\dfrac{3}{7} \quad | \quad \times 10\\\\ 2y+5 &=&\dfrac{30}{7} \quad | \quad -5\\\\ 2y&=&\dfrac{30}{7}-5*\dfrac{7}{7}=-\dfrac{5}{7}\\\\ y&=&-\dfrac{5}{14} \end{array} }$$

$$(-2-\sqrt{5}i)(2-\sqrt{5}i)=-9$$

I'm trying to take an integral bounded below by zero and above by alpha of the sqrt(1-x^2) :

$$\boxed{\int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx \quad ?}$$

Substitute x only:   x = sin(z)   and    dx = cos(z) dz

$$\int\limits_{x=0}^{x=\alpha } \underbrace{\sqrt{1-\sin^2{(z)}}}_{\cos{(z)}} \overbrace{\cos{(z)} \ dz }^{\ dx}=\int\limits_{x=0}^{x=\alpha }\cos^2{(z)} \ dz$$

Product rule (uv)' = u'v+uv'

u =sin(z)     v =cos(z)

u'=cos(z)     v'=-sin(z)

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'}=\cos{(z)}*\cos{(z)}+\sin{(z)}(-\sin{(z)})=\cos^2{(z)}-\sin^2{(z)}}$$

$$\boxed{\sin^2{(z)}=1-\cos^2{(z)} }$$

$$\textstyle{\left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'} =cos^2{(z)}-(1-cos^2{(z)}}) =-1+2\cos^2{(z)}$$

$$2\cos^2{(z)}=1+ \left \textcolor[rgb]{1,0,0}{(} \sin{(z)}\cos{(z)} \right\textcolor[rgb]{1,0,0}{)'} \quad | \quad \int$$

$$2\int{\cos^2{(z)}}\ dz = \underbrace{\int{\ dz }}_z + \sin{(z)}\cos{(z)}$$

$$\boxed{\int{\cos^2{(z)}}\ dz =\dfrac{1}{2} \left( z+\sin{(z)}\cos{(z)} \right)}$$

Back substitute:

$$\\z=\sin^{-1}{(x)}\\ \sin{(z)}=x\\ \cos{(z)}=\sqrt{1-x^2}$$

$$\begin{array}{rcl} \int\limits_{x=0}^{x=\alpha } \sqrt{1-x^2}\ dx &=& \left[ \dfrac{1}{2} \left( \sin^{-1}{(x)}+x\sqrt{1-x^2} \right) } \right]^{x=\alpha}_{x=0}\\\\ &=&\frac{1}{2} \left(\; \sin^{-1}{(\alpha)}+\alpha\sqrt{1-\alpha^2} \;\right) \end{array}$$

Tank A measuring 40 cm by 20 cm by 80 cm,is filled with water to the brim.Tank B is an empty rectangular container with a base area of 60 cm by 40 cm.What is the volume of water that must be poured into Tank B from Tank A such that they have the same height in the end?

$$\\20cm*40cm*(80cm-h)=40cm*60cm*h\\ 20cm*(80cm-h)=60cm*h \quad | \quad :20cm\\ 80cm-h=3h\\ 4h=80cm\\ \boxed{h=20cm}$$

What is the volume of water that must be poured into Tank B from Tank A such that they have the same height in the end?

$$\textcolor[rgb]{1,0,0}{V=}40cm*60cm*h=40*60*20\;cm^3=\textcolor[rgb]{1,0,0}{48000\;cm^3}$$

An empty tank has a capacity of 55.2 litres.Water flows from a Tap A at 3.6 litres per mimute and 2.4 litres per minute from a Tap B.Tap A was turned on first.Tap B was turned on 2 minutes later.The taps were turned off at the same time when the tank was completely filled without overflowing.How much water flowed from Tap B ?

The tank after 2 min. has an empty capacity of $$55.2 - \dfrac{3.6\;l}{min}*2\;min=55.2\;l - 7.2\;l=48\;l$$

$$\left( \dfrac{3.6\;l}{min}+\dfrac{2.4\;l}{min} \right) \times t_{min} = 48\;l \quad | \quad t_{min} = \mbox{time Tab B is running}$$

$$t_{min}=\dfrac{48\;l}{3.6\;l+2.4\;l}*min=\dfrac{48\;\not{l}}{6\;\not{l}}*min=8\;min$$

How much water flowed from Tap B ?

$$=\dfrac{2.4\;l}{min}*8\;min=\textcolor[rgb]{1,0,0}{19.2\;l}$$

Log₃9^1/3 ?

$$\log_3{(9^\frac{1}{3})}=\log_3{(3^\frac{2}{3})}=\frac{2}{3}*\underbrace{\log_3{(3)}}_{=1}=\frac{2}{3}$$

$$\begin{array}{rcccccccl} ( 2x^4 &+& 3x^3 &-& 17x^2 &-& 27x &-& 9 ) : (x^2-2x-3) = 2x^2 +7x+3\\ -(2x^4&-&4x^3&-&6x^2)&&&&\\ --&-&-&-&-&&&&\\ &&7x^3&-&11x^2&-&27x&&\\ &-(&7x^3&-&14x^2&-&21x)&&\\ &&-&-&-&-&-&&\\ &&&&3x^2&-&6x&-&9\\ &&&&-(3x^2&-&6x&-&9)\\ &&&&&&&&0\\ \end{array}$$

$$\begin{array}{rcl} 2x^4 + 3x^3 -17x^2 -27x -9 &=& (x+1)(x-3) (ax+b)(x+c) \\ &=&(x+1)(x-3)(2x^2+7x+3) \end{array}$$

$$\begin{array}{rcccrcc} (ax+b)(x+c)&=&2x^2&+&7x&+&3 \\ (ax+b)(x+c)&=&ax^2&+&(ca+b)x&+&bc \\ \end{array} \\\\ \mbox{Coefficient comparison:} \\ a=2\\ ca+b=7\\ bc=3$$

$$\\2c+b=7 \qquad \Rightarrow \quad b=7-2c\\ bc=3\qquad \Rightarrow \quad (7-2c)c=3\qquad \Rightarrow \quad 2c^2-7c+3=0$$

$$c_{1,2}=\frac{7\pm\sqrt{49-4*2*3}}{2*2}=\frac{7\pm\sqrt{25}}{4}=\frac{7\pm5}{4}\\ c=c_1=\frac{12}{4}=3\\ c=c_2= \frac{2}{4}=\frac{1}{2}\\ b=b_1=7-2c_1=7-2*3=1\\ b=b_2=7-2c_2=7-2*\frac{1}{2}=6$$

1.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_1)(x+c_1)=(x+1)(x-3)(2x+1)(x+3)$$

2.

$$\\2x^4+3x^3-17x^2-27x-9=(x+1)(x-3)(2x+b_2)(x+c_2)=(x+1)(x-3)(2x+6)(x+\frac{1}{2})$$

$$\begin{array}{lcl} 2x^4+3x^3-17x^2-27x-9&=&(\underline{x}+1)(\underline{x}-3)(\underline{ax}+b)(\underline{x}+c) \\ 2x^4&=&x*x*ax*x\\ 2x^4&=&ax^4 \quad \mbox{(Coefficient comparison 2=a)} \end{array} \\ \boxed{a=2}$$

$$\begin{array}{rcl} 2x^4+3x^3-17x^2-27x-9&=&(x+\underline{1})(x-\underline{3})(ax+\underline{b})(x+\underline{c}) \\ -9&=&1*(-3)*b*c\\ -9&=&-3bc \end{array} \\ \boxed{c=\frac{3}{b}}$$

$$\begin{array}{lcl} (x+1)(x-3)(ax+b)(x+c)&=& (x+1)(x-3)(2x+b)(x+\frac{3}{b}) \quad | \quad x+\frac{3}{b}=0 \quad | *b \quad bx+3=0\\ &=&(x+1)(x-3)(2x+b)(bx+3) \end{array}$$

$$\begin{array}{lcl} 2x^4+3x^3-17x^2-27x-9&=&(\underline{x}+1)(\underline{x}-3)(\underline{2x}+b)(\underline{bx}+3) \\ 2x^4&=&x*x*2x*bx\\ 2x^4&=&2bx^4 \quad \mbox{(Coefficient comparison 2=2b)} \end{array} \\ \boxed{b=1}$$

$$\boxed{c=\frac{3}{b}=\frac{3}{1}=3}$$

$$\boxed{(x+1)(x-3)(ax+b)(x+c)=(x+1)(x-3)(2x+1)(x+3)}$$

Du kannst mit der p,q Formel hier nur die imaginären Nullstellen berechnen. Die einzige relle Lösung haben wir ja schon mit x=2. Die p,q Formel liefert hier eine negative Zahl unter dem Wurzelzeichen. Das bedeutet, das die gegebene Funktion die x-Achse nicht weiter schneidet.

Die Funktion $$-x^3+2x^2-x+2=0$$  schneidet die x-Achse (Nullstelle) nur einmal.

x-k = x+2/3   |  k = -2/3

$$\\(15x^4+10x^3-15x^2+11):(x+\frac{2}{3})=\underbrace{15x^3-15x+10}_{q(x)}+\underbrace{\frac{13}{3}\times\frac{1}{x+\frac{2}{3}}}_{p(x)}\\\\ \frac{(15x^4+10x^3-15x^2+11)}{x+\frac{2}{3}} = q(x)+p(x)$$

$$15x^4+10x^3-15x^2+11= (x+\frac{2}{3})q(x)+\underbrace{(x+\frac{2}{3}) p(x)}_{r(x)=\frac{13}{3}}$$

k=-2/3 -> f(-2/3) = r(x) = 13/3