Forum serwisu

\(\frac{x^3-6x^2+x-6}{x}=0\)

\(D_{f}:R \setminus {0}\)

\(x^3-6x^2+x-6=0\\
x^2(x+1)-6(x+1)=0\\
(x^2-6)(x+1)=0\\
(x-\sqrt{6})(x+\sqrt{6})(x+1)=0\\
x-\sqrt{6}=0 \vee x+\sqrt{6}=0 \vee x+1=0\\
x=\sqrt{6} \vee x=-\sqrt{6} \vee x=-1\)

a)
\(cos\alpha = \frac{\frac{2}{3}h_{p}}{b} = \frac{\frac{2}{3} \cdot \frac{a\sqrt{3}}{2}}{4} = \frac{\sqrt{3}}{2}=30^o\)

b)

\(sin\alpha = \frac{H}{b}\)

\(\frac{1}{2}=\frac{H}{4}\\
H=2\)

\(V=\frac{1}{3} \cdot \frac{a^2\sqrt{3}}{4} \cdot H = \frac{1}{3} \cdot \frac{36\sqrt{3}}{4}\cdot 2 = 6\sqrt{3}\)

\pi r^2= 16\\ r^2=\frac{16}{\pi}\\ r=\frac{4\sqrt{\pi}}{\pi} \pi r l =20 l=\frac{20}{\pi r} = \frac{20}{\pi \cdot \frac{4\sqrt{\pi}}{\pi}} = \frac{5\sqrt{\pi}}{\pi} h=\sqrt{l^2-r^2} = \sqrt{\frac{25}{\pi} - \frac{16}{\pi}} = \frac{3\sqrt{\pi}}{\pi} V=\frac{1}{3} \pi r^2 \cdot h = \frac{1}{3}\pi \cd...

\begin{bmatrix}1&-1&2&3&4\\2&1&-1&2&0\\-1&2&1&1&3\\1&5&-8&-5&-12\\3&-7&8&9&13\end{bmatrix} w_{2}-2w_{1}, w_{3}+w_{1}, w_{4}-w_{1}, w_{5}-3w_{1} = \begin{bmatrix}1&-1&2&3&4\\0&3&-5&-4&-8\\...

\(\sqrt{x^2-4x-21}>x-3 \setminus ()^2\)

\(x^2-4x-21> x^2-6x+9\)

\(2x>30\)

\(x>15\)

\(S_{n}=a_{1}+(n-1)r\)

\(S_{10}= -5 + (9 \cdot 2) =13\)

2.

\(a=6\)

\(P_{pc} = a^2\sqrt{3} = 36\sqrt{3} \ j^2\)

\(V=\frac{a^3\sqrt{2}}{12} = \frac{216\sqrt{2}}{12} =18\sqrt{2} \ j^3\)

1. D=6 sin30^o = \frac{H}{D}\\ \frac{1}{2} = \frac{H}{6}\\ H=3 cos30^o = \frac{d_{p}}{D}\\ \frac{\sqrt{3}}{2}=\frac{d_{p}}{6}\\ d_{p}=3\sqrt{3} d_{p}=a\sqrt{2}\\ 3\sqrt{3}=a\sqrt{2}\\ a=\frac{3\sqrt{6}}{2} P_{pc}=2a^2 + 4aH = 2 \cdot \(\frac{3\sqrt{6}}{2}\)^2 + 4 \cdot \frac{3\sqrt{6}}{2} \cdot 3 = ...

2. cos \alpha = -\frac{1}{4} sin^ \alpha + cos^2 \alpha =1\\ sin^2 \alpha =1-cos^2 \alpha \\ sin^2 \alpha = 1-\frac{1}{16}=\frac{15}{16}\\ sin \alpha = -\frac{\sqrt{15}}{4} \vee sin \alpha =\frac{\sqrt{15}}{4} poniewaz w II ćwiartce sin jest dodatni więc eliminujemy rozwiazanie ujemne tg \alpha = \f...

1a)

\(\frac{\frac{\sqrt{3}-2}{5}-\frac{\sqrt{3}+2}{5}}{\frac{\sqrt{3}-2}{5}\cdot \frac{\sqrt{3}+2}{5}} =\frac{\frac{\sqrt{3}-2-\sqrt{3}-2}{5}}{\frac{3-4}{25}} = \frac{\frac{-4}{5}}{\frac{-1}{25}} = -\frac{4}{5} \cdot (-25) = 20\)

2.
a)
\(250 + 250 \cdot 0,23 =250 + 57,50 = 307,50\)

b)
\(307,50 - 100%\\
57,50 - x\)


\(x=\frac{57,50 \cdot 100}{307,50} \approx 16,9%\)

3. f'(x)=\frac{cosy \cdot \sqrt{x^2+y^2} - xcosy \cdot \frac{x}{\sqrt{x^2+y^2}}}{x^2+y^2}=\frac{\frac{(x^2+y^2)cosy}{\sqrt{x^2+y^2}} - \frac{x^2cosy}{\sqrt{x^2+y^2}}}{x^2+y^2} = \frac{y^2cosy}{(x^2+y^2)\sqrt{x^2+y^2}} f'(y)=\frac{-xsiny \cdot \sqrt{x^2+y^2} - xcosy \cdot \frac{y}{\sqrt{x^2+y^2}}}{x^...

1.
\(z'(x) = \frac{1}{2\sqrt{x^4+y^2}}\cdot 4x^3 = \frac{2x^3}{\sqrt{x^4+y^2}}\)

\(z'(y) = \frac{1}{2\sqrt{x^4+y^2}}\cdot 2y = \frac{y}{\sqrt{x^4+y^2}}\)

\int x cosx dx = \begin{bmatrix} f' = cosx&g=x\\f=sinx&g'=1\end{bmatrix} = xsinx - \int sinxdx = xsinx + cosx +C \int e^x sinx dx = \begin{bmatrix}f'=e^x&g=sinx\\f=e^x&g'=cosx\end{bmatrix} = e^xsinx - \int e^x cosxdx = \begin{bmatrix}f'=e^x&g=cosx\\f=e^x&g' = -sinx\end{bmatr...