heureka

what are factors of 2x^4-50x^2?

$$\begin{array}{rcl}
\small{
2\cdot x^4-50\cdot x^2 \\
&=& 2\cdot x^2\cdot (x^2-25)\\
&=& 2\cdot x^2\cdot (x^2-5^2) \\
&=& 2 \cdot x^2 \cdot (x-5)(x+5)\\
&=& 2 \cdot x\cdot x \cdot (x-5)(x+5)\\
2\cdot x^4-50\cdot x^2 &=& 2 \cdot (x-0)(x+0)(x-5)(x+5)\\
}
\end{array}$$

write an explicit definition for the sequence -4,1,6,11,....

$$\\\small{\text{arithmetical sequence: $d = 5$ and $a_1 = -4$ }}\\\\
\small{
\begin{array}{rclclcr}
a_1 &=& -4 \\
a_2 &=& a_1 + 1\cdot 5 &=& -4 + 5 &=& 1\\
a_3 &=& a_1 + 2\cdot 5 &=& -4 + 10 &=& 6\\
a_4 &=& a_1 + 3\cdot 5 &=& -4 + 15 &=& 11\\
\cdots \\
\mathbf{a_n }& \mathbf{=} & \mathbf{a_1 + (n-1)\cdot d \\\\
a_n &=& -4 + (n-1) \cdot 5 \\
a_n &=& -4 + 5\cdot n - 5 \\
a_n &=& -4-5 + 5\cdot n \\
\mathbf{a_n }& \mathbf{=} & \mathbf{-9 + 5\cdot n } \\
\\
\hline
\end{array}
}\\$$

$$\\\small{\text{check:}}\\
\small{\text{$
\begin{array}{rcl}
a_4 &=& -9+5\cdot 4\\
a_4 &=& -9 + 20 \\
a_4 &=& 11 \qquad \text{ okay } \\\\
\end{array}
$}}$$

Given a geometric sequence where the 5th term = 162, and the 8th term = -4374, determine the first three terms of the sequence.

I am unclear how to do this when I was not given the first term or the common ratio. Please help!!

$$\begin{array}{rclr}
\small{ \text{geometric sequence: } a_n &=& a_1\cdot r^{n-1} }\\\\
\small{ \text{we have: } a_5 &=& a_1\cdot r^{5-1} = a_1 \cdot r^4 = & 162 }\\
\small{ \text{and we have: } a_8 &=& a_1\cdot r^{8-1} = a_1 \cdot r^7 = & -4374 }
\end{array}\\\\
\small{\text{If we divide }}
a_5 \small{\text{ and } a_8, \small{\text{we can calculate the ratio r }}$$

$$\small{\text{
$
\begin{array}{rclr}
\dfrac{ a_8 } { a_5 } &=& \dfrac{a_1\cdot r^7}{a_1 \cdot r^4} = \dfrac{-4374 }{ 162} \\\\
\dfrac{a_1\cdot r^7}{a_1 \cdot r^4} &=& \dfrac{-4374 }{ 162} \\\\
\dfrac{r^7}{r^4} &=& \dfrac{-4374 }{ 162} \\\\
r^{7-4} &=& \dfrac{-4374 }{ 162} \\\\
r^{3} &=& \dfrac{-4374 }{ 162} \\\\
r^{3} &=& -27 \qquad | \qquad \sqrt[3]{}\\\\
\mathbf{r} &\mathbf{=}&\mathbf{ -3 }
\end{array}
$}}$$

$$\small{\text{Now we can calculate the first term }} a_1
\small{\text{ with } a_5 \small{\text{ or } a_8$$

$$\small{\text{
$
\begin{array}{rclr}
a_5 = a_1 \cdot r^4 &=& 162 \\\\
a_1 \cdot (-3)^4 &=& 162 \\\\
a_1 \cdot 81 &=& 162 \qquad | \qquad :81\\\\
\mathbf{a_1} &\mathbf{=}& \mathbf{2} \\\\
\end{array}
$}}$$

$$\\\small{\text{check:}}\\
\small{\text{
$
\begin{array}{rclr}
a_8 &=& a_1 \cdot r^7 \\\\
a_8 &=& 2\cdot (-3)^7 \\\\
a_8 &=& 2\cdot( -2187 )\\\\
\mathbf{a_8} &\mathbf{=}& \mathbf{-4374} \qquad \text{ okay } \\\\
\end{array}
$}}$$

write a recursive definition for the sequence 8,6,4,2,...

$$\small{
\begin{array}{rclclcl}
a_1 &=& 8 \\
a_2 &=& a_1 - 2 &=& 8 - 2 &=& 6 \\
a_3 &=& a_2 - 2 &=& 6 - 2 &=& 4 \\
a_4 &=& a_3 - 2 &=& 4 - 2 &=& 2 \\
\cdots \\
\mathbf{a_n }& \mathbf{=} & \mathbf{a_{n-1}-2} \\
\\
\hline
\end{array}
}\\$$

Yo

Welchen Winkel schließen die beiden Winkelhalbierenden zwischen x- und y-Achse bzw. zwischen y- und z-Achse ein?

$$\vec{v}_{xy}=\begin{pmatrix} 1\\1\\0 \end{pmatrix}\qquad
\vec{v}_{yz}=\begin{pmatrix} 0\\1\\1 \end{pmatrix}\\\\\\
\tan{(\varphi)}=\dfrac{ \sin{(\varphi)} } { \cos{(\varphi)} }
\small{= \dfrac{
\left| \vec{v}_{xy} \times \vec{v}_{yz} \right|
}
{ \vec{v}_{xy} \cdot \vec{v}_{yz} }
}
\small{= \dfrac{
\left| \begin{pmatrix} 1\\1\\0 \end{pmatrix} \times \begin{pmatrix} 0\\1\\1 \end{pmatrix} \right|
}
{ \begin{pmatrix} 1\\1\\0 \end{pmatrix} \cdot \begin{pmatrix} 0\\1\\1 \end{pmatrix} }
}
\small
{= \dfrac{
\left| \begin{pmatrix} 1 &1 &1 \\1 & 1 & 0\\ 0 & 1 & 1 \end{pmatrix} \right|
}
{ 1\cdot 0 + 1 \cdot 1 + 0 \cdot 1 }
}
\small
{= \dfrac{
\left| \begin{pmatrix} 1 \\ -1 \\ 1 \end{pmatrix} \right|
}
{ 1 }
}\\\\
\small{
\tan{(\varphi)}= \sqrt{ 1^2 +(-1)^2+1^2 )} = \sqrt{3}
}\\\\
\small{
\varphi= \arctan{ (\sqrt{3}) } = 60 \ensurement{^{\circ}}
}$$

x^4-5x^2+4

$$x^4-5x^2+4
=(x^2-1)\cdot (x^2 -4 )
=(x-1)\cdot (x+1) \cdot (x-2 ) \cdot (x+2)$$

What's 0 divided by 0 equal?

$$\\
\left
\begin{array}{c}
\cdots 0^3 \rightarrow 0^2 \rightarrow 0^1 \rightarrow 0^0 = 0\\\\
\cdots 3^0 \rightarrow 2^0 \rightarrow 1^0 \rightarrow 0^0 = 1\\\\
\end{array}
\right\} \qquad 0^0 \mathrm{ ~is~indeterminate}$$

Explain why (–4x)^0 = 1, but –4x^0 = –4.

I. (–4x)^0 = 1

$$1=\dfrac{(-4x)^1}{(-4x)^1}=(-4x)^1\cdot (-4x)^{-1}=(-4x)^{1-1}=(-4x)^0$$

II. –4x^0 = –4

$$-4 = -4\cdot 1 = -4 \cdot \dfrac { x^1 } { x^1 } = -4 \cdot x^1 \cdot x^{-1} = -4 \cdot x^{1-1} = -4 \cdot x^0$$

what is the 20th term if the first term is 4 and the common difference is -1/3

$$\small{
\text{d} = -\dfrac13 \qquad a_1 = 4
}\\\\
\small{
\text{arithmetical sequence: } a_n = a_1 + (n-1)\cdot d
}\\\\
\begin{array}{rclclcr}
\small{a_{20} &=& 4 + (20-1)\cdot \left( -\dfrac13 \right)\\\\
\small{a_{20} &=& 4 + 19\cdot \left(-\dfrac13 \right)\\\\
\small{a_{20} &=& 4 - \dfrac{19}{3} \\\\
\small{a_{20} &=& -2.\overline{3} \\\\
\end{array}$$