heureka

what two numbers would multiply to -1040 & add up to be 14

$$\\x^2+px+q=0 \quad
\begin{array}{rcrcr}
x_1+x_2&=&14&=&-p\\
x_1x_2&=&-1040&=&q\\
\end{array} \quad
x^2-14x-1040=0$$

$$\underbrace{
\left(
x-\frac{14}{2}
\right)^2}
_{x^2-14x+\frac{14^2}{4}}
-\frac{14^2}{4}-1040=0$$

$$\left(
x-7\right)^2=1040 + 49$$

$$\left(
x-7\right)^2=1089 \quad | \quad \pm \sqrt{}$$

$$x-7=\pm 33$$

$$x_1 = 33 + 7 = 40$$

$$x_2 = -33 + 7 = -26$$

tg(x)= ???? cos(x)

$$tg{(x)} = \sqrt { \frac{1-\cos{(2x)}}{1+\cos{(2x)}}}$$

Example:

$$tg{(45)} = \sqrt { \frac{1-\cos{(90)}}{1+\cos{(90)}}} = \sqrt{\frac{1}{1}} = 1 \quad ok!$$

The line that passes through (1,5) and (4,-4) 

Point 1: $$x_1 = 1, y_1=5$$

Point 2: $$x_2=4, y_2=-4$$

$$m=\frac{y_2-y_1}{x_2-x_1}=\frac{y-y_1}{x-x_1}$$

$$\begin{array}{rcl}
(x-x_1)(y_2-y_1) & = & (y-y_1)(x_2-x_1)\\
x(y_2-y_1)-x_(y_2-y_1) & = & y(x_2-x_1) - y_1(x_2-x_1)\\
y(x_2-x_1) &=& x(y_2-y_1)-x_1(y_2-y_1) + y_1(x_2-x_1) \quad | \quad : (x_2-x_1) \\\\
y &=& \left(
\frac{y_2-y_1}{x_2-x_1}
\right)x -x_1 \left(
\frac{y_2-y_1}{x_2-x_1}\right) + y_1\\\\
y &=& \underbrace{
\left(
\frac{y_2-y_1}{x_2-x_1} \right)}_{m}
x +
\underbrace{
\frac{y_1x_2-x_1y_2}{x_2-x_1} }_{b}
\end{array}$$

$$y =\left(
\frac{y_2-y_1}{x_2-x_1} \right)
x +
\frac{y_1x_2-x_1y_2}{x_2-x_1}$$

$$y =\left(
\frac{-4-5}{4-1} \right)
x +
\frac{5*4-1(-4)}{4-1}$$

$$y =\left(
\frac{-9}{3} \right)
x +
\frac{20+4 }{3}$$

$$y =-3x +8$$

7/8 + 1/7t = 2.... solve equation no decimals please

$$\begin{array}{rcl}
\frac{7}{8}
+
\frac{1}{7}t & = & 2 \quad | \quad +\frac{9}{8}\\\\
\frac{16}{8}
+
\frac{1}{7}t & = & 2 +\frac{9}{8}\\\\
\frac{1}{7}t & = & \frac{9}{8} \quad | \quad *\frac{7}{1}\\\\
t & = & \frac{7*9}{8} \\\\
t & = & \frac{63}{8} \\\\
t & = & \frac{64-1}{8} \\\\
t & = & 8-\frac{1}{8}
\end{array}$$

Hi CPhill,

when the argument in sin is positiv then,  cos can be positive (I.) or negativ(II.):

Hi Melody,

no

cos(sin^-1(5/x))

$$\cos{(\sin^{-1}{( \frac{5}{x} )})} = \pm \sqrt{ 1 - \left(
\frac{5}{x} \right)^2 }$$

or

$$\cos{(\sin^{-1}{( \dfrac{z}{1} )})} = \pm \sqrt{ 1 - \left(z
\right)^2 }$$

if pn=29cm and mn=13cm, then pm=?

0--------------------0--------------------0

p                          n                         m

<-- pn = 29 cm --><-- mn = 13 cm -->

<------------- pm = 42 cm ------------->

pm = 42 cm

Wie löse ich diese Aufgabe: 2^2-3x=0,25

$$\begin{array}{rcl}
2^{2-3x} & = & 0,25 \quad | \quad 0,25 = \frac{1}{4} \\\\
2^{2-3x} & = & \frac{1}{4}\\\\
2^{2-3x} & = & \frac{1}{2^2}\\\\
2^{2-3x} & = & 2^{-2}\\\\
2-3x&=&-2\\\\
3x&=&4
\end{array}$$

$$x=\dfrac{4}{3}=1.33333333333$$

the perimeter of a rectangle is 44cm. the length is 7 more then twice the width. what is the length/width?

width of the rectangle: W

length of the rectangle: L

the perimeter: $$p = 2 * ( L+W ) = 44\;cm \quad \mbox{or} \quad \boxed{L + W = 22 \;cm \qquad (1)}$$

the length is 7 more then twice the width: $$L = 7\;cm + 2W \quad \mbox{or} \quad \boxed{L - 2W = 7\;cm \qquad (2)}$$ 

 $$\begin{array}{lrcl}
(1) - (2):&\\
&\underbrace{L-L}_{=0}+\underbrace{W-(-2W)}_{=3W}&=&\underbrace{22 cm - 7cm}_{=15\;cm}\\
&3W&=&15\;cm \quad | \quad : 3\\
&W&=&5\;cm
\end{array}$$

$$\begin{array}{rcl}
L + W &=& 22\;cm\\
L &=& 22 \;cm- W \\
L&=& 22\;cm- 5 \;cm\\
L &=& 17 \;cm
\end{array}$$