heureka

1/(x+3)-1/(x-3)=3

\(\begin{array}{|rcll|} \hline 3 &=& \frac{1}{x+3}-\frac{1}{x-3} \quad & | \quad \cdot (x+3)(x-3) \\ 3\cdot (x+3)(x-3) &=& \frac{ (x+3)(x-3)}{x+3}-\frac{ (x+3)(x-3)}{x-3} \\ 3\cdot (x+3)(x-3) &=& (x-3) - (x+3) \\ 3\cdot (x+3)(x-3) &=& x- 3 - x-3 \\ 3\cdot (x+3)(x-3) &=& -6 \quad & | \quad : 3 \\ (x+3)(x-3) &=& -2 \\ x^2-9 &=& -2 \quad & | \quad +9 \\ x^2 &=& 7 \quad & | \quad \text{square root both sides} \\ \mathbf{x} & \mathbf{=} & \mathbf{\pm\sqrt{ 7 } }\\ \hline \end{array}\)

Need Help Please

\(\begin{array}{|rcll|} \hline y^{\circ} &=& 180^{\circ}-(33^{\circ}+109^{\circ}) \\ y^{\circ} &=& 180^{\circ}-142^{\circ} \\ y^{\circ} &=& 38^{\circ} \\ \hline \end{array}\)

corresponding angle: \(x^{\circ} = 33^{\circ} \)

Opposite angles are equal: \(z^{\circ} = 109^{\circ} \)

Need Help

Diagonals of a Parallelogram

The diagonals of a parallelogram bisect each other.

In other words the diagonals intersect each other at the half-way point.

\(\begin{array}{|lrcll|} \hline (1) & DH &=& HF \\ & x+1 &=& 3y \\ & x &=& 3y - 1 \\\\ (2) & GH &=& HE \\ & 3x-4 &=& 5y+1 \\ & 3(3y-1) - 4 &=& 5y+1 \\ & 9y-3 - 4 &=& 5y+1 \\ & 9y-7 &=& 5y+1 \quad & | \quad +7-5y\\ & 4y &=& 8 \quad & | \quad : 4 \\ & \mathbf{y} & \mathbf{=} & \mathbf{2} \\\\ (1) & x &=& 3y - 1 \\ & x &=& 3\cdot 2 - 1 \\ & x &=& 6 - 1 \\ & \mathbf{x} & \mathbf{=} & \mathbf{5} \\ \hline \end{array}\)

A quadratic equation has roots 0 and -6 and a maximum at(-3,4). 

how would you find the equation of the relation?

The quadratic equation is in general:
\(\begin{array}{|rcll|} \hline y &=& a\cdot (x-x_1)\cdot (x-x_2) \quad & | \quad x_1 \text{ and } x_2 \text { are roots}\\ \hline \end{array}\)

Let \(x_1 = 0\) and \(x_2 = -6\)

\(\begin{array}{|rcll|} \hline y &=& a\cdot (x-0)\cdot [x-(-6)] \\ y &=& a\cdot x\cdot (x+6) \\ \hline \end{array} \)

The maximum is (-3,4):

\(\begin{array}{|rcll|} \hline y &=& a\cdot x\cdot (x+6) \\ 4 &=& a\cdot (-3)\cdot (-3+6) \\ 4 &=& a\cdot (-3)\cdot 3 \\ 4 &=& a\cdot (-9) \quad & | \quad : (-9)\\ -\frac49 &=& a \\ \mathbf{a} & \mathbf{=} & \mathbf{ -\frac49 } \\ \hline \end{array}\)

The quadratic equation is:

\(\begin{array}{|rcll|} \hline y &=& a\cdot x\cdot (x+6) \quad & | \quad a =-\frac49 \\ y &=& -\frac49\cdot x\cdot (x+6) \\ y &=& -\frac49\cdot (x^2+6x) \\ y &=& -\frac49\cdot x^2 - \frac49\cdot 6x \\ \mathbf{y} & \mathbf{=} & \mathbf{-\frac49 x^2 - \frac83 x} \\ \hline \end{array}\)

see:

Use only 7, 5, 6, and 3 to create an expression that equals 75 and you can only use each number once

\(\begin{array}{|rcll|} \hline 75 &=& (3*7-6)*5 \\ 75 &=& (5+7)*6+3 \\ \hline \end{array}\)

4.181818181818181... as a fraction

\(\begin{array}{|lrcll|} \hline (1) & 100 \times 4.\overline{18} &=& 418.\overline{18} \\ (2) & 1\times 4.\overline{18} &=& 4.\overline{18} \\ \hline (1)-(2) & 100 \times 4.\overline{18} - 1\times 4.\overline{18} &=& 418.\overline{18} - 4.\overline{18} \\ & 99\times 4.\overline{18} &=& 414 \\ & 4.\overline{18} &=& \frac{414}{99} \\ \hline \end{array} \)

How do i find the surface area?

\(\begin{array}{|rcll|} \hline V(s) &=& \left( 0.4547\cdot s^{\frac12} \right)^3 \quad & | \quad \text{cube root both sides} \\ \sqrt[3]{V(s)} &=& 0.4547\cdot s^{\frac12} \quad & | \quad : 0.4547 \\ \frac{\sqrt[3]{V(s)}}{0.4547} &=& s^{\frac12} \quad & | \quad \text{square both sides} \\ \left( \frac{\sqrt[3]{V(s)}}{0.4547} \right)^2 &=& s^{\frac22} \\ \left( \frac{\sqrt[3]{V(s)}}{0.4547} \right)^2 &=& s^1 \\ \left( \frac{\sqrt[3]{V(s)}}{0.4547} \right)^2 &=& s \\ \mathbf{s} & \mathbf{=} & \mathbf{\left( \frac{\sqrt[3]{V(s)}}{0.4547} \right)^2} \\ \hline \end{array} \)

\(\begin{array}{|rcll|} \hline s &=& \left( \frac{\sqrt[3]{V(s)}}{0.4547} \right)^2 \quad & | \quad V(s) = 2500\ m^3 \qquad s \text{ in } m^2\\ s &=& \left( \frac{\sqrt[3]{2500\ m^3}}{0.4547} \right)^2 \\ s &=& \left( \frac{13.5720880830\ m}{0.4547} \right)^2 \\ s &=& ( 29.8484453111 \ m )^2 \\ s &=& 890.929687492\ m^2 \\ \mathbf{s} & \mathbf{\approx} & \mathbf{ 890.93\ m^2 } \\ \hline \end{array} \)

How do i find the height?

\(\begin{array}{|rcll|} \hline h &=& 62\cdot \sqrt[3]{t} + 76 \quad & | \quad -76 \\ h - 76 &=& 62\cdot \sqrt[3]{t} \quad & | \quad :62 \\ \frac{h - 76}{62} &=& \sqrt[3]{t} \\ \frac{h - 76}{62} &=& t^{\frac13} \quad & | \quad \text{cube both sides} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^{\frac33} \\ \left({\frac{h - 76}{62}} \right)^3 &=& t^1 \\ \left({\frac{h - 76}{62}} \right)^3 &=& t \\ \mathbf{t} & \mathbf{=} & \mathbf{ \left({\frac{h - 76}{62}} \right)^3 } \\ \hline \end{array}\)

\(\begin{array}{|rcll|} \hline t &=& \left({\frac{h - 76}{62}} \right)^3\quad & | \quad h = 231\ cm \qquad t \text{ in years}\\ t &=& \left({\frac{231 - 76}{62}}\right)^3 \\ t &=& \left({\frac{155}{62}}\right)^3 \\ t &=& {2.5}^3 \\ \mathbf{t} & \mathbf{=} & \mathbf{15.625 \text{ years}} \\ \hline \end{array}\)

Erstellen Sie ohne Berechnung folgenden mathematischen Term:

Vom Produkt der beiden Zahlen vier und sieben wird die Summe aus drei Viertel und zwei Achtel subtrahiert?

\(4\times 7 - \left( \frac34+ \frac28 \right)\)

Wir müssen mit m3 rechnen, aber ich hab es noch nicht ganz kapiert und wir bekommen dabei keine hilfe mehr.

Bitte helft mir.

\(m^3 \times m^3 = m^{3+3} = m^6 \)

1) 30 m³  x  534 m³ = 16020 m⁶

2) 66 m³  x 932 m³  =  61512 m⁶

3) 7372 m³ x 1232 m³  = 9.082×10⁶ m⁶ = 9753156 m⁶