heureka

Hallo ich benötige hilfe bei der Aufgabe :

x/x+75 = 4/99 X = ?

Ist von zweiten Stahlensatz .

$\begin{array}{|rcll|} \hline \dfrac{x}{x+75} &=& \frac{4}{99} \\ 99x &=& 4\cdot(x+75)\\ 99x &=& 4x + 300 \quad & | \quad -4x \\ 99x -4x&=& 300 \\ 95x&=& 300 \quad & | \quad : 95 \\ x&=& \frac{300}{95} \\ x&=& \frac{60}{19} \\ x &=& 3.15789473684 \\ \hline \end{array}$

p(x)=4x/9-c,find c if x=1,2,3,4,5,

$\begin{array}{|rcll|} \hline p(x) &=& prim(x) \\ \hline p(1) &=& 2 \\ p(2) &=& 3 \\ p(3) &=& 5 \\ p(4) &=& 7 \\ p(5) &=& 11 \\ p(6) &=& 13 \\ p(7) &=& 17 \\ p(8) &=& 19 \\ p(9) &=& 23 \\ p(10) &=& 29 \\ \dots \\ \hline \end{array}$

$\begin{array}{|rcll|} \hline p(x) &=& \frac{4x}{9-c} \\ c &=& 9-\frac{4x}{p(x)} \\ \hline \end{array}$

$\begin{array}{r|c} \hline x & c \\ \hline 1 & 9-\frac{4\cdot 1}{2} = 7 \\ 2 & 9-\frac{4\cdot 2}{3} = \frac{19}{3} \\ 3 & 9-\frac{4\cdot 3}{5} = \frac{33}{5} \\ 4 & 9-\frac{4\cdot 4}{7} = \frac{47}{7} \\ 5 & 9-\frac{4\cdot 5}{11} = \frac{79}{11} \\ \hline \end{array}$

Forty-five students take piano lessons.
The ratio of the numbers of students who take piano lessons to violin lessons is 15:8.
The ratio of the numbers of students who take violin lessons to clarinet lessons is 8:9.

a) How many students take violin lessons?

$\begin{array}{|rcll|} \hline && 45\ \text{piano} \cdot \frac{8}{15} \frac{\text{violin}}{\text{piano}} \\ &=& \frac{45\cdot 8}{15}\ \text{violin} \\ &=& 24\ \text{violin} \\ \hline \end{array}$

b) How many students take clarinet lessons?

$\begin{array}{|rcll|} \hline && 24\ \text{violin} \cdot \frac{9}{8} \frac{\text{clarinet}}{\text{violin}} \\ &=& \frac{24\cdot 9}{8}\ \text{clarinet} \\ &=& 27\ \text{clarinet} \\ \hline \end{array}$

On a hockey team of 45 players, only 9 play at any given time.

How many different groups of people could be on the ice?

formula is:      nCr = 𝑛!/(𝑛−𝑟)!𝑟!

$\begin{array}{|rcll|} \hline && nCr = \frac{n!}{(n-r)!r!} \qquad n= 45,\ r= 9 \\ \hline && \frac{45!}{(45-9)!9!} \\\\ &=& \frac{45!}{36!9!} \\\\ &=& \frac{36!\cdot 37\cdot 38\cdot 39\cdot 40\cdot 41\cdot 42\cdot 43\cdot 44\cdot 45}{36!9!} \\\\ &=& \frac{37\cdot 38\cdot 39\cdot 40\cdot 41\cdot 42\cdot 43\cdot 44\cdot 45}{9!} \\\\ &=& \frac{37\cdot 38\cdot 39\cdot 40\cdot 41\cdot 42\cdot 43\cdot 44\cdot 45}{9\cdot 8\cdot 7\cdot 6\cdot 5\cdot 4\cdot 3\cdot 2\cdot 1} \\\\ &=& 37\cdot \frac{38}{2}\cdot \frac{39}{3}\cdot \frac{40}{8}\cdot 41\cdot \frac{42}{6\cdot 7}\cdot 43\cdot \frac{44}{4}\cdot \frac{45}{5\cdot 9} \\\\ &=& 37\cdot 19\cdot 13\cdot 5\cdot 41\cdot \frac{42}{42}\cdot 43\cdot 11\cdot \frac{45}{45} \\\\ &=& 37\cdot 19\cdot 13\cdot 5\cdot 41\cdot 43\cdot 11 \\\\ &=& \mathbf{886163135} \\ \hline \end{array}$

Evaluate cos(162º). Round your answer to 2 decimal places

$\cos(162^{\circ}) = -0.95105651630$

Rounding to 2 Decimal Places -0.95

you got the right answer.

Out of 20 job seekers, 12 have degrees and 10 have some other qualification. 

If 2 have no qualification how many have both a degree and some other qualification?

two-way table job seekers:

$\begin{array}{|l|r|r|r|} \hline &\text{Degrees} &\text{no degrees}& \text{Total} \\ \hline \text{Qualification } & 4 & 6 & 10 \\ \text{no qualification} & 8 & 2 & 10 \\ \hline \text{Total} & 12 & 8 & 20 \\ \hline \end{array}$

Please help me!

i) Find the equation of BC

slope BA: $\begin{array}{|rcll|} \hline m = \frac{y_A-y_B}{x_A-x_B} \\ \hline \end{array}$

slope BC: $\begin{array}{|rcll|} \hline m_{\perp} = -\frac{1}{m} &=& -\frac{x_A-x_B}{y_A-y_B} \\ \hline \end{array}$

Equation of BC:

$\begin{array}{|rcll|} \hline m_{\perp} = \frac{y -y_B}{x -x_B} &=& -\frac{x_A-x_B}{y_A-y_B} \\ & \dots & \\ y &=& -\left(\frac{x_A-x_B}{y_A-y_B}\right)\cdot (x-x_B) + y_B \\ && x_B = -2 \quad y_B = 8 \qquad x_A = 2 \quad y_A = 14 \\ &=& -\left[\frac{2-(-2)}{14-8}\right]\cdot [x-(-2)] + 8 \\ &=& -\frac{4}{6}\cdot (x+2) + 8 \\ &=& -\frac{2}{3}\cdot (x+2) + 8 \\ \mathbf{y} & \mathbf{=} & \mathbf{-\frac{2}{3}\cdot x + \frac{20}{3}} \\ \hline \end{array}$

ii) Find the coordinates of D

coordinates of C ( $y_C=0$ )

$\begin{array}{|rcll|} \hline 0 & = & -\frac{2}{3}\cdot x_C + \frac{20}{3} \\ \frac{2}{3}\cdot x_C &=& \frac{20}{3} \\ x_C &=& \frac{20}{3}\cdot \frac{3}{2} \\ \mathbf{x_C} & \mathbf{=} & \mathbf{10} \\ \hline \end{array}$

C is (10,0)

coordinates of D:

$\begin{array}{|rcll|} \hline x_D &=& x_C + (x_A-x_B) \\ x_D &=& 10 + [2-(-2)] \\ x_D &=& 10 + 4 \\ \mathbf{x_D} & \mathbf{=} & \mathbf{14} \\\\ y_D &=& y_C + (y_A-y_B) \\ y_D &=& 0 + (14-8) \\ \mathbf{y_D} & \mathbf{=} & \mathbf{6} \\ \hline \end{array}$

D is (14,6)

iii) Find the angle that CD makes with the positive x axis.

$\begin{array}{|rcll|} \hline \varphi &=& \arctan(m) \quad & | \quad m = \frac{y_A-y_B}{x_A-x_B} = \frac{14-8}{2-(-2)}=\frac{6}{4}=\frac32 \\ \varphi &=& \arctan(\frac32) \\ \varphi &=& \arctan(1.5) \\ \mathbf{\varphi} & \mathbf{=} & \mathbf{56.3099324740^{\circ} } \\ \hline \end{array}$

I have a 3-4-5 triangle and they want me to find h, how do I find h?

In a right triangle:

$\begin{array}{|rcll|} \hline h &=& \frac{ab}{c}\\ &=& \frac{3\cdot 4}{5} \\ &=& \frac{12}{5} \\ \mathbf{h} & \mathbf{=} & \mathbf{2.4} \\ \hline \end{array}$

What is the equation (in graphing form) for the parabola with vertex (2,4) that passes through the point (6,10)?

60,20€ 19% DAVON

$\begin{array}{|rcll|} \hline && 60,20€ \cdot 19\ \% \\ &=& 60,20€ \cdot \frac{19}{100} \\ &=& 0,6020€ \cdot 19 \\ &=& 11,44€ \\ \hline \end{array}$